Suppose the stochastic process of the asset price \(S_t\) follows a geometric Brownian motion: \[\,d S_t = 0.2 S_t \,dt + 0.3 S_t \,dW_t^\mathcal{P}\] where \(W_t^P\) is a standard Brownian motion on the real world probability space, denoted by \((\omega, \mathcal{F},\mathcal{P})\). A European call option written on the stock with parameters: exercise price \(K = 15\), maturity time \(T=2\), \(t=0.5\) (year), the risk-free interest rate is \(r = 0.06\), \(S_t=12\).
Let \(\widetilde W_t = W_t + \theta t\), with \(\theta\) being a constant is an adapted process with respect to the filtration \(()\mathcal{F}_t)_{t \geq 0}\). Find \(\theta\) such that the discounted \((e^{−0.07t}S_t)_{t \ge 0}\) is a martingale process with respect to \(\widetilde W_t\).
Derive the Black-Scholes-Merton formula and then compute the option price.
Let \(\mathcal{P}\) and \(\mathcal{Q}\) be the physical and risk neutral measures on \((\omega,\mathcal{F})\) respectively. Let \(\{W_t^\mathcal{P}\}_{t \geq 0}\) be the standard Brownian motion on \(\mathcal{P}\). For \(\{S_t\}_{t \geq 0}\) be the process of asset price, we have: \[dSt = \mu S_t \,dt + \sigma S_t \,dW_t^\mathcal{P}\]
With \(f(t,x)=e^{-rt}x\) then we can calculate: \[\begin{align*} f_t&=-re^{-rt}x \\ f_x&=e^{-rt} \\ f_{tt}&=r^2e^{-rt}x \\ f_{xx}&=0 \\ f_{tx}&=-re^{-rt} \end{align*}\]
Apply Ito-Doeblin formula, we have: \[\begin{align*} d(e^{-rt}S_t) &= -re^{-rt}S_t\,dt+e^{-rt}\,dS_t -re^{-rt}\,dt \,dS_t \\ &= -re^{-rt}S_t\,dt+e^{-rt}(\mu S_t \,dt + \sigma S_t \,dW_t^\mathcal{P}) -re^{-rt}\,dt(\mu S_t \,dt + \sigma S_t \,dW_t^\mathcal{P}) \\ &= e^{-rt}S_t[(\mu-r) \,dt+\sigma \,d W_t^{\mathcal{P}}] -\mu re^{-rt}St\,dt^2 + \sigma S_t \,dt\,dW_t^\mathcal{P})\\ &= e^{-rt}S_t[\mu-r \,dt+\sigma (\,d W_t^{\mathcal{Q}}-\theta_t) \,dt] \\ &= e^{-rt}S_t[(\mu-r-\sigma \theta_t) \,dt+\sigma \,d W_t^{\mathcal{Q}} \,dt] \end{align*}\]
In order to have the process \(\{e^{−rt}S_t\}_{t \geq 0}\) be the martigale process, we choose \(\theta_t\) such that: \[\mu-r-\sigma \theta_t=0 \rightarrow \theta_t=\frac{\mu-r}{\sigma} \] then \[E^\mathcal{P}[e^{\frac{1}{2} \int_0^T \theta_t^2\,dt}]<\infty\]
By the Girsanov’s theorem, there is a risk neutral measure \(\mathcal{Q}\) defined by the Radon-Nykodym \[E \left[\frac{\,d \mathcal{Q}}{\,d \mathcal{P}} |\mathcal{F}_t\right]=Z_t=e^{-\int_0^t \theta_s \,d W_s^\mathcal{P}-\frac{1}{2} \theta_s^2 \,ds}\] such that \(W_t^\mathcal{Q}=W_t^\mathcal{P}+\int_0^t\theta_s \,ds\) is the standard Brownian motion under \(\mathcal{Q}\). \[W_t^\mathcal{Q}=W_t^\mathcal{P}+\int_0^t\theta_s \,ds \\ \rightarrow \,dW_t^\mathcal{Q}=\,dW_t^\mathcal{P}+\theta_t\,dt\]
Since \(e^{-rt}S_t\) is a martingale under \(\mathcal{Q}\) and \(\mathcal{Q}\) is an equivalent martingale measure to \(\mathcal{P}\), we can write: \[\begin{align*} \,dS_t &= \mu S_t \,dt + \sigma S_t \,dW_t^\mathcal{P} \\ &=\mu S_t \,dt + \sigma S_t (\,dW_t^\mathcal{Q}-\theta_t \,dt) \\ &=\mu S_t \,dt + \sigma S_t (\,dW_t^\mathcal{Q}-\frac{\mu-r}{\sigma} \,dt) \\ &=\left(\mu-\sigma\cdot\frac{\mu-r}{\sigma} \right)S_t \,dt + \sigma S_t \,dW_t^\mathcal{Q} \\ &=rS_t \,dt + \sigma S_t \,dW_t^\mathcal{Q} \\ \end{align*}\]
Let \(Y_t=\ln S_t\) and \(f(t,x)=\ln x\), we have: \[\begin{align*} f_t&=0 \\ f_x&=\frac{1}{x} \\ f_{tt}&=0 \\ f_{xx}&=-\frac{1}{x^2} \\ f_{tx}&=0 \\ \end{align*}\]
Applying Ito Doeblin formula, we have: \[\begin{align*} \,d(Y_t)&=\frac{1}{S_t} \,dS_t-\frac{1}{2}\frac{1}{S_t^2}\,dS_t^2 \\ &=\frac{1}{S_t} (rS_t \,dt + \sigma S_t \,dW_t^\mathcal{Q})-\frac{1}{2}\frac{1}{S_t^2}(rS_t \,dt + \sigma S_t \,dW_t^\mathcal{Q})^2 \\ &=(r-\frac{1}{2} \cdot \sigma^2)\,dt + \sigma \,dW_t^\mathcal{Q} \\ \,d(\ln S_t)&=(r-\frac{1}{2} \cdot \sigma^2)\,dt + \sigma \,dW_t^\mathcal{Q} \\ \int_t^T \,d(\ln S_s)&=\int_t^T(r-\frac{1}{2} \cdot \sigma^2)\,ds + \int_t^T\sigma \,dW_s^\mathcal{Q} \\ \ln \left( \frac{S_T}{S_t} \right)&=(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^\mathcal{Q}-W_t^\mathcal{Q}) \\ S_T&=e^{\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})} \end{align*}\]
We have \[\begin{align*} \ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}}) &\sim \mathcal{N}(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t),\sigma^2(T-t)) \\ e^{\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})} &\sim \ln \mathcal{N}(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t),\sigma^2(T-t)) \\ (S_T|S_t) &\sim \ln \mathcal{N}(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t),\sigma^2(T-t)) \\ \end{align*}\]
The probability density function of \((S_T|S_t)\) is \[f_{S_T|S_t}(x)=\frac{1}{\sigma x\sqrt{2\pi(T-t)}}e^{-\frac{1}{2} \frac{(\ln x-(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows a geometric Brownian motion:
\[\,d S_t = 0.05 \,dt + 0.3 \,d W_t^{\mathcal{P}},\]
where \(W_t^{\mathcal{P}}\) is a standard Brownian motion on the real world probability space, denoted by \((\omega, \mathcal{F},\mathcal{P})\). A European digital call option written on the stock with parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\), \(t = 0.5\), \(T = 1\), \(S_t = 9\).
Let \(W_t^Q = W_t^P + \int_0^t \theta_s \,d s\), with \((\theta_t)_{t \geq 0}\) is an adapted process with \((\mathcal{F}_t)_{t \geq 0}\). \[\begin{align*} d(e^{−rt}S_t) &= −re^{−rt}S_t \,dt + e^{−rt}\,d S_t \\ &= −re^{−rt}S_t \,dt + e^{−rt}(\mu \,dt + \sigma \,dW_t^P) \\ &= e^{−rt}((\mu − rS_t) \,dt + \sigma (W_t^Q - \theta_t \,d t)) \\ &= e^{−rt}((\mu − rS_t -\sigma\theta_t) \,dt + \sigma W_t^Q) \\ \end{align*}\]
We choose \(\theta_t = \frac{\mu -r S_t}{\sigma}\) then \(E^{\mathcal{P}} \left(e^{\frac{1}{2} \int_0^T \theta_t^2 \,dt} \right)< \infty\) and \(Z_t = e^{\int_0^t \theta_s \,d W_s- \frac{1}{2} \int_0^t \theta_s^2 \,ds}\) is a positive martingale. From Girsanov’s theorem, there exist a risk-neutral measure \(\mathcal{Q}\) defined by \(E\left[\frac{\,d \mathcal{Q}}{\,d \mathcal{P}} | \mathcal{F}_t \right] = Z_t\) such that \(W_t^{\mathcal{Q}}\) is a Standard Brownian motion under \(\mathcal{Q}\). As \(d(e^{−rt}S_t) = \sigma e^{−rt} \,d W_t^{\mathcal{Q}}\), \(e^{−rt}S_t\) is a martingale under \(\mathcal{Q}\) and \(\mathcal{Q}\) is an equivalent martingale measure to \(\mathcal{P}\)
By taking the integral both sides of the above equality from t to T, we obtain: \[\begin{align*} \int_t^T d(e^{−rt}S_t) &= \int_t^T\sigma e^{−ru} \,d W_t^{\mathcal{Q}} \\ e^{−rt} S_T - e^{−rt} S_t &= \int_t^T\sigma e^{−ru} \,d W_t^{\mathcal{Q}} \\ S_T&= e^{r(T-t)} S_t + \int_t^T \sigma e^{r(T-u)} \,d W_t^{\mathcal{Q}} \\ & \sim \mathcal{N} \left(e^{r(T-t)} S_t , \left( \int_t^T \sigma e^{r(T-u)} \,d W_t^{\mathcal{Q}}\right)^2 \right) \\ & \sim \mathcal{N} \left( e^{r(T-t)} S_t , \frac{\sigma^2}{2r}(e^{2r(T-t)}-1) \right) \end{align*}\]
The probability density function of \(S_T|S_t\) can be written as: \[f_{S_T|s_t}(y) = \frac{1}{v \sqrt{2\pi}} e^{-\frac{(y-m)^2}{2v^2}}\] where \(m=e^{r(T-t)} S_t\) and \(v^2=\frac{\sigma^2}{2r}(e^{2r(T-t)}-1)\)
Suppose the stochastic process of the asset price \(S_t\) follows a geometric Brownian motion: \[\,d S_t = 0.2 S_t \,dt + 0.3 S_t \,dW_t^\mathcal{P}\] where \(W_t^P\) is a standard Brownian motion on the real world probability space, denoted by \((\omega, \mathcal{F},\mathcal{P})\). A European call option written on the stock with parameters: exercise price \(K = 15\), maturity time \(T=2\), \(t=0.5\) (year), the risk-free interest rate is \(r = 0.06\), \(S_t=12\).
Let \(\widetilde W_t = W_t + \theta t\), with \(\theta\) being a constant is an adapted process with respect to the filtration \(()\mathcal{F}_t)_{t \geq 0}\). Find \(\theta\) such that the discounted \((e^{−0.07t}S_t)_{t \ge 0}\) is a martingale process with respect to \(\widetilde W_t\).
Derive the Black-Scholes-Merton formula and then compute the option price.
Let \(\mathcal{P}\) and \(\mathcal{Q}\) be the physical and risk neutral measures on \((\omega,\mathcal{F})\) respectively. Let \(\{W_t^\mathcal{P}\}_{t \geq 0}\) be the standard Brownian motion on \(\mathcal{P}\). For \(\{S_t\}_{t \geq 0}\) be the process of asset price, we have: \[dSt = \mu S_t \,dt + \sigma S_t \,dW_t^\mathcal{P}\]
With \(f(t,x)=e^{-rt}x\) then we can calculate: \[\begin{align*} f_t&=-re^{-rt}x \\ f_x&=e^{-rt} \\ f_{tt}&=r^2e^{-rt}x \\ f_{xx}&=0 \\ f_{tx}&=-re^{-rt} \end{align*}\]
Apply Ito-Doeblin formula, we have: \[\begin{align*} d(e^{-rt}S_t) &= -re^{-rt}S_t\,dt+e^{-rt}\,dS_t -re^{-rt}\,dt \,dS_t \\ &= -re^{-rt}S_t\,dt+e^{-rt}(\mu S_t \,dt + \sigma S_t \,dW_t^\mathcal{P}) -re^{-rt}\,dt(\mu S_t \,dt + \sigma S_t \,dW_t^\mathcal{P}) \\ &= e^{-rt}S_t[(\mu-r) \,dt+\sigma \,d W_t^{\mathcal{P}}] -\mu re^{-rt}St\,dt^2 + \sigma S_t \,dt\,dW_t^\mathcal{P})\\ &= e^{-rt}S_t[\mu-r \,dt+\sigma (\,d W_t^{\mathcal{Q}}-\theta_t) \,dt] \\ &= e^{-rt}S_t[(\mu-r-\sigma \theta_t) \,dt+\sigma \,d W_t^{\mathcal{Q}} \,dt] \end{align*}\]
In order to have the process \(\{e^{−rt}S_t\}_{t \geq 0}\) be the martigale process, we choose \(\theta_t\) such that: \[\mu-r-\sigma \theta_t=0 \rightarrow \theta_t=\frac{\mu-r}{\sigma} \] then \[E^\mathcal{P}[e^{\frac{1}{2} \int_0^T \theta_t^2\,dt}]<\infty\]
By the Girsanov’s theorem, there is a risk neutral measure \(\mathcal{Q}\) defined by the Radon-Nykodym \[E \left[\frac{\,d \mathcal{Q}}{\,d \mathcal{P}} |\mathcal{F}_t\right]=Z_t=e^{-\int_0^t \theta_s \,d W_s^\mathcal{P}-\frac{1}{2} \theta_s^2 \,ds}\] such that \(W_t^\mathcal{Q}=W_t^\mathcal{P}+\int_0^t\theta_s \,ds\)
is the standard Brownian motion under \(\mathcal{Q}\). \[W_t^\mathcal{Q}=W_t^\mathcal{P}+\int_0^t\theta_s \,ds \\ \rightarrow \,dW_t^\mathcal{Q}=\,dW_t^\mathcal{P}+\theta_t\,dt\]
Since \(e^{-rt}S_t\) is a martingale under \(\mathcal{Q}\) and \(\mathcal{Q}\) is an equivalent martingale measure to \(\mathcal{P}\), we can write: \[\begin{align*} \,dS_t &= \mu S_t \,dt + \sigma S_t \,dW_t^\mathcal{P} \\ &=\mu S_t \,dt + \sigma S_t (\,dW_t^\mathcal{Q}-\theta_t \,dt) \\ &=\mu S_t \,dt + \sigma S_t (\,dW_t^\mathcal{Q}-\frac{\mu-r}{\sigma} \,dt) \\ &=\left(\mu-\sigma\cdot\frac{\mu-r}{\sigma} \right)S_t \,dt + \sigma S_t \,dW_t^\mathcal{Q} \\ &=rS_t \,dt + \sigma S_t \,dW_t^\mathcal{Q} \\ \end{align*}\]
Let \(Y_t=\ln S_t\) and \(f(t,x)=\ln x\), we have: \[\begin{align*} f_t&=0 \\ f_x&=\frac{1}{x} \\ f_{tt}&=0 \\ f_{xx}&=-\frac{1}{x^2} \\ f_{tx}&=0 \\ \end{align*}\]
