Today we going use all the results we learned in the previous blogs. What’s the point of theorems if we don’t use them!! That’s what we are gonna do today…
Let \(\mathbb{W(t)\ , t \geq 0}\) be a Brownian motion, let \(\mathbb{F(t)\ , t \geq 0}\) be an associated filtration. Let \(\mathbb{\alpha(t)}\) and \(\mathbb{\sigma(t)}\) be adapted processes. Define the Ito Process as-
\[\mathbb{X(t)}\ = \mathbb{\int_{0}^{t}\sigma(s)\ dW(s)}\ + \mathbb{\int_{0}^{t}(\alpha(s)-\frac{1}{2}\sigma^2(s))\ ds}\]
In differential form, we can write this as- \[\mathbb{dX(t)}\ = \mathbb{\sigma(t)\ dW(t)}\ + \mathbb{(\alpha(t)-\frac{1}{2}\sigma^2(t))\ dt}\]
which implies, \[\mathbb{dX(t)\ dX(t)}\ = \mathbb{\sigma^2(t)\ dW(t)\ dW(t)}\ =\ \mathbb{\sigma^2(t)\ dt}\]
Now consider an asset pricing process given by- \[\mathbb{S(t)}\ =\ \mathbb{S(0)\ e^{X(t)}}\ =\ \mathbb{S(0)\ \exp{\{\mathbb{\int_{0}^{t}\sigma(s)\ dW(s)}\ + \mathbb{\int_{0}^{t}(\alpha(s)-\frac{1}{2}\sigma^2(s))\ ds}}\}}\] where \(\mathbb{S(0)}\) is a non-random and positive number.
What is the Ito- Doeblin formula for \(\mathbb{S(t)}\)?
We may write \(\mathbb{S(t)\ =\ f(X(t))}\); where \(\mathbb{f(x)\ =\ S(0)\ e^x}\)
This implies- \[\mathbb{f'(x)\ =\ S(0)\ e^x}; \qquad \mathbb{f''(x)\ =\ S(0)\ e^x}\]
So, using the Ito-Doeblin formula we can write-
\[\begin{aligned} \mathbb{dS(t)}\ &=\ \mathbb{df(X(t))} \\ &=\ \mathbb{f'(X(t))\ dX(t)}\ +\ \frac{1}{2}\ \mathbb{f''(X(t))\ dX(t)\ dX(t)} \\ &=\ \mathbb{S(0)\ e^{X(t)}\ dX(t)}\ +\ \frac{1}{2}\ \mathbb{S(0)\ e^{X(t)}\ dX(t)\ dX(t)} \\ &=\ \mathbb{S(t)\ dX(t)}\ +\ \frac{1}{2}\ \mathbb{S(t)\ dX(t)\ dX(t)} \\ &=\ \mathbb{S(t)\ \{\mathbb{\sigma(t)\ dW(t)}\ + \mathbb{(\alpha(t)-\frac{1}{2}\sigma^2(t))\ dt}\}}\ +\ \frac{1}{2}\ \mathbb{S(t)\ \sigma^2(t) dt} \\ &=\ \mathbb{\alpha(t)\ S(t)\ dt}\ +\ \mathbb{S(t)}\ \mathbb{\sigma(t)\ dW(t)} \\ \end{aligned}\]
The asset price \(\mathbb{S(t)}\) instantaneous mean rate of return = \(\mathbb{\alpha(t)}\) and volatility = \(\mathbb{\sigma(t)}\). Here both the instantaneous mean rate of return and the volatility are time-varying and random.
In this model- \(\mathbb{S(t)}\) always takes positive value. It is driven by a single Brownian motion, \(\mathbb{W(t)}\).
If we consider \(\mathbb{\alpha(t)} = \alpha\) and \(\mathbb{\sigma(t)} = \sigma\), both constants; \(\mathbb{S(t)}\) becomes-
\[\begin{aligned} \mathbb{S(t)}\ &=\ \mathbb{S(0)\ \exp{\{\mathbb{\int_{0}^{t}\sigma\ dW(s)}\ + \mathbb{\int_{0}^{t}(\alpha-\frac{1}{2}\sigma^2)\ ds}}\}} \\ &=\ \mathbb{S(0)\ \exp{\{\mathbb{\sigma\ W(t)}\ + \mathbb{(\alpha-\frac{1}{2}\sigma^2)\ t}}\}} \\ \end{aligned}\]
The exponent part- \(\mathbb{\sigma\ W(t)}\ + \mathbb{(\alpha-\frac{1}{2}\sigma^2)\ t} \sim \mathbb{N((\alpha-\frac{1}{2}\sigma^2)\ t\ ,\ \sigma^2\ t )}\)
So \(\mathbb{S(t)}\) follows log-normal distribution.
Now since Brownian motion is a martingale, you may think that the mean rate of return of \(\mathbb{S(t)}\) must be- \(\mathbb{\alpha-\frac{1}{2}\sigma^2}\).
The mistake you have made here is- though \(\mathbb{W(t)}\) is a martingale, \(\ \mathbb{S(0)\ e^{\sigma W(t)}}\) is not. In order to correct for this, one must subtract \(\mathbb{\frac{1}{2}\sigma^2t}\) in the exponential;
The process \(\mathbb{S(0)\ e^{\{\sigma W(t)-\frac{1}{2}\sigma^2t\}}}\) is a martingale.