Applying Ito Doeblin formula, we have: \[\begin{align*} \,d(Y_t)&=\frac{1}{S_t} \,dS_t-\frac{1}{2}\frac{1}{S_t^2}\,dS_t^2 \\ &=\frac{1}{S_t} (rS_t \,dt + \sigma S_t \,dW_t^\mathcal{Q})-\frac{1}{2}\frac{1}{S_t^2}(rS_t \,dt + \sigma S_t \,dW_t^\mathcal{Q})^2 \\ &=(r-\frac{1}{2} \cdot \sigma^2)\,dt + \sigma \,dW_t^\mathcal{Q} \\ \,d(\ln S_t)&=(r-\frac{1}{2} \cdot \sigma^2)\,dt + \sigma \,dW_t^\mathcal{Q} \\ \int_t^T \,d(\ln S_s)&=\int_t^T(r-\frac{1}{2} \cdot \sigma^2)\,ds + \int_t^T\sigma \,dW_s^\mathcal{Q} \\ \ln \left( \frac{S_T}{S_t} \right)&=(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^\mathcal{Q}-W_t^\mathcal{Q}) \\ S_T&=e^{\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})} \end{align*}\]
Let \(\mathcal{P}\) and \(\mathcal{Q}\) be the physical and risk neutral measures on \((\omega,\mathcal{F})\) respectively. Let \(\{W_t^\mathcal{P}\}_{t \geq 0}\) be the standard Brownian motion on \(\mathcal{P}\). For \(\{S_t\}_{t \geq 0}\) be the process of asset price, we have: \[dSt = \mu S_t \,dt + \sigma S_t \,dW_t^\mathcal{P}\]
With \(f(t,x)=e^{-rt}x\) then we can calculate: \[\begin{align*} f_t&=-re^{-rt}x \\ f_x&=e^{-rt} \\ f_{tt}&=r^2e^{-rt}x \\ f_{xx}&=0 \\ f_{tx}&=-re^{-rt} \end{align*}\]
Apply Ito-Doeblin formula, we have: \[\begin{align*} d(e^{-rt}S_t) &= -re^{-rt}S_t\,dt+e^{-rt}\,dS_t -re^{-rt}\,dt \,dS_t \\ &= -re^{-rt}S_t\,dt+e^{-rt}(\mu S_t \,dt + \sigma S_t \,dW_t^\mathcal{P}) -re^{-rt}\,dt(\mu S_t \,dt + \sigma S_t \,dW_t^\mathcal{P}) \\ &= e^{-rt}S_t[(\mu-r) \,dt+\sigma \,d W_t^{\mathcal{P}}] -\mu re^{-rt}St\,dt^2 + \sigma S_t \,dt\,dW_t^\mathcal{P})\\ &= e^{-rt}S_t[\mu-r \,dt+\sigma (\,d W_t^{\mathcal{Q}}-\theta_t) \,dt] \\ &= e^{-rt}S_t[(\mu-r-\sigma \theta_t) \,dt+\sigma \,d W_t^{\mathcal{Q}} \,dt] \end{align*}\]
In order to have the process \(\{e^{−rt}S_t\}_{t \geq 0}\) be the martigale process, we choose \(\theta_t\) such that: \[\mu-r-\sigma \theta_t=0 \rightarrow \theta_t=\frac{\mu-r}{\sigma} \] then \[E^\mathcal{P}[e^{\frac{1}{2} \int_0^T \theta_t^2\,dt}]<\infty\]
By the Girsanov’s theorem, there is a risk neutral measure \(\mathcal{Q}\) defined by the Radon-Nykodym \[E \left[\frac{\,d \mathcal{Q}}{\,d \mathcal{P}} |\mathcal{F}_t\right]=Z_t=e^{-\int_0^t \theta_s \,d W_s^\mathcal{P}-\frac{1}{2} \theta_s^2 \,ds}\] such that \(W_t^\mathcal{Q}=W_t^\mathcal{P}+\int_0^t\theta_s \,ds\)
is the standard Brownian motion under \(\mathcal{Q}\). \[W_t^\mathcal{Q}=W_t^\mathcal{P}+\int_0^t\theta_s \,ds \\ \rightarrow \,dW_t^\mathcal{Q}=\,dW_t^\mathcal{P}+\theta_t\,dt\]
Since \(e^{-rt}S_t\) is a martingale under \(\mathcal{Q}\) and \(\mathcal{Q}\) is an equivalent martingale measure to \(\mathcal{P}\), we can write: \[\begin{align*} \,dS_t &= \mu S_t \,dt + \sigma S_t \,dW_t^\mathcal{P} \\ &=\mu S_t \,dt + \sigma S_t (\,dW_t^\mathcal{Q}-\theta_t \,dt) \\ &=\mu S_t \,dt + \sigma S_t (\,dW_t^\mathcal{Q}-\frac{\mu-r}{\sigma} \,dt) \\ &=\left(\mu-\sigma\cdot\frac{\mu-r}{\sigma} \right)S_t \,dt + \sigma S_t \,dW_t^\mathcal{Q} \\ &=rS_t \,dt + \sigma S_t \,dW_t^\mathcal{Q} \\ \end{align*}\]
Let \(Y_t=\ln S_t\) and \(f(t,x)=\ln x\), we have: \[\begin{align*} f_t&=0 \\ f_x&=\frac{1}{x} \\ f_{tt}&=0 \\ f_{xx}&=-\frac{1}{x^2} \\ f_{tx}&=0 \\ \end{align*}\]
Applying Ito Doeblin formula, we have: \[\begin{align*} \,d(Y_t)&=\frac{1}{S_t} \,dS_t-\frac{1}{2}\frac{1}{S_t^2}\,dS_t^2 \\ &=\frac{1}{S_t} (rS_t \,dt + \sigma S_t \,dW_t^\mathcal{Q})-\frac{1}{2}\frac{1}{S_t^2}(rS_t \,dt + \sigma S_t \,dW_t^\mathcal{Q})^2 \\ &=(r-\frac{1}{2} \cdot \sigma^2)\,dt + \sigma \,dW_t^\mathcal{Q} \\ \,d(\ln S_t)&=(r-\frac{1}{2} \cdot \sigma^2)\,dt + \sigma \,dW_t^\mathcal{Q} \\ \int_t^T \,d(\ln S_s)&=\int_t^T(r-\frac{1}{2} \cdot \sigma^2)\,ds + \int_t^T\sigma \,dW_s^\mathcal{Q} \\ \ln \left( \frac{S_T}{S_t} \right)&=(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^\mathcal{Q}-W_t^\mathcal{Q}) \\ S_T&=e^{\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})} \end{align*}\]
We have \[\begin{align*} \ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}}) &\sim \mathcal{N}(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t),\sigma^2(T-t)) \\ e^{\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})} &\sim \ln \mathcal{N}(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t),\sigma^2(T-t)) \\ (S_T|S_t) &\sim \ln \mathcal{N}(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t),\sigma^2(T-t)) \\ \end{align*}\]
The probability density function of \((S_T|S_t)\) is \[f_{S_T|S_t}(x)=\frac{1}{\sigma x\sqrt{2\pi(T-t)}}e^{-\frac{1}{2} \frac{(\ln x-(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \]
From Feyman-Kac formula we have: \[ C(S_t,t)=e^{-r(T-t)} \mathbb{E}^\mathcal{Q}[\max(S_T-K,0)|\mathcal{F}_t] \]
Applying the moment generating function, we have: \[\begin{align*} C(S_t,t)&=e^{-r(T-t)}\int_{-\infty}^\infty \max(x-K,0) f_{S_T|S_t}(x) \,dx \\ &=e^{-r(T-t)}\int_K^\infty (x-K) \frac{1}{\sigma x\sqrt{2\pi}\sqrt{(T-t)}}e^{-\frac{1}{2} \frac{(\ln x-(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \,dx \end{align*}\]
\[\begin{align*} C(S_t,t)&=e^{-r(T-t)}\int_K^\infty (x-K) \frac{1}{\sigma x\sqrt{2\pi}\sqrt{(T-t)}}e^{-\frac{1}{2} \frac{(\ln x-(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \,dx \\ &=\frac{e^{-r(T-t)}}{\sigma \sqrt{2\pi}\sqrt{(T-t)}} \int_K^\infty (x-K)e^{-\frac{1}{2} \frac{(\ln x-\ln S_t -(r-\frac{1}{2} \cdot \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \frac{1}{x}\,dx \end{align*}\]
Let \(u=\frac{\ln x-\ln S_t -(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}\), we have \[\begin{align*} \ln \left( \frac{x}{S_t} \right)&=\sigma u\sqrt{T-t}+(r-\frac{1}{2} \cdot \sigma^2)(T-t) \\ x&=S_te^{\sigma u\sqrt{T-t}+(r-\frac{1}{2} \cdot \sigma^2)(T-t)} \end{align*}\]
\[\begin{align*} \,du&=\frac{1}{\sigma x\sqrt{T-t}}\,dx \\ \sigma\sqrt{T-t}\,du&=\frac{1}{x}\,dx \end{align*}\]
\[\begin{align*} u_K&=\frac{\ln K-\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}} \\ &=\frac{\ln \left( \frac{K}{S_t} \right)+(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}} \end{align*}\]
The value of the call option becomes: \[\begin{align*} C(S_t,t)&=\frac{e^{-r(T-t)}}{\sigma \sqrt{2\pi} \sqrt{T-t}} \int_K^\infty (x-K)e^{-\frac{1}{2} \frac{(\ln x-\ln S_t -(r-\frac{1}{2} \cdot \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \frac{1}{x}\,dx \\ &=\frac{e^{-r(T-t)}}{\sigma \sqrt{2\pi}\sqrt{T-t}} \int_{u_K}^\infty (S_te^{\sigma u\sqrt{T-t}+(r-\frac{1}{2} \cdot \sigma^2)(T-t)}-K)e^{-\frac{1}{2} u^2} \sigma\sqrt{T-t}\,du \\ &=\frac{e^{-r(T-t)}}{\sigma\sqrt{2\pi}\sqrt{T-t}} \sigma\sqrt{T-t} \int_{u_K}^\infty (S_te^{\sigma u\sqrt{T-t}+(r-\frac{1}{2} \cdot \sigma^2)(T-t)}e^{-\frac{1}{2} u^2}-Ke^{-\frac{1}{2} u^2}) \,du \\ &=\frac{e^{-r(T-t)}}{\sqrt{2\pi}} \int_{u_K}^\infty (S_te^{\sigma u\sqrt{T-t}+(r-\frac{1}{2} \cdot \sigma^2)(T-t)-\frac{1}{2} u^2}-Ke^{-\frac{1}{2} u^2}) \,du \\ &=\frac{1}{\sqrt{2\pi}} \int_{u_K}^\infty S_te^{\sigma u\sqrt{T-t}-\frac{1}{2} \cdot \sigma^2(T-t)-\frac{1}{2} u^2}\,du \\ &-\frac{e^{-r(T-t)}}{\sqrt{2\pi}} \int_{u_K}^\infty Ke^{-\frac{1}{2} u^2} \,du \\ &=I_1 - I_2 \\ \end{align*}\]
\[\begin{align*} I_1 &=S_t \cdot \frac{1}{\sqrt{2\pi}} \int_{u_K}^\infty e^{-\frac{1}{2} u^2+\sigma u\sqrt{T-t}-\frac{1}{2} \cdot \sigma^2(T-t)}\,du \\ &=S_t \cdot \frac{1}{\sqrt{2\pi}} \int_{u_K}^\infty e^{-\frac{1}{2} (u-\sigma\sqrt{T-t})^2}\,du \\ &=S_t \cdot \frac{1}{\sqrt{2\pi}} \int_{u_K}^\infty e^{-\frac{1}{2} (u-\sigma\sqrt{T-t})^2}\,d(u-\sigma\sqrt{T-t}) \\ &=S_t \cdot \frac{1}{\sqrt{2\pi}} \int_{u_K-\sigma\sqrt{T-t}}^\infty e^{-\frac{1}{2} u^2}\,du \\ &= S_t \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\sigma\sqrt{T-t}-u_K} e^{-\frac{1}{2} u^2}\,du \\ &=S_t\cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\sigma\sqrt{T-t}-\frac{\ln \left( \frac{K}{S_t} \right)-(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}} e^{-\frac{1}{2} u^2}\,du \\ &=S_t\cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\frac{\sigma^2(T-t)-\ln \left( \frac{K}{S_t} \right)-(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}} e^{-\frac{1}{2} u^2}\,du \\ &=S_t\cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\frac{\ln \left( \frac{S_t}{K} \right)+(r+\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}} e^{-\frac{1}{2} u^2}\,du \\ &=S_t \cdot N \left({\frac{\ln \left( \frac{S_t}{K} \right)+(r+\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}}\right) \end{align*}\]
\[\begin{align*} I_2 &=Ke^{-r(T-t)} \cdot \frac{1}{\sqrt{2\pi}} \int_{u_K}^\infty e^{-\frac{1}{2} u^2} \,du \\ &=Ke^{-r(T-t)} \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{-u_K} e^{-\frac{1}{2} u^2} \,du \\ &=Ke^{-r(T-t)} \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{-\frac{\ln \left( \frac{K}{S_t} \right)-(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}} e^{-\frac{1}{2} u^2} \,du \\ &=Ke^{-r(T-t)} \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\frac{\ln \left( \frac{S_t}{K} \right)+(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}} e^{-\frac{1}{2} u^2} \,du \\ &=Ke^{-r(T-t)} \cdot N \left({\frac{\ln \left( \frac{S_t}{K} \right)+(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}}\right) \end{align*}\]
Answer: The value of the call option is: \[\begin{align*} C(S_t,t) = &S_t \cdot N \left({\frac{\ln \left( \frac{S_t}{K} \right)+(r+\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}}\right) - Ke^{-r(T-t)} \cdot N \left({\frac{\ln \left( \frac{S_t}{K} \right)+(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}}\right) \end{align*}\]
Suppose the stochastic process of the asset price \(S_t\) follows a geometric Brownian motion: \[\,d S_t = 0.2 S_t \,dt + 0.3 S_t \,dW_t^\mathcal{P}\] where \(W_t^P\) is a standard Brownian motion on the real world probability space, denoted by \((\omega, \mathcal{F},\mathcal{P})\). A European put option written on the stock with parameters: exercise price \(K = 15\), maturity time \(T=2\), \(t=0.5\) (year), the risk-free interest rate is \(r = 0.06\), \(S_t=12\).
Let \(\widetilde W_t = W_t + \theta t\), with \(\theta\) being a constant is an adapted process with respect to the filtration \(()\mathcal{F}_t)_{t \geq 0}\). Find \(\theta\) such that the discounted \((e^{−0.07t}S_t)_{t \ge 0}\) is a martingale process with respect to \(\widetilde W_t\).
Derive the Black-Scholes-Merton formula and then compute the option price.
\[S_T=e^{\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})}\]
We have \[\begin{align*} \ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}}) &\sim \mathcal{N}(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t),\sigma^2(T-t)) \\ e^{\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})} &\sim \log\mathcal{N}(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t),\sigma^2(T-t)) \\ (S_T|S_t) &\sim \log\mathcal{N}(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t),\sigma^2(T-t)) \\ \end{align*}\]
The probability density function of \((S_T|S_t)\) is \[f_{S_T|S_t}(x)=\frac{1}{\sigma x\sqrt{2\pi(T-t)}}e^{-\frac{1}{2} \frac{(\ln x-(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \]
From Feyman-Kac formula we have: \[P(S_t,t)=e^{-r(T-t)} \mathbb{E}^\mathcal{Q}[\max(K-S_T,0)|\mathcal{F}_t] \]
Applying the moment generating function, we have: \[\begin{align*} P(S_t,t)&=e^{-r(T-t)}\int_{0}^\infty \max(K-x,0) f_{S_T|S_t}(x) \,dx \\ &=e^{-r(T-t)}\int_{-\infty}^\infty \max(K-x,0) f_{S_T|S_t}(x) \,dx \\ &=e^{-r(T-t)}\int_{-\infty}^K (K-x) \frac{1}{\sigma x\sqrt{2\pi}\sqrt{(T-t)}}e^{-\frac{1}{2} \frac{(\ln x-(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \,dx \\ &=e^{-r(T-t)}\int_{-\infty}^K (K-x) \frac{1}{\sigma x\sqrt{2\pi}\sqrt{(T-t)}}e^{-\frac{1}{2} \frac{(\ln x-(\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \,dx \\ &=\frac{e^{-r(T-t)}}{\sigma \sqrt{2\pi}\sqrt{(T-t)}} \int_{-\infty}^K (K-x)e^{-\frac{1}{2} \frac{(\ln x-\ln S_t -(r-\frac{1}{2} \cdot \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \frac{1}{x}\,dx \end{align*}\]
Let \(u=\frac{\ln x-\ln S_t -(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}\), we have \[\begin{align*} \ln \left( \frac{x}{S_t} \right)&=\sigma u\sqrt{T-t}+(r-\frac{1}{2} \cdot \sigma^2)(T-t) \\ x&=S_te^{\sigma u\sqrt{T-t}+(r-\frac{1}{2} \cdot \sigma^2)(T-t)} \end{align*}\]
\[\begin{align*} \,du&=\frac{1}{\sigma x\sqrt{T-t}}\,dx \\ \sigma\sqrt{T-t}\,du&=\frac{1}{x}\,dx \end{align*}\]
\[\begin{align*} u_K&=\frac{\ln K-\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}} \\ &=\frac{\ln \left( \frac{K}{S_t} \right)+(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}} \end{align*}\]
The value of the put option becomes: \[\begin{align*} P(S_t,t)&=\frac{e^{-r(T-t)}}{\sigma \sqrt{2\pi} \sqrt{T-t}} \int_{-\infty}^K (K-x)e^{-\frac{1}{2} \frac{(\ln x-\ln S_t -(r-\frac{1}{2} \cdot \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \frac{1}{x}\,dx \\ &=\frac{e^{-r(T-t)}}{\sigma \sqrt{2\pi}\sqrt{T-t}} \int_{-\infty}^{u_K} (-S_te^{\sigma u\sqrt{T-t}+(r-\frac{1}{2} \cdot \sigma^2)(T-t)}+K)e^{-\frac{1}{2} u^2} \sigma\sqrt{T-t}\,du \\ &=\frac{e^{-r(T-t)}\sqrt{T-t}}{\sqrt{2\pi}\sqrt{T-t}} \int_{-\infty}^{u_K} (-S_te^{\sigma u\sqrt{T-t}+(r-\frac{1}{2} \cdot \sigma^2)(T-t)}e^{-\frac{1}{2} u^2}+Ke^{-\frac{1}{2} u^2}) \,du \\ &=\frac{e^{-r(T-t)}}{\sqrt{2\pi}} \int_{-\infty}^{u_K} (-S_te^{\sigma u\sqrt{T-t}+(r-\frac{1}{2} \cdot \sigma^2)(T-t)-\frac{1}{2} u^2}+Ke^{-\frac{1}{2} u^2}) \,du \\ &=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{u_K} -S_te^{\sigma u\sqrt{T-t}-\frac{1}{2} \cdot \sigma^2(T-t)-\frac{1}{2} u^2}\,du \\ &+\frac{e^{-r(T-t)}}{\sqrt{2\pi}} \int_{-\infty}^{u_K} Ke^{-\frac{1}{2} u^2} \,du \\ &= I_1+I_2 \end{align*}\]
\[\begin{align*} I_1&=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{u_K} -S_te^{\sigma u\sqrt{T-t}-\frac{1}{2} \cdot \sigma^2(T-t)-\frac{1}{2} u^2}\,du \\ &=-S_t \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{u_K} e^{-\frac{1}{2} (u-\sigma\sqrt{T-t})^2}\,du \\ &=-S_t \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{u_K} e^{-\frac{1}{2} (u-\sigma\sqrt{T-t})^2}\,d(u-\sigma\sqrt{T-t}) \\ &=-S_t \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{u_K-\sigma\sqrt{T-t}} e^{-\frac{1}{2} u^2}\,du \\ &=-S_t\cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\frac{\ln \left( \frac{K}{S_t} \right)-(r-\frac{1}{2} \cdot \sigma^2)(T-t)-\sigma^2(T-t)}{\sigma\sqrt{T-t}}} e^{-\frac{1}{2} u^2}\,du \\ &=-S_t\cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{-\frac{\ln \left( \frac{S_t}{K} \right)+(r+\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}} e^{-\frac{1}{2} u^2}\,du \\ &=-S_t \cdot N \left({-\frac{\ln \left( \frac{S_t}{K} \right)+(r+\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}}\right) \\ \end{align*}\]
\[\begin{align*} I_2&=Ke^{-r(T-t)} \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{u_K} e^{-\frac{1}{2} u^2} \,du \\ &=Ke^{-r(T-t)} \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\frac{\ln \left( \frac{K}{S_t} \right)-(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}} e^{-\frac{1}{2} u^2} \,du \\ &=Ke^{-r(T-t)} \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{-\left(\frac{\ln \left( \frac{S_t}{K} \right)+(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}\right)} e^{-\frac{1}{2} u^2} \,du \\ &=Ke^{-r(T-t)} \cdot N \left(-{\frac{\ln \left( \frac{S_t}{K} \right)+(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}}\right) \end{align*}\]
Answer: The value of the put option is: \[\begin{align*} P(S_t,t) &= -S_t \cdot N \left({-\frac{\ln \left( \frac{S_t}{K} \right)+(r+\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}}\right) + Ke^{-r(T-t)} \cdot N \left(-{\frac{\ln \left( \frac{S_t}{K} \right)+(r-\frac{1}{2} \cdot \sigma^2)(T-t)}{\sigma\sqrt{T-t}}}\right)\\ \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows a geometric Brownian motion:
\[\,d S_t = 0.05 \,dt + 0.3 \,d W_t^{\mathcal{P}},\]
where \(W_t^{\mathcal{P}}\) is a standard Brownian motion on the real world probability space, denoted by \((\omega, \mathcal{F},\mathcal{P})\). A European digital call option written on the stock with parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\), \(t = 0.5\), \(T = 1\), \(S_t = 9\).
Let \(W_t^Q = W_t^P + \int_0^t \theta_s \,d s\), with \((\theta_t)_{t \geq 0}\) is an adapted process with \((\mathcal{F}_t)_{t \geq 0}\). \[\begin{align*} d(e^{−rt}S_t) &= −re^{−rt}S_t \,dt + e^{−rt}\,d S_t \\ &= −re^{−rt}S_t \,dt + e^{−rt}(\mu \,dt + \sigma \,dW_t^P) \\ &= e^{−rt}((\mu − rS_t) \,dt + \sigma (W_t^Q - \theta_t \,d t)) \\ &= e^{−rt}((\mu − rS_t -\sigma\theta_t) \,dt + \sigma W_t^Q) \\ \end{align*}\]
We choose \(\theta_t = \frac{\mu -r S_t}{\sigma}\) then \(E^{\mathcal{P}} \left(e^{\frac{1}{2} \int_0^T \theta_t^2 \,dt} \right)< \infty\) and \(Z_t = e^{\int_0^t \theta_s \,d W_s- \frac{1}{2} \int_0^t \theta_s^2 \,ds}\) is a positive martingale. From Girsanov’s theorem, there exist a risk-neutral measure \(\mathcal{Q}\) defined by \(E\left[\frac{\,d \mathcal{Q}}{\,d \mathcal{P}} | \mathcal{F}_t \right] = Z_t\) such that \(W_t^{\mathcal{Q}}\) is a Standard Brownian motion under \(\mathcal{Q}\). As \(d(e^{−rt}S_t) = \sigma e^{−rt} \,d W_t^{\mathcal{Q}}\), \(e^{−rt}S_t\) is a martingale under \(\mathcal{Q}\) and \(\mathcal{Q}\) is an equivalent martingale measure to \(\mathcal{P}\)
By taking the integral both sides of the above equality from t to T, we obtain: \[\begin{align*} \int_t^T d(e^{−rt}S_t) &= \int_t^T\sigma e^{−ru} \,d W_t^{\mathcal{Q}} \\ e^{−rt} S_T - e^{−rt} S_t &= \int_t^T\sigma e^{−ru} \,d W_t^{\mathcal{Q}} \\ S_T&= e^{r(T-t)} S_t + \int_t^T \sigma e^{r(T-u)} \,d W_t^{\mathcal{Q}} \\ & \sim \mathcal{N} \left(e^{r(T-t)} S_t , \left( \int_t^T \sigma e^{r(T-u)} \,d W_t^{\mathcal{Q}}\right)^2 \right) \\ & \sim \mathcal{N} \left( e^{r(T-t)} S_t , \frac{\sigma^2}{2r}(e^{2r(T-t)}-1) \right) \end{align*}\]
The probability density function of \(S_T|S_t\) can be written as: \[f_{S_T|s_t}(y) = \frac{1}{v \sqrt{2\pi}} e^{-\frac{(y-m)^2}{2v^2}}\] where \(m=e^{r(T-t)} S_t\) and \(v^2=\frac{\sigma^2}{2r}(e^{2r(T-t)}-1)\)
From the Feyman-Kac formula, where \(E^{\mathcal{Q}}\) is the expectation under the risk-neutral measure Q, the price of the European put option: \[\begin{align*} C(S_t,t) &= e^{-r(T-t)}E^{\mathcal{Q}} [\max(S_t-K,0)|\mathcal{F}_t] \\ &= e^{-r(T-t)} \int_{-\infty}^{\infty}\max(y-K,0) f_{S_T|s_t}(y) \,dy \\ &= e^{-r(T-t)} \int_{K}^{\infty} \frac{y-K}{v \sqrt{2\pi}} e^{-\frac{(y-m)^2}{2v^2}} \,dy \\ &= e^{-r(T-t)} \int_{K}^{\infty} \frac{y-K}{\sqrt{2\pi}} e^{-\frac{(y-m)^2}{2v^2}} \frac{1}{v}\,dy \end{align*}\]
Using the variable transformation \(u = \frac{y-m}{v}\) then \(y=m+uv\) and \(\,du = \frac{1}{v} \,dy\), we obtain: \[\begin{align*} C(S_t,t) &= e^{-r(T-t)} \int_{\frac{K-m}{v}}^{\infty} \frac{(m+uv)-K}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= e^{-r(T-t)}\int_{\frac{K-m}{v}}^{\infty} \frac{m-K}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du + e^{-r(T-t)}\int_{\frac{K-m}{v}}^{\infty} \frac{uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= I_1 + I_2 \end{align*}\]
With the notation \[\begin{align*} m&=e^{r(T-t)} S_t \\ v&=\sigma \sqrt{\frac{e^{2r(T-t)}-1}{2r}} \\ \hat \sigma &= \sigma \sqrt{\frac{1-e^{-2r(T-t)}}{2r}} \\ d &= \frac{S_t-Ke^{-r(T-t)}}{ \hat \sigma} \\ N(x) &= \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}y^2} \,dy \\ N'(x) &=\frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} \end{align*}\]
We have \[\begin{align*} I_1 &= e^{-r(T-t)}\int_{\frac{K-m}{v}}^{\infty} \frac{m-K}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= e^{-r(T-t)} (m-K) \int_{-\infty}^{\frac{m-K}{v}} \frac{1}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= e^{-r(T-t)} (m-K) N \left( \frac{m-K}{v} \right) \\ &= e^{-r(T-t)} (e^{r(T-t)} S_t-K) N \left( \frac{e^{r(T-t)} S_t-K}{\sigma \sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) \\ &= (S_t-Ke^{-r(T-t)}) N \left( \frac{ S_t-Ke^{-r(T-t)}}{\sigma \sqrt{\frac{1-e^{-2r(T-t)}}{2r}}} \right)\\ I_2 &= e^{-r(T-t)}\int_{\frac{K-m}{v}}^{\infty} \frac{uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= ve^{-r(T-t)}\int_{\frac{K-m}{v}}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,d \left(\frac{u^2}{2} \right) \\ &= ve^{-r(T-t)}\int_{\frac{1}{2} \left(\frac{K-m}{v} \right)^2}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-x} \,dx \\ &= ve^{-r(T-t)} \frac{1}{\sqrt{2\pi}} \cdot (-e^{-x}) \Biggr|_{\frac{1}{2} \left(\frac{K-m}{v} \right)^2}^{\infty} \\ &= ve^{-r(T-t)} \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{1}{2} \left(\frac{m-K}{v} \right)^2} \\ &= \sigma \sqrt{\frac{e^{2r(T-t)}-1}{2r}}e^{-r(T-t)} \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{1}{2} \left(\frac{m-K}{v} \right)^2} \\ &= \sigma e^{r(T-t)} \sqrt{\frac{1-e^{2r(T-t)}}{2r}}e^{-r(T-t)} N' \left( \frac{ S_t-Ke^{-r(T-t)}}{\sigma \sqrt{\frac{1-e^{-2r(T-t)}}{2r}}} \right) \end{align*}\]
Answer: The value of the call option is: \[\begin{align*} C(S_t,t) &= (S_t-Ke^{-r(T-t)}) N \left( \frac{ S_t-Ke^{-r(T-t)}}{\sigma \sqrt{\frac{1-e^{-2r(T-t)}}{2r}}} \right) + \sigma \sqrt{\frac{1-e^{2r(T-t)}}{2r}} N' \left( \frac{ S_t-Ke^{-r(T-t)}}{\sigma \sqrt{\frac{1-e^{-2r(T-t)}}{2r}}} \right) \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows a geometric Brownian motion:
\[\,d S_t = 0.05 \,dt + 0.3 \,d W_t^{\mathcal{P}},\]
where \(W_t^{\mathcal{P}}\) is a standard Brownian motion on the real world probability space, denoted by \((\omega, \mathcal{F},\mathcal{P})\). A European digital put option written on the stock with parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\), \(t = 0.5\), \(T = 1\), \(S_t = 9\).
\[\begin{align*} S_T&= e^{r(T-t)} S_t + \int_t^T \sigma e^{r(T-u)} \,d W_t^{\mathcal{Q}} \\ & \sim \mathcal{N} \left(e^{r(T-t)} S_t , \left( \int_t^T \sigma e^{r(T-u)} \,d W_t^{\mathcal{Q}}\right)^2 \right) \\ & \sim \mathcal{N} \left( e^{r(T-t)} S_t , \frac{\sigma^2}{2r}(e^{2r(T-t)}-1) \right) \end{align*}\]
The probability density function of \(S_T|S_t\) can be written as: \[f_{S_T|s_t}(y) = \frac{1}{v \sqrt{2\pi}} e^{-\frac{(y-m)^2}{2v^2}}\] where \(m=e^{r(T-t)} S_t\) and \(v^2=\frac{\sigma^2}{2r}(e^{2r(T-t)}-1)\)
From the Feyman-Kac formula, where \(E^{\mathcal{Q}}\) is the expectation under the risk-neutral measure Q, the price of the European put option: \[\begin{align*} P(S_t,t) &= e^{-r(T-t)}E^{\mathcal{Q}} [\max(K-S_t,0)|\mathcal{F}_t] \\ &= e^{-r(T-t)} \int_{-\infty}^{\infty}\max(K-y,0) f_{S_T|s_t}(y) \,dy \\ &= e^{-r(T-t)} \int_{-\infty}^{K} \frac{K-y}{v \sqrt{2\pi}} e^{-\frac{(y-m)^2}{2v^2}} \,dy \\ &= e^{-r(T-t)} \int_{-\infty}^{K} \frac{K-y}{\sqrt{2\pi}} e^{-\frac{(y-m)^2}{2v^2}} \frac{1}{v}\,dy \end{align*}\]
Using the variable transformation \(u = \frac{y-m}{v}\) then \(y=m+uv\) and \(\,du = \frac{1}{v} \,dy\), we obtain: \[\begin{align*} P(S_t,t) &= e^{-r(T-t)} \int_{-\infty}^{\frac{K-m}{v}} \frac{K-(m+uv)}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= e^{-r(T-t)}\int_{-\infty}^{\frac{K-m}{v}} \frac{K-m}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du + \left(- e^{-r(T-t)}\int_{-\infty}^{\frac{K-m}{v}} \frac{uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \right) \\ &= I_1 + I_2 \end{align*}\]
With the notation \[\begin{align*} m&=e^{r(T-t)} S_t \\ v&=\sigma \sqrt{\frac{e^{2r(T-t)}-1}{2r}} \\ \hat \sigma &= \sigma \sqrt{\frac{1-e^{-2r(T-t)}}{2r}} \\ d &= \frac{S_t-Ke^{-r(T-t)}}{ \hat \sigma} \\ N(x) &= \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}y^2} \,dy \\ N'(x) &=\frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} \end{align*}\]
We have \[\begin{align*} I_1 &= e^{-r(T-t)}\int_{-\infty}^{\frac{K-m}{v}} \frac{K-m}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= e^{-r(T-t)} (K-m) \int_{-\infty}^{\frac{K-m}{v}} \frac{1}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= e^{-r(T-t)} (K-m) N \left( -\frac{m-K}{v} \right) \\ &= e^{-r(T-t)} (K-e^{r(T-t)} S_t) N \left( -\frac{e^{r(T-t)} S_t-K}{\sigma \sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) \\ &= -(S_t-Ke^{-r(T-t)}) N \left( -\frac{ S_t-Ke^{-r(T-t)}}{\sigma \sqrt{\frac{1-e^{-2r(T-t)}}{2r}}} \right)\\ I_2 &= -e^{-r(T-t)}\int_{-\infty}^{\frac{K-m}{v}} \frac{uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= ve^{-r(T-t)}\int_{-\infty}^{\frac{K-m}{v}} \frac{1}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,d \left(-\frac{u^2}{2} \right) \\ &= ve^{-r(T-t)}\int_{-\infty}^{-\frac{1}{2} \left(\frac{K-m}{v} \right)^2} \frac{1}{\sqrt{2\pi}} e^{x} \,dx \\ &= ve^{-r(T-t)} \frac{1}{\sqrt{2\pi}} \cdot (e^{x}) \Biggr|_{-\infty}^{-\frac{1}{2} \left(\frac{K-m}{v} \right)^2} \\ &= ve^{-r(T-t)} \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{1}{2} \left(\frac{m-K}{v} \right)^2} \\ &= \sigma \sqrt{\frac{e^{2r(T-t)}-1}{2r}}e^{-r(T-t)} \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{1}{2} \left(-\frac{m-K}{v} \right)^2} \\ &= \sigma e^{r(T-t)} \sqrt{\frac{1-e^{2r(T-t)}}{2r}}e^{-r(T-t)} N' \left( -\frac{ S_t-Ke^{-r(T-t)}}{\sigma \sqrt{\frac{1-e^{-2r(T-t)}}{2r}}} \right) \end{align*}\]
Answer: The value of the put option is: \[\begin{align*} P(S_t,t) &= -(S_t-Ke^{-r(T-t)}) N \left( -\frac{ S_t-Ke^{-r(T-t)}}{\sigma \sqrt{\frac{1-e^{-2r(T-t)}}{2r}}} \right) + \sigma \sqrt{\frac{1-e^{2r(T-t)}}{2r}} N' \left( -\frac{ S_t-Ke^{-r(T-t)}}{\sigma \sqrt{\frac{1-e^{-2r(T-t)}}{2r}}} \right)\\ \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows a geometric Brownian motion:
\[\,d S_t = 0.05 S_t \,dt + 0.3 S_t \,d W_t^{\mathcal{P}},\]
where \(W_t^{\mathcal{P}}\) is a standard Brownian motion on the real world probability space, denoted by \((\omega, \mathcal{F},\mathcal{P})\). A European digital call option written on the stock with parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\), \(t = 0.5\), \(T = 1\), \(S_t = 9\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\),\(T\) and \(S_t\) for the fair price of the option?