Proof: Let \(\mathbb{Z(t) = \exp{\{\sigma W(t)-\frac{1}{2}\sigma^2t\}} }\). Let \(\mathbb{W(t),\ t \geq 0}\) be a Brownian motion with filtration \(\mathbb{F(t),\ t \geq 0}\) and \(\sigma\) is a constant.
For \(\mathbb{0 \leq s \leq t}\) we can write-
\[\begin{aligned} & \mathbb{E[Z(t)|F(s)]} \\ =\ & \mathbb{E[\exp{\{\sigma W(t)-\frac{1}{2}\sigma^2t\}}|F(s)]} \\ =\ & \mathbb{E[\exp{\{\sigma (W(t)-W(s))}\}\ \exp{\{\sigma\ W(s)-\frac{1}{2}\sigma^2t\}}|F(s)]} \\ =\ & \mathbb{\exp{\{\sigma\ W(s)-\frac{1}{2}\sigma^2t\}}\ E[\exp{\{\sigma (W(t)-W(s))}\}\ |F(s)]} \\ =\ & \mathbb{\exp{\{\sigma\ W(s)-\frac{1}{2}\sigma^2t\}}\ \exp{\{\frac{1}{2}\sigma^2(t-s)\}}} \\ =\ & \mathbb{\exp{\{\sigma\ W(s)-\frac{1}{2}\sigma^2s\}}} \\ =\ & \mathbb{Z(s)} \qquad \text{Q.E.D.} \end{aligned}\]
So \(\mathbb{S(0)\ e^{\{\sigma W(t)-\frac{1}{2}\sigma^2t\}}}\) is a martingale.
Now if we add \(\mathbb{\alpha t}\) in th exponential- we get \(\mathbb{S(t)}\), same process with mean rate of return \(\mathbb{\alpha}\).
Theorem: Ito Integral of a deterministic integrand
Let \(\mathbb{W(s),\ s \geq 0}\) be a Brownian motion, and let \(\mathbb{\Delta(s)}\) be a non-random function of time.
Define, \[\mathbb{I(t)}\ =\ \mathbb{\int_{0}^{t}\Delta(s)\ dW(s)}\]
For each \(\mathbb{t \geq 0}\), the random variable \(\mathbb{I(t)}\) is normally distributed with expected value zero and variance \(\mathbb{\int_{0}^{t}\Delta^2(s)\ ds}\)
Proof: Proving expectation and variance is easy. Harder part is to prove the Normality. Let us start with the easy one.
Previously I have proved that- the Ito Integral is a martingale. i.e.
\[\mathbb{E[I(t)|F(0)]}\ =\ \mathbb{I(0)}\ =\ 0\]
Ito Isometry tells us-
\[\mathbb{Var[I(t)|F(0)]}\ =\ \mathbb{E[I^2(t)|F(0)]}\ =\ \mathbb{\int_{0}^{t}\Delta^2(u)\ du}\]
Now the best way to prove that- any random variable follows some distribution is check it’s Moment Generating Function (if it exists) or Characteristic Function.
If \(\mathbb{I(t)}\ \sim \mathbb{N(0,\ \int_{0}^{t}\Delta^2(u)\ du)}\), then the Moment Generating Function will be-
\[\begin{aligned} \mathbb{M_{I(t)}(u)}\ =\ &\mathbb{E\ [\exp{\{u\ I(t) \}}]}\ =\ \mathbb{\exp{\{\frac{1}{2}u^2 \int_{0}^{t}\Delta^2(s)\ ds \}}} \\ \implies & \mathbb{E\ [\exp{\{u\ I(t) - \frac{1}{2}u^2 \int_{0}^{t}\Delta^2(s)\ ds \}}]}\ =\ 1 \\ \implies & \mathbb{E\ [\exp{\{\int_{0}^{t}u \Delta(s)\ dW(s) - \frac{1}{2} \int_{0}^{t}(u \Delta(s))^2\ ds \}}]}\ =\ 1 \\ \end{aligned}\]
We just proved that- \(\mathbb{Z(t) = \exp{\{\sigma W(t)-\frac{1}{2}\sigma^2t\}} }\) is a martingale. Similar to that- \[\mathbb{\exp{\{\int_{0}^{t}u \Delta(s)\ dW(s) - \frac{1}{2} \int_{0}^{t}(u \Delta(s))^2\ ds \}}}\]
is also a martingale.
It is generalized geometric Brownian motion, with \(\alpha = 0\) and \(\mathbb{\sigma(s) = u\ \Delta(s)}\).
\[\mathbb{E\ [\exp{\{\int_{0}^{t}u \Delta(s)\ dW(s) - \frac{1}{2} \int_{0}^{t}(u \Delta(s))^2\ ds \}}|\ F(0) ]}\ =\ 1\qquad \text{Q.E.D.}\]
I hope you liked it. In the next blog we are going to see another awesome example. It’s called Vasicek interest rate model introduced by Oldrich Vasicek.