\[S_T=e^{\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})}\]
From the Feynman–Kac formula the digital call option is given by: \[C_d(S_t,t) = e^{−r(T−t)}E^Q [\mathcal{I}_{S_T > K} |Ft],\] where \(E^Q\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{S_T} = \begin{cases} 1 & \text{if } S_t > K\\ 0 & \text{if } S_t \leq K \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain \[\begin{align*} C_d(S_t,t) &= e^{−r(T−t)}E^Q [\mathcal{I}_{S_T > K} | \mathcal{F}_t] \\ &= e^{−r(T−t)} [1 \cdot \mathcal{Q}(S_T > K | \mathcal{F}_t) + 0 \cdot \mathcal{Q}(S_T \leq K | \mathcal{F}_t)] \\ &= e^{−r(T−t)} \cdot \mathcal{Q}(S_T > K | \mathcal{F}_t) \\ &= e^{−r(T−t)} \cdot \mathcal{Q}(e^{\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})} > K | \mathcal{F}_t) \\ &= e^{−r(T−t)} \cdot \mathcal{Q} \left(\frac{W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}}}{\sqrt{T-t}} > \frac{\ln \left( \frac{K}{S_t}\right) - \left(r-\frac{1}{2}\sigma^2 \right)(T-t)}{\sigma \sqrt{T-t}} \right) \\ &= e^{−r(T−t)} \cdot \mathcal{Q} \left(\mathcal{Z} < -\frac{\ln \left( \frac{K}{S_t}\right) - \left(r-\frac{1}{2}\sigma^2 \right)(T-t)}{\sigma \sqrt{T-t}} \right) \\ &= e^{−r(T−t)} \cdot N \left(\frac{\ln \left( \frac{S_t}{K}\right) + \left(r-\frac{1}{2}\sigma^2 \right)(T-t)}{\sigma \sqrt{T-t}} \right) \end{align*}\] where \(N(x) = \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}y^2}\,dy\)
Answer: The value of the call option is: \[\begin{align*} C(S_t,t) &= e^{−r(T−t)} \cdot N \left(\frac{\ln \left( \frac{S_t}{K}\right) + \left(r-\frac{1}{2}\sigma^2 \right)(T-t)}{\sigma \sqrt{T-t}} \right) \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows a geometric Brownian motion:
\[\,d S_t = 0.05 S_t \,dt + 0.34 S_t \,d W_t^{\mathcal{P}},\]
where \(W_t^{\mathcal{P}}\) is a standard Brownian motion on the real world probability space, denoted by \((\omega, \mathcal{F},\mathcal{P})\). A European digital put option written on the stock with parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\), \(t = 0.5\), \(T = 1\), \(S_t = 9\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\),\(T\) and \(S_t\) for the fair price of the option?
\[S_T=e^{\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})}\]
From the Feynman–Kac formula the digital call option is given by: \[C_d(S_t,t) = e^{−r(T−t)}E^Q [\mathcal{I}_{S_T > K} |Ft],\] where \(E^Q\). is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{S_T} = \begin{cases} 1 & \text{if } S_t > K\\ 0 & \text{if } S_t \leq K \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain \[\begin{align*} P_d(S_t,t) &= e^{−r(T−t)}E^Q [\mathcal{I}_{S_T< K} | \mathcal{F}_t] \\ &= e^{−r(T−t)} [1 \cdot \mathcal{Q}(S_T < K | \mathcal{F}_t) + 0 \cdot \mathcal{Q}(S_T \geq K | \mathcal{F}_t)] \\ &= e^{−r(T−t)} \cdot \mathcal{Q}(S_T < K | \mathcal{F}_t) \\ &= e^{−r(T−t)} \cdot \mathcal{Q}(e^{\ln S_t +(r-\frac{1}{2} \cdot \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})} < K | \mathcal{F}_t) \\ &= e^{−r(T−t)} \cdot \mathcal{Q} \left(\frac{W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}}}{\sqrt{T-t}} < \frac{\ln \left( \frac{K}{S_t}\right) - \left(r-\frac{1}{2}\sigma^2 \right)(T-t)}{\sigma \sqrt{T-t}} \right) \\ &= e^{−r(T−t)} \cdot \mathcal{Q} \left(\mathcal{Z} < \frac{\ln \left( \frac{K}{S_t}\right) - \left(r-\frac{1}{2}\sigma^2 \right)(T-t)}{\sigma \sqrt{T-t}} \right) \\ &= e^{−r(T−t)} \cdot N \left(-\frac{\ln \left( \frac{S_t}{K}\right) + \left(r-\frac{1}{2}\sigma^2 \right)(T-t)}{\sigma \sqrt{T-t}} \right) \end{align*}\] where \(N(x) = \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}y^2}\,dy\) )(T-t)}{}$.
Answer: The value of the put option is: \[\begin{align*} P(S_t,t) &= e^{−r(T−t)} \cdot N \left(-\frac{\ln \left( \frac{S_t}{K}\right) + \left(r-\frac{1}{2}\sigma^2 \right)(T-t)}{\sigma \sqrt{T-t}} \right) \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows an arithmetic Brownian motion: \[\,d S_t = 0.05 \,dt + 0.3 \,dW_t^{\mathcal{Q}},\] where \(\mathcal{Q}\) is a standard Brownian motion on a risk-neutral probability space, denoted by \((\omega, \mathcal{F}, \mathcal{Q})\). A European digital call option written on the stock with parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\), \(t = 0.5\), \(T = 1\), \(S_t = 9\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\),\(T\) and \(S_t\) and then compute the fair price.
\[\,dS_t =\mu \,dt + \sigma\,d W_t^{\mathcal{Q}}\]
Integrating both sides from \(t\) to \(T\), we have: \[\begin{align*} \int_t^T \,d S_s &= \int_t^T \mu \,d s + \int_t^T \sigma \,d W_s^{\mathcal{Q}} \\ \rightarrow S_T &= S_t + \mu(T-t) + \sigma(W_T^{\mathcal{Q}} - W_t^{\mathcal{Q}}) \\ &\sim \mathcal{N}(S_t+\mu(T-t),\sigma^2(T-t)) \end{align*}\]
From Feyman-Kac formula, we have: \[C_d(S_t,t) = e^{-r(T-t)}E^{\mathcal{Q}}[\mathcal{I}_{S_T > K}|\mathcal{F}_t],\] where \(E^{\mathcal{Q}}\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{S_T} = \begin{cases} 1 & \text{if } S_t > K\\ 0 & \text{if } S_t \leq K \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain: \[\begin{align*} C_d(S_t,t) &= e^{−r(T−t)}E^Q [\mathcal{I}_{S_T > K} | \mathcal{F}_t] \\ &= e^{−r(T−t)} [1 \cdot \mathcal{Q}(S_T > K | \mathcal{F}_t) + 0 \cdot \mathcal{Q}(S_T \leq K | \mathcal{F}_t)] \\ &= e^{−r(T−t)} \cdot \mathcal{Q}(S_T > K | \mathcal{F}_t) \\ &= e^{−r(T−t)} \cdot \mathcal{Q}(S_t+ \mu(T-t) +\sigma(W_T^{\mathcal{Q}} - W_t^{\mathcal{Q}}) > K | \mathcal{F}_t) \\ &= e^{−r(T−t)} \cdot \mathcal{Q} \left(\frac{W_T^{\mathcal{Q}} - W_t^{\mathcal{Q}}}{\sqrt{T-t}} > \frac{K -(S_t+ \mu(T-t))}{\sigma \sqrt{T-t}} | \mathcal{F}_t \right) \\ &= e^{−r(T−t)} N \left( -\frac{K -(S_t+ \mu(T-t))}{\sigma \sqrt{T-t}} | \mathcal{F}_t \right) \\ \end{align*}\]
Answer: The value of the call option is: \[\begin{align*} C(S_t,t) &= e^{−r(T−t)} N \left( -\frac{K -(S_t+ \mu(T-t))}{\sigma \sqrt{T-t}} | \mathcal{F}_t \right) \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows an arithmetic Brownian motion: \[dS_t = 0.05\,dt + 0.3\,dW_t^{\mathcal{Q}},\] where \(\mathcal{Q}\) is a standard Brownian motion on a risk-neutral probability space, denoted by \((\omega, \mathcal{F}, \mathcal{Q})\). A European digital put option written on the stock with parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\), \(t = 0.5\), \(T = 1\), \(S_t = 7\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\),\(T\) and \(S_t\) and then compute the fair price.
\[\,dS_t =\mu \,dt + \sigma\,d W_t^{\mathcal{Q}}\]
Integrating both sides from \(t\) to \(T\), we have: \[\begin{align*} \int_t^T \,d S_s &= \int_t^T \mu \,d s + \int_t^T \sigma \,d W_s^{\mathcal{Q}} \\ \rightarrow S_T &= S_t + \mu(T-t) + \sigma(W_T^{\mathcal{Q}} - W_t^{\mathcal{Q}}) \\ &\sim \mathcal{N}(S_t+\mu(T-t),\sigma^2(T-t)) \end{align*}\]
From Feyman-Kac formula, we have: \[P_d(S_t,t) = e^{-r(T-t)}E^{\mathcal{Q}}[\mathcal{I}_{S_T > K}|\mathcal{F}_t],\] where \(E^{\mathcal{Q}}\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{S_T} = \begin{cases} 1 & \text{if } S_t > K\\ 0 & \text{if } S_t \leq K \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain: \[\begin{align*} P_d(S_t,t) &= e^{−r(T−t)}E^Q [\mathcal{I}_{S_T < K} | \mathcal{F}_t] \\ &= e^{−r(T−t)} [1 \cdot \mathcal{Q}(S_T < K | \mathcal{F}_t) + 0 \cdot \mathcal{Q}(S_T \leq K | \mathcal{F}_t)] \\ &= e^{−r(T−t)} \cdot \mathcal{Q}(S_T < K | \mathcal{F}_t) \\ &= e^{−r(T−t)} \cdot \mathcal{Q}(S_t+ \mu(T-t) +\sigma(W_T^{\mathcal{Q}} - W_t^{\mathcal{Q}}) < K | \mathcal{F}_t) \\ &= e^{−r(T−t)} \cdot \mathcal{Q} \left(\frac{W_T^{\mathcal{Q}} - W_t^{\mathcal{Q}}}{\sqrt{T-t}} < \frac{K -(S_t+ \mu(T-t))}{\sigma \sqrt{T-t}} | \mathcal{F}_t \right) \\ &= e^{−r(T−t)} N \left(\frac{K -(S_t+ \mu(T-t))}{\sigma \sqrt{T-t}} | \mathcal{F}_t \right) \end{align*}\]
Answer: The value of the put option is: \[\begin{align*} P(S_t,t) &= e^{−r(T−t)} N \left(\frac{K -(S_t+ \mu(T-t))}{\sigma \sqrt{T-t}} | \mathcal{F}_t \right) \\ \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows an arithmetic Brownian motion: \[dS_t = 0.05 \,dt + 0.3\,d W_t^{\mathcal{P}},\] where \(W_t^{\mathcal{P}}\) is a standard Brownian motion on the real world probability space, denoted by \((\omega, \mathcal{F},\mathcal{P})\). A European digital call option written on the stock with parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\), \(t = 0.5\), \(T = 1\), \(S_t = 9\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\),\(T\) and \(S_t\) and then compute the fair price.
\[\begin{align*} S_T&= e^{r(T-t)} S_t + \int_t^T \sigma e^{r(T-u)} \,d W_t^{\mathcal{Q}} \\ & \sim \mathcal{N} \left(e^{r(T-t)} S_t , \left( \int_t^T \sigma e^{r(T-u)} \,d W_t^{\mathcal{Q}}\right)^2 \right) \\ & \sim \mathcal{N} \left( e^{r(T-t)} S_t , \frac{\sigma^2}{2r}(e^{2r(T-t)}-1) \right) \end{align*}\]
From the Feynman–Kac formula the digital call option is given by: \[C_d(S_t,t) = e^{−r(T−t)}E^Q [\mathcal{I}_{S_T > K} |Ft],\] where \(E^Q\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{S_T} = \begin{cases} 1 & \text{if } S_t > K\\ 0 & \text{if } S_t \leq K \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain \[\begin{align*} C_d(S_t,t) &= e^{−r(T−t)}E^Q [\mathcal{I}_{S_T > K} | \mathcal{F}_t] \\ &= e^{−r(T−t)} [1 \cdot \mathcal{Q}(S_T > K | \mathcal{F}_t) + 0 \cdot \mathcal{Q}(S_T \leq K | \mathcal{F}_t)] \\ &= e^{−r(T−t)} \cdot \mathcal{Q}(S_T > K | \mathcal{F}_t) \\ &= e^{−r(T−t)} \cdot \mathcal{Q} \left( \frac{S_T-e^{r(T-t)} S_t}{\frac{\sigma^2}{2r}(e^{2r(T-t)}-1)} > \frac{K-e^{r(T-t)} S_t}{\frac{\sigma^2}{2r}(e^{2r(T-t)}-1)} | \mathcal{F}_t \right) \\ &= e^{−r(T−t)} \cdot \mathcal{Q} \left( \mathcal{Z} < -\frac{K-e^{r(T-t)} S_t}{\frac{\sigma^2}{2r}(e^{2r(T-t)}-1)} | \mathcal{F}_t \right) \\ &= e^{−r(T−t)} \cdot N \left( -\frac{K-e^{r(T-t)} S_t}{\frac{\sigma^2}{2r}(e^{2r(T-t)}-1)} \right) \\ \end{align*}\] where \(N(x) = \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}y^2}\,dy\) and \(d = \frac{e^{r(T-t)} S_t - K}{\frac{\sigma^2}{2r}(e^{2r(T-t)}-1)}\).
Answer: The value of the call option is: \[\begin{align*} C(S_t,t) &= e^{−r(T−t)} \cdot N \left( -\frac{K-e^{r(T-t)} S_t}{\frac{\sigma^2}{2r}(e^{2r(T-t)}-1)} \right) \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows an arithmetic Brownian motion: \[dS_t = 0.05 \,dt + 0.3\,d W_t^{\mathcal{P}},\] where \(W_t^{\mathcal{P}}\) is a standard Brownian motion on the real world probability space, denoted by \((\omega, \mathcal{F},\mathcal{P})\). A European digital call option written on the stock with parameters: exercise price \(K = 32\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.08\), \(t = 0.5\), \(T = 1\), \(S_t =26.9\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\),\(T\) and \(S_t\) and then compute the fair price.
\[\begin{align*} S_T&= e^{r(T-t)} S_t + \int_t^T \sigma e^{r(T-u)} \,d W_t^{\mathcal{Q}} \\ & \sim \mathcal{N} \left(e^{r(T-t)} S_t , \left( \int_t^T \sigma e^{r(T-u)} \,d W_t^{\mathcal{Q}}\right)^2 \right) \\ & \sim \mathcal{N} \left( e^{r(T-t)} S_t , \frac{\sigma^2}{2r}(e^{2r(T-t)}-1) \right) \end{align*}\]
From Feyman-Kac formula, we have: \[P_d(S_t,t) = e^{-r(T-t)}E^{\mathcal{Q}}[\mathcal{I}_{S_T < K}|\mathcal{F}_t],\] where \(E^{\mathcal{Q}}\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{S_T} = \begin{cases} 1 & \text{if } S_t < K\\ 0 & \text{if } S_t \geq K \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain: \[\begin{align*} P_d(S_t,t) &= e^{−r(T−t)}E^Q [\mathcal{I}_{S_T < K} | \mathcal{F}_t] \\ &= e^{−r(T−t)} [1 \cdot \mathcal{Q}(S_T < K | \mathcal{F}_t) + 0 \cdot \mathcal{Q}(S_T \geq K | \mathcal{F}_t)] \\ &= e^{−r(T−t)} \cdot \mathcal{Q}(S_T < K | \mathcal{F}_t) \\ &= e^{−r(T−t)} \cdot \mathcal{Q}(S_t+ \mu(T-t) +\sigma(W_T^{\mathcal{Q}} - W_t^{\mathcal{Q}}) < K | \mathcal{F}_t) \\ &= e^{−r(T−t)} \cdot \mathcal{Q} \left(\frac{W_T^{\mathcal{Q}} - W_t^{\mathcal{Q}}}{\sqrt{T-t}} < \frac{K -(S_t+ \mu(T-t))}{\sigma \sqrt{T-t}} | \mathcal{F}_t \right) \\ &= e^{−r(T−t)} N \left( \frac{K-e^{r(T-t)} S_t}{\frac{\sigma^2}{2r}(e^{2r(T-t)}-1)} \right) \end{align*}\]
Answer: The value of the put option is: \[\begin{align*} P(S_t,t) &= e^{−r(T−t)} \cdot N \left(\frac{K-e^{r(T-t)} S_t}{\frac{\sigma^2}{2r}(e^{2r(T-t)}-1)} \right) \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows a geometric Brownian motion: \[\,d S_t = 0.05 S_t \,d t + 0.3 S_t \,d W_t^P,\] where \(W_t^P\) is a standard Brownian motion on the real world probability space, denoted by \((\omega,\mathcal{F},\mathcal{P})\). Asset-or-nothing call option written on the stock with parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\).
Compute the fair price when \(t = 0.5\), \(T = 1\), \(S_t = 9\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\),\(T\) and \(S_t\) for the fair price of the option?
\[S_T = e^{\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})}\]
We have \[\begin{align*} \ln S_t +(r-\frac{1}{2} \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}}) &\sim \mathcal{N}(\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t),\sigma^2(T-t)) \\ e^{\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})} &\sim \ln \mathcal{N}(\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t),\sigma^2(T-t)) \\ (S_T|S_t) &\sim \ln \mathcal{N}(\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t),\sigma^2(T-t)) \\ \end{align*}\]
The probability density function of \((S_T|S_t)\) is \[f_{S_T|S_t}(x)=\frac{1}{\sigma x\sqrt{2\pi(T-t)}}e^{-\frac{1}{2} \frac{(\ln x-(\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t)))^2}{\sigma^2(T-t)}}\]
From the Feynman–Kac formula, we have: \[C_a(S_t,t) = e^{−r(T−t)}E^Q [S_T\mathcal{I}_{S_T > K} |F_t],\] where \(E^Q\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{S_T} = \begin{cases} 1 & \text{if } S_t > K\\ 0 & \text{if } S_t \leq K \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain \[\begin{align*} C_a(S_t,t) &= e^{−r(T−t)}E^Q [S_T\mathcal{I}_{S_T > K} | \mathcal{F}_t] \\ &= e^{−r(T−t)} \int_K^{\infty} \frac{x}{\sigma x\sqrt{2\pi(T-t)}}e^{-\frac{1}{2} \frac{(\ln x-(\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \,d x \\ &= \frac{1}{\sqrt{2\pi}} \int_K^{\infty} e^{−r(T−t)} \cdot e^{-\frac{1}{2} \frac{(\ln x-(\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \cdot \frac{1}{\sigma \sqrt{(T-t)}} \,d x \\ \end{align*}\]
Let \(u=\frac{\ln x-\ln S_t -(r-\frac{1}{2} \sigma^2)(T-t)}{\sigma\sqrt{T-t}}\), we have \[\begin{align*} \ln \left( \frac{x}{S_t} \right)&=\sigma u\sqrt{T-t}+(r-\frac{1}{2} \sigma^2)(T-t) \\ x&=S_te^{\sigma u\sqrt{T-t}+(r-\frac{1}{2} \sigma^2)(T-t)} \end{align*}\]
\[\begin{align*} \,du&=\frac{1}{\sigma x\sqrt{T-t}}\,dx \\ \frac{1}{\sigma \sqrt{(T-t)}} \,d x &= x \,d u \\ &=S_te^{\sigma u\sqrt{T-t}+(r-\frac{1}{2} \sigma^2)(T-t)} \,du \end{align*}\]
\[\begin{align*} u_K&=\frac{\ln K-\ln S_t -(r-\frac{1}{2} \sigma^2)(T-t)}{\sigma\sqrt{T-t}} \\ &=\frac{\ln \left( \frac{K}{S_t} \right)-(r-\frac{1}{2} \sigma^2)(T-t)}{\sigma\sqrt{T-t}} \\ \rightarrow -u_K + \sigma \sqrt{T-t} &= \frac{\ln \left( \frac{S_t}{K} \right)+(r-\frac{1}{2} \sigma^2)(T-t)}{\sigma\sqrt{T-t}} + \sigma \sqrt{T-t} \\ &= \frac{\ln \left( \frac{S_t}{K} \right)+(r+\frac{1}{2} \sigma^2)(T-t)}{\sigma\sqrt{T-t}} \end{align*}\]
The value of the call option becomes: \[\begin{align*} C_a(S_t,t)&= \int_{u_k}^{\infty} e^{−r(T−t)} \cdot e^{-\frac{1}{2}u^2}\cdot S_te^{\sigma u\sqrt{T-t}+(r-\frac{1}{2} \sigma^2)(T-t)} \,du \\ &= S_t \int_{u_k}^{\infty} e^{-\frac{1}{2}u^2+u \sigma \sqrt{T-t}-\frac{1}{2} \cdot\sigma^2(T-t)} \,du \\ &= S_t \int_{u_k}^{\infty} e^{-\frac{1}{2}(u-\sigma\sqrt{T-t})^2} \,du \\ &= S_t \int_{u_k -\sigma\sqrt{T-t}}^{\infty} e^{-\frac{1}{2}u^2} \,du \\ &= S_t \int_{-\infty}^{-u_k +\sigma\sqrt{T-t}} e^{-\frac{1}{2}u^2} \,du \\ &= S_t \cdot N \left( \frac{\ln \left( \frac{S_t}{K} \right)+(r+\frac{1}{2} \sigma^2)(T-t)}{\sigma\sqrt{T-t}}\right) \end{align*}\] where \(N(x) = \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}y^2}\,dy\) ^2)(T-t)}{}$.
Answer: The value of the call option is: \[\begin{align*} C(S_t,t) &= S_t \cdot N \left( \frac{\ln \left( \frac{S_t}{K} \right)+(r+\frac{1}{2} \sigma^2)(T-t)}{\sigma\sqrt{T-t}}\right) \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows a geometric Brownian motion: \[ \,d S_t = 0.05 S_t \,dt + 0.3 S_t \,d W_t^P,\] where \(W_t^P\) is a standard Brownian motion on the real world probability space, denoted by \((\omega,\mathcal{F},\mathcal{P})\). Asset-or-nothing put option written on the stock with parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\), \(t = 0.5\), \(T = 1\), \(S_t = 7\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\),\(T\) and \(S_t\) for the fair price of the option?
Compute the fair price when \(t = 0.5\), \(T = 1\), \(S_t = 9\).
\[S_T = e^{\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})}\]
We have \[\begin{align*} \ln S_t +(r-\frac{1}{2} \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}}) &\sim \mathcal{N}(\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t),\sigma^2(T-t)) \\ e^{\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t)+\sigma(W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})} &\sim \ln \mathcal{N}(\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t),\sigma^2(T-t)) \\ (S_T|S_t) &\sim \ln \mathcal{N}(\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t),\sigma^2(T-t)) \\ \end{align*}\]
The probability density function of \((S_T|S_t)\) is \[f_{S_T|S_t}(x)=\frac{1}{\sigma x\sqrt{2\pi(T-t)}}e^{-\frac{1}{2} \frac{(\ln x-(\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t)))^2}{\sigma^2(T-t)}}\]
From the Feynman–Kac formula, we have: \[P_a(S_t,t) = e^{−r(T−t)}E^Q [S_T\mathcal{I}_{S_T < K} |F_t],\] where \(E^Q\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{S_T} = \begin{cases} 1 & \text{if } S_t < K\\ 0 & \text{if } S_t \geq K \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain \[\begin{align*} P_a(S_t,t) &= e^{−r(T−t)}E^Q [S_T\mathcal{I}_{S_T < K} | \mathcal{F}_t] \\ &= e^{−r(T−t)} \int_{-\infty}^{K} \frac{x}{\sigma x\sqrt{2\pi(T-t)}}e^{-\frac{1}{2} \frac{(\ln x-(\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \,d x \\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{K} e^{−r(T−t)} \cdot e^{-\frac{1}{2} \frac{(\ln x-(\ln S_t +(r-\frac{1}{2} \sigma^2)(T-t)))^2}{\sigma^2(T-t)}} \cdot \frac{1}{\sigma \sqrt{(T-t)}} \,d x \\ \end{align*}\]
Let \(u=\frac{\ln x-\ln S_t -(r-\frac{1}{2} \sigma^2)(T-t)}{\sigma\sqrt{T-t}}\), we have \[\begin{align*} \ln \left( \frac{x}{S_t} \right)&=\sigma u\sqrt{T-t}+(r-\frac{1}{2} \sigma^2)(T-t) \\ x&=S_te^{\sigma u\sqrt{T-t}+(r-\frac{1}{2} \sigma^2)(T-t)} \end{align*}\]
\[\begin{align*} \,du&=\frac{1}{\sigma x\sqrt{T-t}}\,dx \\ \frac{1}{\sigma \sqrt{(T-t)}} \,d x &= x \,d u \\ &=S_te^{\sigma u\sqrt{T-t}+(r-\frac{1}{2} \sigma^2)(T-t)} \,du \end{align*}\]
\[\begin{align*} u_K&=\frac{\ln K-\ln S_t -(r-\frac{1}{2} \sigma^2)(T-t)}{\sigma\sqrt{T-t}} \\ &=\frac{\ln \left( \frac{K}{S_t} \right)-(r-\frac{1}{2} \sigma^2)(T-t)}{\sigma\sqrt{T-t}} \\ \rightarrow u_K - \sigma \sqrt{T-t} &= -\frac{\ln \left( \frac{S_t}{K} \right)+(r-\frac{1}{2} \sigma^2)(T-t)}{\sigma\sqrt{T-t}} - \sigma \sqrt{T-t} \\ &= -\frac{\ln \left( \frac{S_t}{K} \right)+(r+\frac{1}{2} \sigma^2)(T-t)}{\sigma\sqrt{T-t}} \end{align*}\]
The value of the call option becomes: \[\begin{align*} P_a(S_t,t)&= \int_{-\infty}^{u_k} e^{−r(T−t)} \cdot e^{-\frac{1}{2}u^2}\cdot S_te^{\sigma u\sqrt{T-t}+(r-\frac{1}{2} \sigma^2)(T-t)} \,du \\ &= S_t \int_{-\infty}^{u_k} e^{-\frac{1}{2}u^2+u \sigma \sqrt{T-t}-\frac{1}{2} \cdot\sigma^2(T-t)} \,du \\ &= S_t \int_{-\infty}^{u_k} e^{-\frac{1}{2}(u-\sigma\sqrt{T-t})^2} \,du \\ &= S_t \int_{-\infty}^{u_k -\sigma\sqrt{T-t}} e^{-\frac{1}{2}u^2} \,du \\ &= S_t \cdot N \left( -\frac{\ln \left( \frac{S_t}{K} \right)+(r+\frac{1}{2} \sigma^2)(T-t)}{\sigma\sqrt{T-t}}\right) \end{align*}\] where \(N(x) = \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}y^2}\,dy\)
Answer: The value of the put option is: \[\begin{align*} P(S_t,t) &= S_t \cdot N \left( -\frac{\ln \left( \frac{S_t}{K} \right)+(r+\frac{1}{2} \sigma^2)(T-t)}{\sigma\sqrt{T-t}}\right) \end{align*}\]
Suppose the asset price process \((S_t)+{t \geq 0}\) follows an arithmetic Brownian motion: \[ \,d S_t = 0.05 \,d t + 0.3 \,d W_t^Q,\] where \(W_t^Q\) is a standard Brownian motion on a risk-neutral probability space, denoted by \((\omega,\mathcal{F},\mathcal{Q})\). Asset-or-nothing call option written on the stock with parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\).
Compute the fair price when \(t = 0.5\), \(T = 1\), \(S_t = 9\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\),\(T\) and \(S_t\) for the fair price of the option?
\[\,dS_t =\mu \,dt + \sigma\,d W_t^{\mathcal{Q}}\]
Integrating both sides from \(t\) to \(T\), we have: \[\begin{align*} \int_t^T \,d S_s &= \int_t^T \mu \,d s + \int_t^T \sigma \,d W_s^{\mathcal{Q}} \\ \rightarrow S_T &= S_t + \mu(T-t) + \sigma(W_T^{\mathcal{Q}} - W_t^{\mathcal{Q}}) \\ &\sim \mathcal{N}(S_t+\mu(T-t),\sigma^2(T-t)) \end{align*}\]
The probability density function of \(S_T|S_t\) can be written as: \[f_{S_T|s_t}(x) = \frac{1}{v \sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}}\] where \(m=S_t+\mu(T-t)\) and \(v^2=\sigma^2(T-t)\)
From Feynman-Kac formula, we have: \[C_a(S_t,t) = e^{-r(T-t)}E^{\mathcal{Q}}[S_T \mathcal{I}_{S_T > K}|\mathcal{F}_t],\] where \(E^{\mathcal{Q}}\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{S_T} = \begin{cases} 1 & \text{if } S_t > K\\ 0 & \text{if } S_t \leq K \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain: \[\begin{align*} C_a(S_t,t) &= e^{−r(T−t)}E^Q [S_T \mathcal{I}_{S_T > K} | \mathcal{F}_t] \\ &= e^{−r(T−t)} \int_K^{\infty} xf_{S_T|s_t}(x) \,d x \\ &= e^{−r(T−t)} \int_K^{\infty} \frac{x}{v \sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}} \,d x \\ &= e^{-r(T-t)} \int_{K}^{\infty} \frac{x}{v \sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}} \,dx \\ &= e^{-r(T-t)} \int_{K}^{\infty} \frac{x}{\sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}} \frac{1}{v}\,dx \end{align*}\]
Using the variable transformation \(u = \frac{x-m}{v}\) then \(x=m+uv\) and \(\,du = \frac{1}{v} \,dx\), we obtain: \[\begin{align*} C_a(S_t,t) &= e^{-r(T-t)} \int_{\frac{K-m}{v}}^{\infty} \frac{m+uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= e^{-r(T-t)}\int_{\frac{K-m}{v}}^{\infty} \frac{m}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du + e^{-r(T-t)}\int_{\frac{K-m}{v}}^{\infty} \frac{uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= e^{-r(T-t)}(I_1 + I_2) \end{align*}\]
With the notation \[\begin{align*} m&=S_t+\mu(T-t) \\ v^2&=\sigma^2(T-t) \\ v&=\sigma\sqrt{T-t} \\ d &= \frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \\ N(x) &= \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} \,dx \\ N'(x) &=\frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} \end{align*}\]
We have \[\begin{align*} I_1 &= \int_{\frac{K-m}{v}}^{\infty} \frac{m}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= m \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{-\frac{K-m}{v}} e^{-\frac{u^2}{2}} \,du \\ &= m \cdot N \left( -\frac{K-m}{v} \right) \\ &= (S_t+\mu(T-t)) \cdot N \left( -\frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right) \\ I_2 &= \int_{\frac{K-m}{v}}^{\infty} \frac{uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= v\int_{\frac{K-m}{v}}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,d \left(\frac{u^2}{2} \right) \\ &= v\int_{\frac{1}{2} \left(\frac{K-m}{v} \right)^2}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-x} \,dx \\ &= v \cdot \frac{1}{\sqrt{2\pi}} \cdot (-e^{-x}) \Biggr|_{\frac{1}{2} \left(\frac{K-m}{v} \right)^2}^{\infty} \\ &= \sigma\sqrt{T-t} \cdot \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{1}{2} \left(\frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right)^2} \\ &= \sigma \sqrt{T-t} \cdot N' \left( \frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right) \\ C_a(S_t,t) &= e^{-r(T-t)}(I_1 + I_2) \\ &= e^{-r(T-t)} \left( (S_t+\mu(T-t)) \cdot N \left( -\frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right) + \sigma \sqrt{T-t} \cdot N' \left( \frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right) \right) \end{align*}\]
Answer: The value of the call option is: \[\begin{align*} C(S_t,t) &= e^{-r(T-t)} \left( (S_t+\mu(T-t)) \cdot N \left( -\frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right) + \sigma \sqrt{T-t} \cdot N' \left( \frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right) \right) \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows an arithmetic Brownian motion: \[ \,d S_t = 0.05 \,d t + 0.3 \,d W_t^Q,\] where \(W_t^Q\) is a standard Brownian motion on a risk-neutral probability space, denoted by \((\omega,\mathcal{F},\mathcal{P})\). Asset-or-nothing put option written on the stock with parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\), \(t = 0.5\), \(T = 1\), \(S_t = 7\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of t,T and St and then compute the fair price.
Compute the fair price when \(t = 0.5\), \(T = 1\), \(S_t = 9\).
\[\,dS_t =\mu \,dt + \sigma\,d W_t^{\mathcal{Q}}\]
Integrating both sides from \(t\) to \(T\), we have: \[\begin{align*} \int_t^T \,d S_s &= \int_t^T \mu \,d s + \int_t^T \sigma \,d W_s^{\mathcal{Q}} \\ \rightarrow S_T &= S_t + \mu(T-t) + \sigma(W_T^{\mathcal{Q}} - W_t^{\mathcal{Q}}) \\ &\sim \mathcal{N}(S_t+\mu(T-t),\sigma^2(T-t)) \end{align*}\]
The probability density function of \(S_T|S_t\) can be written as: \[f_{S_T|s_t}(x) = \frac{1}{v \sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}}\] where \(m=S_t+\mu(T-t)\) and \(v^2=\sigma^2(T-t)\)
From Feynman-Kac formula, we have: \[P_a(S_t,t) = e^{-r(T-t)}E^{\mathcal{Q}}[S_T \mathcal{I}_{S_T < K}|\mathcal{F}_t],\] where \(E^{\mathcal{Q}}\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{S_T} = \begin{cases} 1 & \text{if } S_t < K\\ 0 & \text{if } S_t \geq K \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain: \[\begin{align*} P_a(S_t,t) &= e^{−r(T−t)}E^Q [S_T \mathcal{I}_{S_T < K} | \mathcal{F}_t] \\ &= e^{−r(T−t)} \int_{-\infty}^{K} xf_{S_T|s_t}(x) \,d x \\ &= e^{−r(T−t)} \int_{-\infty}^{K} \frac{x}{v \sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}} \,d x \\ &= e^{-r(T-t)} \int_{-\infty}^{K} \frac{x}{v \sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}} \,dx \\ &= e^{-r(T-t)} \int_{-\infty}^{K} \frac{x}{\sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}} \frac{1}{v}\,dx \end{align*}\]
Using the variable transformation \(u = \frac{x-m}{v}\) then \(x=m+uv\) and \(\,du = \frac{1}{v} \,dx\), we obtain: \[\begin{align*} P_a(S_t,t) &= e^{-r(T-t)} \int_{-\infty}^{\frac{K-m}{v}} \frac{m+uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= e^{-r(T-t)}\int_{-\infty}^{\frac{K-m}{v}} \frac{m}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du + e^{-r(T-t)}\int_{-\infty}^{\frac{K-m}{v}} \frac{uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= e^{-r(T-t)}(I_1 + I_2) \end{align*}\]
With the notation \[\begin{align*} m&=S_t+\mu(T-t) \\ v^2&=\sigma^2(T-t) \\ v&=\sigma\sqrt{T-t} \\ d &= \frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \\ N(x) &= \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} \,dx \\ N'(x) &=\frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} \end{align*}\]
We have \[\begin{align*} I_1 &= \int_{-\infty}^{\frac{K-m}{v}} \frac{m}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= m \cdot N \left( \frac{K-m}{v} \right) \\ &= (S_t+\mu(T-t)) N \left( \frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right) \\ I_2 &= \int_{-\infty}^{\frac{K-m}{v}} \frac{uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= -v\int_{-\infty}^{\frac{K-m}{v}} \frac{1}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,d \left(-\frac{u^2}{2} \right) \\ &= -v\int_{-\infty}^{-\frac{1}{2} \left(\frac{K-m}{v} \right)^2} \frac{1}{\sqrt{2\pi}} e^{x} \,dx \\ &= -v \cdot \frac{1}{\sqrt{2\pi}} \cdot (e^{x}) \Biggr|_{-\infty}^{-\frac{1}{2} \left(\frac{K-m}{v} \right)^2} \\ &= -\sigma\sqrt{T-t} \cdot \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{1}{2} \left(\frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right)^2} \\ &= -\sigma \sqrt{T-t} \cdot N' \left( \frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right) \\ &= -\sigma \sqrt{T-t} \cdot N' \left( -\frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right) \\ P_a(S_t,t) &= e^{-r(T-t)}(I_1 + I_2) \\ &= e^{-r(T-t)} \left( (S_t+\mu(T-t)) \cdot N \left( \frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right) - \sigma \sqrt{T-t} \cdot N' \left( -\frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right) \right) \end{align*}\]
Answer: The value of the put option is: \[\begin{align*} P(S_t,t) &= e^{-r(T-t)} \left( (S_t+\mu(T-t)) \cdot N \left( \frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right) - \sigma \sqrt{T-t} \cdot N' \left( -\frac{K-(S_t+\mu(T-t))}{\sigma \sqrt{T-t}} \right) \right) \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows an arithmetic Brownian motion: \[\,d S_t = 0.05 \,d t + 0.3 \,d W_t^P,\] where \(W_t^P\) is a standard Brownian motion on the real world probability space, denoted by \((\omega,\mathcal{F},\mathcal{P})\). Asset-or-nothing call option parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\), \(T\) and \(S_t\) for the fair price of the option?
\[\begin{align*} S_T&= e^{r(T-t)} S_t + \int_t^T \sigma e^{r(T-u)} \,d W_t^{\mathcal{Q}} \\ & \sim \mathcal{N} \left(e^{r(T-t)} S_t , \left( \int_t^T \sigma e^{r(T-u)} \,d W_t^{\mathcal{Q}}\right)^2 \right) \\ & \sim \mathcal{N} \left( e^{r(T-t)} S_t , \frac{\sigma^2}{2r}(e^{2r(T-t)}-1) \right) \end{align*}\]
The probability density function of \(S_T|S_t\) can be written as: \[f_{S_T|s_t}(x) = \frac{1}{v \sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}}\] where \(m=e^{r(T-t)} S_t\) and \(v^2=\frac{\sigma^2}{2r}(e^{2r(T-t)}-1)\)
From Feynman-Kac formula, we have: \[C_a(S_t,t) = e^{-r(T-t)}E^{\mathcal{Q}}[S_T \mathcal{I}_{S_T > K}|\mathcal{F}_t],\] where \(E^{\mathcal{Q}}\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{S_T} = \begin{cases} 1 & \text{if } S_t > K\\ 0 & \text{if } S_t \leq K \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain: \[\begin{align*} C_a(S_t,t) &= e^{−r(T−t)}E^Q [S_T \mathcal{I}_{S_T > K} | \mathcal{F}_t] \\ &= e^{−r(T−t)} \int_K^{\infty} xf_{S_T|s_t}(x) \,d x \\ &= e^{−r(T−t)} \int_K^{\infty} \frac{x}{v \sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}} \,d x \\ &= e^{-r(T-t)} \int_{K}^{\infty} \frac{x}{v \sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}} \,dx \\ &= e^{-r(T-t)} \int_{K}^{\infty} \frac{x}{\sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}} \frac{1}{v}\,dx \end{align*}\]
Using the variable transformation \(u = \frac{x-m}{v}\) then \(x=m+uv\) and \(\,du = \frac{1}{v} \,dx\), we obtain: \[\begin{align*} C_a(S_t,t) &= e^{-r(T-t)} \int_{\frac{K-m}{v}}^{\infty} \frac{m+uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= e^{-r(T-t)}\int_{\frac{K-m}{v}}^{\infty} \frac{m}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du + e^{-r(T-t)}\int_{\frac{K-m}{v}}^{\infty} \frac{uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= e^{-r(T-t)}(I_1 + I_2) \end{align*}\]
With the notation \[\begin{align*} m&=e^{r(T-t)} S_t \\ v^2&=\frac{\sigma^2}{2r}(e^{2r(T-t)}-1) \\ v&=\sigma \frac{e^{2r(T-t)}-1}{2r} \\ d &= \frac{K-e^{r(T-t)} S_t}{\sigma \frac{e^{2r(T-t)}-1}{2r}} \\ N(x) &= \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} \,dx \\ N'(x) &=\frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} \end{align*}\]
We have \[\begin{align*} I_1 &= \int_{\frac{K-m}{v}}^{\infty} \frac{m}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= m \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{-\frac{K-m}{v}} e^{-\frac{u^2}{2}} \,du \\ &= m \cdot N \left( -\frac{K-m}{v} \right) \\ &= e^{r(T-t)} S_t \cdot N \left( -\frac{K-e^{r(T-t)} S_t}{\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) \\ I_2 &= \int_{\frac{K-m}{v}}^{\infty} \frac{uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= v\int_{\frac{K-m}{v}}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,d \left(\frac{u^2}{2} \right) \\ &= v\int_{\frac{1}{2} \left(\frac{K-m}{v} \right)^2}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-x} \,dx \\ &= v \cdot \frac{1}{\sqrt{2\pi}} \cdot (-e^{-x}) \Biggr|_{\frac{1}{2} \left(\frac{K-m}{v} \right)^2}^{\infty} \\ &= \sigma\sqrt{T-t} \cdot \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{1}{2} \left(\frac{K-(e^{r(T-t)} S_t)}{\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right)^2} \\ &= \sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}} \cdot N' \left( \frac{K-(e^{r(T-t)} S_t)}{\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) \\ C_a(S_t,t) &= e^{-r(T-t)}(I_1 + I_2) \\ &= e^{-r(T-t)} \left( e^{r(T-t)} S_t \cdot N \left( -\frac{K-e^{r(T-t)} S_t}{\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) + \sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}} \cdot N' \left( \frac{K-e^{r(T-t)} S_t}{\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) \right) \end{align*}\]
Answer: The value of the call option is: \[\begin{align*} C(S_t,t) &= e^{-r(T-t)} \left( e^{r(T-t)} S_t \cdot N \left( -\frac{K-e^{r(T-t)} S_t}{\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) + \sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}} \cdot N' \left( \frac{K-e^{r(T-t)} S_t}{\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) \right) \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows an arithmetic Brownian motion: \[\,d S_t = 0.05 \,d t + 0.3 \,d W_t^P,\] where \(W_t^P\) is a standard Brownian motion on the real world probability space, denoted by \((\omega,\mathcal{F},\mathcal{P})\). Asset-or-nothing put option parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\),\(T\) and \(S_t\) for the fair price of the option?
Compute the fair price when \(t = 0.5\), \(T = 1\), \(S_t = 9\).
\[\begin{align*} S_T&= e^{r(T-t)} S_t + \int_t^T \sigma e^{r(T-u)} \,d W_t^{\mathcal{Q}} \\ & \sim \mathcal{N} \left(e^{r(T-t)} S_t , \left( \int_t^T \sigma e^{r(T-u)} \,d W_t^{\mathcal{Q}}\right)^2 \right) \\ & \sim \mathcal{N} \left( e^{r(T-t)} S_t , \frac{\sigma^2}{2r}(e^{2r(T-t)}-1) \right) \end{align*}\]
The probability density function of \(S_T|S_t\) can be written as: \[f_{S_T|s_t}(x) = \frac{1}{v \sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}}\] where \(m=e^{r(T-t)} S_t\) and \(v^2=\frac{\sigma^2}{2r}(e^{2r(T-t)}-1)\)
From Feynman-Kac formula, we have: \[P_a(S_t,t) = e^{-r(T-t)}E^{\mathcal{Q}}[S_T \mathcal{I}_{S_T < K}|\mathcal{F}_t],\] where \(E^{\mathcal{Q}}\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{S_T} = \begin{cases} 1 & \text{if } S_t < K\\ 0 & \text{if } S_t \geq K \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain: \[\begin{align*} P_a(S_t,t) &= e^{−r(T−t)}E^Q [S_T \mathcal{I}_{S_T < K} | \mathcal{F}_t] \\ &= e^{−r(T−t)} \int_{-\infty}^{K} xf_{S_T|s_t}(x) \,d x \\ &= e^{−r(T−t)} \int_{-\infty}^{K} \frac{x}{v \sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}} \,d x \\ &= e^{-r(T-t)} \int_{-\infty}^{K} \frac{x}{v \sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}} \,dx \\ &= e^{-r(T-t)} \int_{-\infty}^{K} \frac{x}{\sqrt{2\pi}} e^{-\frac{(x-m)^2}{2v^2}} \frac{1}{v}\,dx \end{align*}\]
Using the variable transformation \(u = \frac{x-m}{v}\) then \(x=m+uv\) and \(\,du = \frac{1}{v} \,dx\), we obtain: \[\begin{align*} P_a(S_t,t) &= e^{-r(T-t)} \int_{-\infty}^{\frac{K-m}{v}} \frac{m+uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= e^{-r(T-t)}\int_{-\infty}^{\frac{K-m}{v}} \frac{m}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du + e^{-r(T-t)}\int_{-\infty}^{\frac{K-m}{v}} \frac{uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= e^{-r(T-t)}(I_1 + I_2) \end{align*}\]
With the notation \[\begin{align*} m&=e^{r(T-t)} S_t \\ v^2&=\frac{\sigma^2}{2r}(e^{2r(T-t)}-1) \\ v&=\sigma \frac{e^{2r(T-t)}-1}{2r} \\ d &= \frac{K-e^{r(T-t)} S_t}{\sigma \frac{e^{2r(T-t)}-1}{2r}} \\ N(x) &= \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} \,dx \\ N'(x) &=\frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} \end{align*}\]
We have \[\begin{align*} I_1 &= \int_{-\infty}^{\frac{K-m}{v}} \frac{m}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= m \cdot N \left( \frac{K-m}{v} \right) \\ &= e^{r(T-t)} S_t \cdot N \left( \frac{K-e^{r(T-t)} S_t}{\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) \\ I_2 &= \int_{-\infty}^{\frac{K-m}{v}} \frac{uv}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,du \\ &= -v\int_{-\infty}^{\frac{K-m}{v}} \frac{1}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \,d \left(-\frac{u^2}{2} \right) \\ &= -v\int_{-\infty}^{-\frac{1}{2} \left(\frac{K-m}{v} \right)^2} \frac{1}{\sqrt{2\pi}} e^{x} \,dx \\ &= -v \cdot \frac{1}{\sqrt{2\pi}} \cdot (e^{x}) \Biggr|_{-\infty}^{-\frac{1}{2} \left(\frac{K-m}{v} \right)^2} \\ &= -\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}} \cdot N' \left( \frac{K-(e^{r(T-t)} S_t)}{\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) \\ &= -\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}} \cdot N' \left( -\frac{K-(e^{r(T-t)} S_t)}{\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) \\ P_a(S_t,t) &= e^{-r(T-t)}(I_1 + I_2) \\ &= e^{-r(T-t)} \left( e^{r(T-t)} S_t \cdot N \left( \frac{K-e^{r(T-t)} S_t}{\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) - \sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}} \cdot N' \left( -\frac{K-e^{r(T-t)} S_t}{\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) \right) \end{align*}\]
Answer: The value of the put option is: \[\begin{align*} P(S_t,t) &= e^{-r(T-t)} \left( e^{r(T-t)} S_t \cdot N \left( \frac{K-e^{r(T-t)} S_t}{\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) - \sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}} \cdot N' \left( -\frac{K-e^{r(T-t)} S_t}{\sigma\sqrt{\frac{e^{2r(T-t)}-1}{2r}}} \right) \right) \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows a geometric Brownian motion: \[\,d S_t =0.15 S_t \,dt + 0.4 S_t \,d W_t^{\mathcal{P}},\] where \(W_t^{\mathcal{P}}\) is a standard Brownian motion on the real world probability space, denoted by \((\Omega,\mathcal{F},\mathcal{P})\). An American digital call option written on the stock with parameters: exercise price \(K = 9\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.1\), \(t = 0.5\), \(T = 1\), \(S_t = 8\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\), \(T\) and \(S_t\) and then compute the fair price.
\[\begin{align*} S_T &= e^{\ln S_t + (r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})} \end{align*}\]
Let \(\tau\) be the waiting time (after \(t\)) for the asset price hits \(K\) from below. The payoff of this option is $1 at \(\tau\), if \(\tau \leq T − t\) and \(0\) if \(\tau > T − t\). Mathematically, \(\tau\) can be expressed as: \[\begin{align*} \tau &= \inf\{u-t \geq 0 | S_u = K\} \\ &= \inf\{u-t \geq 0 | e^{\ln S_t + (r-\frac{1}{2}\sigma^2)(u-t) + \sigma (W_u^{\mathcal{Q}}-W_t^{\mathcal{Q}})} = K\} \\ &= \inf\{u-t \geq 0 |W_u^{\mathcal{Q}}-W_t^{\mathcal{Q}} = \frac{1}{\sigma} \ln \frac{K}{S_t} - \frac{1}{\sigma} (r-\frac{1}{2}\sigma^2) \cdot (u-t)\} \\ \end{align*}\]
As \(W_u^{\mathcal{Q}}-W_t^{\mathcal{Q}} \stackrel{\text{d}}{=} W_{u-t}^{\mathcal{Q}}\), the distribution of \(\tau\) is identical with that of the first passage time the Brownian motion \(W_{u-t}^{\mathcal{Q}}\) hits the slope line \(a+b(u-t)\), where \(a = \frac{1}{\sigma} \ln \frac{K}{S_t}>0\) and \(b = - \frac{1}{\sigma} (r-\frac{1}{2}\sigma^2)\). The probability density function of \(\tau\) is: \[f_{\tau}(s) = \frac{a}{\sqrt{2\pi s^3}}e^{-\frac{(a+bs)^2}{2s}}\]
Under the risk-neutral measure, the option price is the expected value of the discounted payoff. Thus, the option price at time \(t\), denoted by \(C_d(S_t,t)\), computed by: \[C_d(S_t,t) = E^{\mathcal{Q}}[e^{-rt} \mathcal{I}_{\tau \leq T-t}|\mathcal{F_t} ]\] where \(E^{\mathcal{Q}}[\cdot ]\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{\tau \leq T-t} = \begin{cases} 1 & \text{if } \tau \leq T-t\\ 0 & \text{if } \tau > T-t \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain: \[\begin{align*} C_d(S_t,t) &= \int_0^{T-t} e^{-rs} f_{\tau} (s) \,d s \\ &= \int_0^{T-t} e^{-rs} \frac{a}{\sqrt{2\pi s^3}}e^{-\frac{(a+bs)^2}{2s}} \,d s \\ &= \int_0^{T-t} \frac{a}{\sqrt{2\pi s^3}}e^{-rs-\frac{(a+bs)^2}{2s}} \,d s \\ &= \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{a}{\sqrt{s^3}}e^{-ab-\frac{(2r+b^2)s^2+a^2}{2s}} \,d s \end{align*}\]
With \(c = \sqrt{2r+b^2}\), we have \[\begin{align*} C_d(S_t,t) &= e^{-ab} \cdot \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{a}{\sqrt{s^3}}e^{-\frac{c^2s^2+a^2}{2s}} \,d s \\ &= e^{-ab} \cdot \left( \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{a-cs}{2\sqrt{s^3}}e^{-\frac{1}{2s}((a+cs)^2-2acs)} \,d s + \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{a+cs}{2\sqrt{s^3}}e^{-\frac{1}{2s}((a-cs)^2+2acs)} \,d s \right)\\ &= e^{-ab+ac} \cdot \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{a-cs}{2\sqrt{s^3}}e^{-\frac{1}{2} \left( \frac{a+cs}{\sqrt{s}}\right)^2} \,d s + e^{-ab-ac} \cdot \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{a+cs}{2\sqrt{s^3}}e^{-\frac{1}{2} \left( \frac{a-cs}{\sqrt{s}}\right)^2} \,d s \\ \end{align*}\]
Variable substitution \[\begin{align*} v(s)=-\frac{a+cs}{\sqrt{s}} \\ \rightarrow \begin{cases} \,d v = \frac{a-cs}{2\sqrt{s^3}} \,d s \\ \lim_{s \to 0} v = -\infty \\ v(T-t)=-\frac{a+c(T-t)}{\sqrt{T-t}} \\ \end{cases} \end{align*}\] \[\begin{align*} w(s)=-\frac{a-cs}{\sqrt{s}} \\ \rightarrow \begin{cases} \,d w = \frac{a+cs}{2\sqrt{s^3}} \,d s \\ \lim_{s \to 0} w = -\infty \\ w(T-t)=-\frac{a-c(T-t)}{\sqrt{T-t}} \\ \end{cases} \end{align*}\]
\[\begin{align*} I &= e^{-ab+ac} \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{-\frac{a+c(T-t)}{\sqrt{T-t}}} e^{-\frac{1}{2}v^2} \,dv + e^{-ab-ac} \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{-\frac{a-c(T-t)}{\sqrt{T-t}}} e^{-\frac{1}{2}w^2} \,dw \\ &= e^{-ab+ac} \cdot N \left(-\frac{a+c(T-t)}{\sqrt{T-t}} \right) + e^{-ab-ac} \cdot N \left(-\frac{a-c(T-t)}{\sqrt{T-t}} \right) \end{align*}\] where \(N(x) = \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}y^2}\,dy\)
Answer: The value of the call option is: \[\begin{align*} C(S_t,t) &= e^{-ab+ac} \cdot N \left(-\frac{a+c(T-t)}{\sqrt{T-t}} \right) + e^{-ab-ac} \cdot N \left(-\frac{a-c(T-t)}{\sqrt{T-t}} \right) \end{align*}\] where \(a = \frac{1}{\sigma} \ln \frac{K}{S_t}>0, b = - \frac{1}{\sigma} (r-\frac{1}{2}\sigma^2), c = \sqrt{2r+b^2}\)
Suppose the asset price process \((S_t)_{t \geq 0}\) follows a geometric Brownian motion: \[\,d S_t =0.15 S_t \,dt + 0.4 S_t \,d W_t^{\mathcal{P}},\] where \(W_t^{\mathcal{P}}\) is a standard Brownian motion on the real world probability space, denoted by \((\Omega,\mathcal{F},\mathcal{P})\). An American digital put option written on the stock with parameters: exercise price \(K = 10\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.1\), \(t = 0.5\), \(T = 1\), \(S_t = 12\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\), \(T\) and \(S_t\) and then compute the fair price.
\[\begin{align*} S_T &= e^{\ln S_t + (r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})} \end{align*}\]
Let \(\tau\) be the waiting time (after \(t\)) for the asset price hits \(K\) from below. The payoff of this option is $1 at \(\tau\), if \(\tau \leq T − t\) and \(0\) if \(\tau > T − t\). Mathematically, \(\tau\) can be expressed as: \[\begin{align*} \tau &= \inf\{u-t \geq 0 | S_u = K\} \\ &= \inf\{u-t \geq 0 | e^{\ln S_t + (r-\frac{1}{2}\sigma^2)(u-t) + \sigma (W_u^{\mathcal{Q}}-W_t^{\mathcal{Q}})} = K\} \\ &= \inf\{u-t \geq 0 |W_u^{\mathcal{Q}}-W_t^{\mathcal{Q}} = \frac{1}{\sigma} \ln \frac{K}{S_t} - \frac{1}{\sigma} (r-\frac{1}{2}\sigma^2) \cdot (u-t)\} \\ \end{align*}\]
As \(W_u^{\mathcal{Q}}-W_t^{\mathcal{Q}} \stackrel{\text{d}}{=} W_{u-t}^{\mathcal{Q}}\), the distribution of \(\tau\) is identical with that of the first passage time the Brownian motion \(W_{u-t}^{\mathcal{Q}}\) hits the slope line \(a+b(u-t)\), where \(a = \frac{1}{\sigma} \ln \frac{K}{S_t}<0\) and \(b = - \frac{1}{\sigma} (r-\frac{1}{2}\sigma^2)\). The probability density function of \(\tau\) is: \[f_{\tau}(s) = \frac{-a}{\sqrt{2\pi s^3}}e^{-\frac{(a+bs)^2}{2s}}\]
Under the risk-neutral measure, the option price is the expected value of the discounted payoff. Thus, the option price at time \(t\), denoted by \(P_d(S_t,t)\), computed by: \[P_d(S_t,t) = E^{\mathcal{Q}}[e^{-rt} \mathcal{I}_{\tau \leq T-t}|\mathcal{F_t} ]\] where \(E^{\mathcal{Q}}[\cdot ]\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{\tau \leq T-t} = \begin{cases} 1 & \text{if } \tau \leq T-t\\ 0 & \text{if } \tau > T-t \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain: \[\begin{align*} P_d(S_t,t) &= \int_0^{T-t} e^{-rs} f_{\tau} (s) \,d s \\ &= \int_0^{T-t} e^{-rs} \frac{-a}{\sqrt{2\pi s^3}}e^{-\frac{(a+bs)^2}{2s}} \,d s \\ &= \int_0^{T-t} \frac{-a}{\sqrt{2\pi s^3}}e^{-rs-\frac{(a+bs)^2}{2s}} \,d s \\ &= \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{-a}{\sqrt{s^3}}e^{-ab-\frac{(2r+b^2)s^2+a^2}{2s}} \,d s \end{align*}\]
With \(c = \sqrt{2r+b^2}\), we have \[\begin{align*} P_d(S_t,t) &= e^{-ab} \cdot \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{-a}{\sqrt{s^3}}e^{-\frac{c^2s^2+a^2}{2s}} \,d s \\ &= e^{-ab} \cdot \left( \frac{1}{\sqrt{2\pi}} \int_0^{T-t} -\frac{a-cs}{2\sqrt{s^3}}e^{-\frac{1}{2s}((a+cs)^2-2acs)} \,d s + \frac{1}{\sqrt{2\pi}} \int_0^{T-t} -\frac{a+cs}{2\sqrt{s^3}}e^{-\frac{1}{2s}((a-cs)^2+2acs)} \,d s \right)\\ &= e^{-ab+ac} \cdot \frac{1}{\sqrt{2\pi}} \int_0^{T-t} -\frac{a-cs}{2\sqrt{s^3}}e^{-\frac{1}{2} \left( \frac{a+cs}{\sqrt{s}}\right)^2} \,d s + e^{-ab-ac} \cdot \frac{1}{\sqrt{2\pi}} \int_0^{T-t} -\frac{a+cs}{2\sqrt{s^3}}e^{-\frac{1}{2} \left( \frac{a-cs}{\sqrt{s}}\right)^2} \,d s \\ \end{align*}\]
Variable substitution \[\begin{align*} v(s)=\frac{a+cs}{\sqrt{s}} \\ \rightarrow \begin{cases} \,d v = -\frac{a-cs}{2\sqrt{s^3}} \,d s \\ \lim_{s \to 0} v = -\infty \\ v(T-t)=\frac{a+c(T-t)}{\sqrt{T-t}} \\ \end{cases} \end{align*}\] \[\begin{align*} w(s)=\frac{a-cs}{\sqrt{s}} \\ \rightarrow \begin{cases} \,d w = -\frac{a+cs}{2\sqrt{s^3}} \,d s \\ \lim_{s \to 0} w = -\infty \\ w(T-t)=\frac{a-c(T-t)}{\sqrt{T-t}} \\ \end{cases} \end{align*}\]
$$\begin{align*} I &= e^{-ab+ac} {-}^{} e{-v2} ,dv + e^{-ab-ac} {-}^{} e{-w2} ,dw \ &= e^{-ab+ac} N ( ) + e^{-ab-ac} N ( ) where \(N(x) = \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}y^2}\,dy\)
Answer: The value of the call option is: \[\begin{align*} C(S_t,t) &= e^{-ab+ac} \cdot N \left(\frac{a+c(T-t)}{\sqrt{T-t}} \right) + e^{-ab-ac} \cdot N \left(\frac{a-c(T-t)}{\sqrt{T-t}} \right) \end{align*}\] where \(a = \frac{1}{\sigma} \ln \frac{K}{S_t}>0, b = - \frac{1}{\sigma} (r-\frac{1}{2}\sigma^2), c = \sqrt{2r+b^2}\)
Suppose the asset price process \((S_t)_{t \geq 0}\) follows a geometric Brownian motion: \[\,d S_t =0.05 \,dt + 0.3 \,d W_t^{\mathcal{P}},\] where \(W_t^{\mathcal{P}}\) is a standard Brownian motion on the real world probability space, denoted by \((\Omega,\mathcal{F},\mathcal{P})\). An American digital call option written on the stock with parameters: exercise price \(K = 10\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\), \(t = 0.5\), \(T = 1\), \(S_t = 9\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\),\(T\) and \(S_t\) and then compute the fair price.
\[S_T = S_t + \mu(T-t) + \sigma(W_T^{\mathcal{Q}} - W_t^{\mathcal{Q}})\]
Let \(\tau\) be the waiting time (after \(t\)) for the asset price hits \(K\) from below. The payoff of this option is $1 at \(\tau\), if \(\tau \leq T − t\) and \(0\) if \(\tau > T − t\). Mathematically, \(\tau\) can be expressed as: \[\begin{align*} \tau &= \inf\{u-t \geq 0 | S_u = K\} \\ &= \inf\{u-t \geq 0 | S_t + \mu(u-t) + \sigma(W_u^{\mathcal{Q}} - W_t^{\mathcal{Q}}) = K\} \\ &= \inf\{u-t \geq 0 |W_u^{\mathcal{Q}}-W_t^{\mathcal{Q}} = \frac{1}{\sigma} (K-S_t) - \frac{\mu}{\sigma} \cdot (u-t)\} \\ \end{align*}\]
As \(W_u^{\mathcal{Q}}-W_t^{\mathcal{Q}} \stackrel{\text{d}}{=} W_{u-t}^{\mathcal{Q}}\), the distribution of \(\tau\) is identical with that of the first passage time the Brownian motion \(W_{u-t}^{\mathcal{Q}}\) hits the slope line \(a+b(u-t)\), where \(a = \frac{1}{\sigma} (K-S_t)>0\) and \(b = - \frac{\mu}{\sigma}\). The probability density function of \(\tau\) is: \[f_{\tau}(s) = \frac{a}{\sqrt{2\pi s^3}}e^{-\frac{(a+bs)^2}{2s}}\]
Under the risk-neutral measure, the option price is the expected value of the discounted payoff. Thus, the option price at time \(t\), denoted by \(C_d(S_t,t)\), computed by: \[C_d(S_t,t) = E^{\mathcal{Q}}[e^{-rt} \mathcal{I}_{\tau \leq T-t}|\mathcal{F_t} ]\] where \(E^{\mathcal{Q}}[\cdot ]\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{\tau \leq T-t} = \begin{cases} 1 & \text{if } \tau \leq T-t\\ 0 & \text{if } \tau > T-t \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain: \[\begin{align*} C_d(S_t,t) &= \int_0^{T-t} e^{-rs} f_{\tau} (s) \,d s \\ &= \int_0^{T-t} e^{-rs} \frac{a}{\sqrt{2\pi s^3}}e^{-\frac{(a+bs)^2}{2s}} \,d s \\ &= \int_0^{T-t} \frac{a}{\sqrt{2\pi s^3}}e^{-rs-\frac{(a+bs)^2}{2s}} \,d s \\ &= \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{a}{\sqrt{s^3}}e^{-ab-\frac{(2r+b^2)s^2+a^2}{2s}} \,d s \end{align*}\]
With \(c = \sqrt{2r+b^2}\), we have \[\begin{align*} C_d(S_t,t) &= e^{-ab} \cdot \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{a}{\sqrt{s^3}}e^{-\frac{c^2s^2+a^2}{2s}} \,d s \\ &= e^{-ab} \cdot \left( \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{a-cs}{2\sqrt{s^3}}e^{-\frac{1}{2s}((a+cs)^2-2acs)} \,d s + \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{a+cs}{2\sqrt{s^3}}e^{-\frac{1}{2s}((a-cs)^2+2acs)} \,d s \right)\\ &= e^{-ab+ac} \cdot \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{a-cs}{2\sqrt{s^3}}e^{-\frac{1}{2} \left( \frac{a+cs}{\sqrt{s}}\right)^2} \,d s + e^{-ab-ac} \cdot \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{a+cs}{2\sqrt{s^3}}e^{-\frac{1}{2} \left( \frac{a-cs}{\sqrt{s}}\right)^2} \,d s \\ \end{align*}\]
Variable substitution \[\begin{align*} v(s)=-\frac{a+cs}{\sqrt{s}} \\ \rightarrow \begin{cases} \,d v = \frac{a-cs}{2\sqrt{s^3}} \,d s \\ \lim_{s \to 0} v = -\infty \\ v(T-t)=-\frac{a+c(T-t)}{\sqrt{T-t}} \\ \end{cases} \end{align*}\] \[\begin{align*} w(s)=-\frac{a-cs}{\sqrt{s}} \\ \rightarrow \begin{cases} \,d w = \frac{a+cs}{2\sqrt{s^3}} \,d s \\ \lim_{s \to 0} w = -\infty \\ w(T-t)=-\frac{a-c(T-t)}{\sqrt{T-t}} \\ \end{cases} \end{align*}\]
\[\begin{align*} I &= e^{-ab+ac} \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{-\frac{a+c(T-t)}{\sqrt{T-t}}} e^{-\frac{1}{2}v^2} \,dv + e^{-ab-ac} \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{-\frac{a-c(T-t)}{\sqrt{T-t}}} e^{-\frac{1}{2}w^2} \,dw \\ &= e^{-ab+ac} \cdot N \left(-\frac{a+c(T-t)}{\sqrt{T-t}} \right) + e^{-ab-ac} \cdot N \left(-\frac{a-c(T-t)}{\sqrt{T-t}} \right) \\ &= e^{-ab+ac} \cdot N \left(d_1 \right) + e^{-ab-ac} \cdot N \left(d_2 \right) \end{align*}\] where \(N(x) = \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}y^2}\,dy\) and \(d_1 = -\frac{a+c(T-t)}{\sqrt{T-t}}\) and \(d_2 = -\frac{a-c(T-t)}{\sqrt{T-t}}\).
Answer: The value of the call option is: \[\begin{align*} C(S_t,t) &= e^{-ab+ac} \cdot N \left(\frac{a+c(T-t)}{\sqrt{T-t}} \right) + e^{-ab-ac} \cdot N \left(\frac{a-c(T-t)}{\sqrt{T-t}} \right) \end{align*}\] where \(a = \frac{1}{\sigma} (K-S_t)>0, b = - \frac{\mu}{\sigma}, c = \sqrt{2r+b^2}\)
Suppose the asset price process \((S_t)_{t \geq 0}\) follows a geometric Brownian motion: \[\,d S_t =0.05 \,dt + 0.3 \,d W_t^{\mathcal{P}},\] where \(W_t^{\mathcal{P}}\) is a standard Brownian motion on the real world probability space, denoted by \((\Omega,\mathcal{F},\mathcal{P})\). An American digital put option written on the stock with parameters: exercise price \(K = 8\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\), \(t = 0.5\), \(T = 1\), \(S_t = 10\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\),\(T\) and \(S_t\) and then compute the fair price.
\[S_T = S_t + \mu(T-t) + \sigma(W_T^{\mathcal{Q}} - W_t^{\mathcal{Q}})\]
Let \(\tau\) be the waiting time (after \(t\)) for the asset price hits \(K\) from below. The payoff of this option is $1 at \(\tau\), if \(\tau \leq T − t\) and \(0\) if \(\tau > T − t\). Mathematically, \(\tau\) can be expressed as: \[\begin{align*} \tau &= \inf\{u-t \geq 0 | S_u = K\} \\ &= \inf\{u-t \geq 0 | S_t + \mu(u-t) + \sigma(W_u^{\mathcal{Q}} - W_t^{\mathcal{Q}}) = K\} \\ &= \inf\{u-t \geq 0 |W_u^{\mathcal{Q}}-W_t^{\mathcal{Q}} = \frac{1}{\sigma} (K-S_t) - \frac{\mu}{\sigma} \cdot (u-t)\} \\ \end{align*}\]
As \(W_u^{\mathcal{Q}}-W_t^{\mathcal{Q}} \stackrel{\text{d}}{=} W_{u-t}^{\mathcal{Q}}\), the distribution of \(\tau\) is identical with that of the first passage time the Brownian motion \(W_{u-t}^{\mathcal{Q}}\) hits the slope line \(a+b(u-t)\), where \(a = \frac{1}{\sigma} (K-S_t)>0\) and \(b = - \frac{\mu}{\sigma}\). The probability density function of \(\tau\) is: \[f_{\tau}(s) = \frac{a}{\sqrt{2\pi s^3}}e^{-\frac{(a+bs)^2}{2s}}\] \[f_{\tau}(s) = \frac{-a}{\sqrt{2\pi s^3}}e^{-\frac{(a+bs)^2}{2s}}\]
Under the risk-neutral measure, the option price is the expected value of the discounted payoff. Thus, the option price at time \(t\), denoted by \(P_d(S_t,t)\), computed by: \[P_d(S_t,t) = E^{\mathcal{Q}}[e^{-rt} \mathcal{I}_{\tau \leq T-t}|\mathcal{F_t} ]\] where \(E^{\mathcal{Q}}[\cdot ]\) is the expectation under the risk-neutral measure \(\mathcal{Q}\) and \[\begin{align*} \mathcal{I}_{\tau \leq T-t} = \begin{cases} 1 & \text{if } \tau \leq T-t\\ 0 & \text{if } \tau > T-t \end{cases} \end{align*}\]
Using the definition of conditional expectation, we obtain: \[\begin{align*} P_d(S_t,t) &= \int_0^{T-t} e^{-rs} f_{\tau} (s) \,d s \\ &= \int_0^{T-t} e^{-rs} \frac{-a}{\sqrt{2\pi s^3}}e^{-\frac{(a+bs)^2}{2s}} \,d s \\ &= \int_0^{T-t} \frac{-a}{\sqrt{2\pi s^3}}e^{-rs-\frac{(a+bs)^2}{2s}} \,d s \\ &= \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{-a}{\sqrt{s^3}}e^{-ab-\frac{(2r+b^2)s^2+a^2}{2s}} \,d s \end{align*}\]
With \(c = \sqrt{2r+b^2}\), we have \[\begin{align*} P_d(S_t,t) &= e^{-ab} \cdot \frac{1}{\sqrt{2\pi}} \int_0^{T-t} \frac{-a}{\sqrt{s^3}}e^{-\frac{c^2s^2+a^2}{2s}} \,d s \\ &= e^{-ab} \cdot \left( \frac{1}{\sqrt{2\pi}} \int_0^{T-t} -\frac{a-cs}{2\sqrt{s^3}}e^{-\frac{1}{2s}((a+cs)^2-2acs)} \,d s + \frac{1}{\sqrt{2\pi}} \int_0^{T-t} -\frac{a+cs}{2\sqrt{s^3}}e^{-\frac{1}{2s}((a-cs)^2+2acs)} \,d s \right)\\ &= e^{-ab+ac} \cdot \frac{1}{\sqrt{2\pi}} \int_0^{T-t} -\frac{a-cs}{2\sqrt{s^3}}e^{-\frac{1}{2} \left( \frac{a+cs}{\sqrt{s}}\right)^2} \,d s + e^{-ab-ac} \cdot \frac{1}{\sqrt{2\pi}} \int_0^{T-t} -\frac{a+cs}{2\sqrt{s^3}}e^{-\frac{1}{2} \left( \frac{a-cs}{\sqrt{s}}\right)^2} \,d s \\ \end{align*}\]
Variable substitution \[\begin{align*} v(s)=\frac{a+cs}{\sqrt{s}} \\ \rightarrow \begin{cases} \,d v = -\frac{a-cs}{2\sqrt{s^3}} \,d s \\ \lim_{s \to 0} v = -\infty \\ v(T-t)=\frac{a+c(T-t)}{\sqrt{T-t}} \\ \end{cases} \end{align*}\] \[\begin{align*} w(s)=\frac{a-cs}{\sqrt{s}} \\ \rightarrow \begin{cases} \,d w = -\frac{a+cs}{2\sqrt{s^3}} \,d s \\ \lim_{s \to 0} w = -\infty \\ w(T-t)=\frac{a-c(T-t)}{\sqrt{T-t}} \\ \end{cases} \end{align*}\]
\[\begin{align*} I &= e^{-ab+ac} \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\frac{a+c(T-t)}{\sqrt{T-t}}} e^{-\frac{1}{2}v^2} \,dv + e^{-ab-ac} \cdot \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\frac{a-c(T-t)}{\sqrt{T-t}}} e^{-\frac{1}{2}w^2} \,dw \\ &= e^{-ab+ac} \cdot N \left(\frac{a+c(T-t)}{\sqrt{T-t}} \right) + e^{-ab-ac} \cdot N \left(\frac{a-c(T-t)}{\sqrt{T-t}} \right) \end{align*}\] where \(N(x) = \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}y^2}\,dy\)
Answer: The value of the call option is: \[\begin{align*} C(S_t,t) &= e^{-ab+ac} \cdot N \left(\frac{a+c(T-t)}{\sqrt{T-t}} \right) + e^{-ab-ac} \cdot N \left(\frac{a-c(T-t)}{\sqrt{T-t}} \right) \end{align*}\] where \(a = \frac{1}{\sigma} (K-S_t)>0, b = - \frac{\mu}{\sigma}, c = \sqrt{2r+b^2}\)
Suppose the asset price process \((S_t)_{t \geq 0}\) follows a geometric Brownian motion: \[\,d S_t = 0.05 S_t \,dt+0.3 S_t \,d W_t^P,\] where \(W_t^P\) is a standard Brownian motion on the real world probability space, denoted by \((\Omega, \mathcal{F}, \mathcal{P})\). Consider a European exotic option written on the stock that has pay-off \[\Phi(S_T)=S_T − K\] at expiry, where \(K\) is a constant and the current stock price \(S_0 = 7\). The parameters are \(K = 10\), maturity time \(T > t\) (year), the risk-free interest rate is \(r = 0.03\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\), \(T\) and \(S_t\) for the fair price of the option?
\[\begin{align*} S_T &= e^{\ln S_t+(r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})} \\ & \sim \log N(\ln S_t+(r-\frac{1}{2}\sigma^2)(T-t),\sigma^2(T-t)) \end{align*}\]
\[\begin{align*} C(S_t,t) &= e^{-r(T-t)} E^{\mathcal{Q}}\left[ S_T-K | \mathcal{F}_t \right] \\ &= e^{-r(T-t)} (e^{\ln S_t+(r-\frac{1}{2}\sigma^2)(T-t)+ \frac{1}{2}\sigma^2(T-t)}-K) \\ &= S_t-Ke^{-r(T-t)} \end{align*}\]
Compute the fair price when \(t = 0.5\), \(T = 1\), \(S_t = 9\).
\[\begin{align*} C(S_t,t) &= S_t-Ke^{-r(T-t)} \\ &= 9-10e^{-0.03(1-0.5)} \\ &= -0.85 \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows a geometric Brownian motion: \[\,d S_t = 0.05 S_t \,dt + 0.3 S_t \, d W_t^P,\] where \(W_t^P\) is a standard Brownian motion on the real world probability space, denoted by \((\Omega, \mathcal{F}, \mathcal{P})\). Consider a European exotic option written on the stock that has pay-off \[\phi(S_T) = \ln \left( \frac{S_T}{S_0} \right) − K\] at expiry, where \(K\) is a constant and the current stock price \(S_0 = 7\). The parameters are \(K = 1\), maturity time \(T > t\) (year), the risk-free interest rate is \(r =0.03\), \(t =0.5\), \(T = 1\), \(S_t =9\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\), \(T\) and \(S_t\) and then compute the fair price.
\[\begin{align*} S_T &= e^{\ln S_t+(r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})} \\ \rightarrow \ln \left( \frac{S_T}{S_0} \right) -K &= \ln S_t+(r-\frac{1}{2}\sigma^2)(T-t)-\ln S_0 - K \end{align*}\]
From the Feynman–Kac formula, we have: \[\begin{align*} C(S_t,t) &= e^{-r(T-t)} E^{\mathcal{Q}}\left[ \ln \left( \frac{S_T}{S_0} \right) -K | \mathcal{F}_t \right] \\ &= e^{-r(T-t)} \left(\ln S_t+(r-\frac{1}{2}\sigma^2)(T-t)-\ln S_0 - K \right) \\ &= e^{-0.03(1-0.5)} \left(\ln 9+(0.03-\frac{1}{2} \cdot 0.3^2)(1-0.5)-\ln 7 - 1 \right) \\ &= -0.74 \end{align*}\]
Suppose the asset price process \((S_t)_{t \geq 0}\) follows a geometric Brownian motion: \[\,d S_t = 0.05 S_t \,dt + 0.3 S_t \, d W_t^P,\] where \(W_t^P\) is a standard Brownian motion on the real world probability space, denoted by \((\Omega, \mathcal{F}, \mathcal{P})\). Consider a European exotic option written on the stock that has pay-off \[\phi(S_T) = \left( \ln \left( \frac{S_T}{S_0} \right) − K \right)^2\] at expiry, where \(K\) is a constant and the current stock price \(S_0 = 7\). The parameters are \(K = 1\), maturity time \(T > t\) (year), the risk-free interest rate is \(r =0.03\), \(t =0.5\), \(T = 1\), \(S_t =9\).
Under the Black-Scholes framework, using the probabilistic approach to derive the formula in terms of \(t\), \(T\) and \(S_t\) and then compute the fair price.
\[S_T = e^{\ln S_t+(r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}})}\]
\[\begin{align*} \left[ \ln \left( \frac{S_T}{S_0} \right) -K \right]^2 &= \left[ \ln \left( S_T \right) -(\ln \left( S_0 \right)+K) \right]^2 \\ &= \ln^2 \left( S_T \right) -2 \ln \left( S_T \right)(\ln \left( S_0 \right)+K) + (\ln \left( S_0 \right)+K)^2 \\ \end{align*}\]
\[\begin{align*} C(S_t,t) &= e^{-r(T-t)} E^{\mathcal{Q}}\left[ \left( \ln \left( \frac{S_T}{S_0} \right) -K \right)^2 | \mathcal{F}_t \right] \\ &= e^{-r(T-t)} \left[ E^{\mathcal{Q}} \left(\ln^2 \left( S_T \right) -2 \cdot \ln \left( S_T \right) \cdot (\ln \left( S_0 \right)+K) \right | \mathcal{F}_t)+(\ln \left( S_0 \right)+K)^2 \right] \\ &= e^{-r(T-t)} \left[ E^{\mathcal{Q}} \left(\left( \ln S_t+(r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}}) \right)^2 -2 \cdot \left( \ln S_t+(r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}}) \right) \cdot (\ln \left( S_0 \right)+K) \right) +(\ln \left( S_0 \right)+K)^2 \right] \\ &= e^{-r(T-t)} \left[ E^{\mathcal{Q}} \left(\left( \ln S_t+(r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}}) \right)^2 \right) -2 \cdot (\ln \left( S_0 \right)+K) \cdot E^{\mathcal{Q}}\left( \ln S_t+(r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^{\mathcal{Q}}-W_t^{\mathcal{Q}}) \right) \right] \\ &= e^{-r(T-t)} \left[ \sigma^2(T-t) + (\ln S_t+(r-\frac{1}{2}\sigma^2)(T-t))^2 -2 \cdot (\ln \left( S_0 \right)+K) \cdot \left( \ln S_t+(r-\frac{1}{2}\sigma^2)(T-t) \right)+(\ln \left( S_0 \right)+K)^2 \right] \\ &= 0.60 \end{align*}\]