Brownian Motion
In this blog, we are going to find the distribution of first time passage for a Brownian motion. We are going to do in the similar manner as we have done in symmetric random walk. We are going to use Optional Sampling theorem for martingales to obtain the distributions.
In symmetric random walk, we started with a martingale- \[S_n = e^{\sigma M_n}(\frac{2}{e^\sigma+e^{-\sigma}})^n \qquad \text{where} \ M_n = \sum_{j=1}^{k}X_j \]
Similarly in this case, we are going to find a martingale, containing Brownian motion in the exponential function. Fix a constant \(\sigma\). For a fixed \(\sigma\) define exponential martingale as- \[Z(t) = \exp\{\sigma .W(t) -\frac{1}{2}\sigma^2t \}\] First we will try to prove that- \(Z(t)\) is a martingale.
Theorem: Let \(W(t), \ t\geq 0\) be a Brownian motion with filtration \(\mathbb{F(t)}, \ t \geq 0\) and let \(\sigma\) be a constant. The process- \[Z(t) = \exp\{\sigma .W(t) -\frac{1}{2}\sigma^2t \} \qquad \text{with} \quad t \geq 0\] is martingale.
Proof: For \(0 \leq s \leq t\), we have-
\[\begin{aligned} &E[Z(t) | \mathbb{F(s)}] \\ = \ &E[\exp\{\sigma .W(t) -\frac{1}{2}\sigma^2t \} | \mathbb{F(s)}] \\ = \ &E[\exp\{\sigma .W(t) - \sigma .W(s) +\sigma .W(s) -\frac{1}{2}\sigma^2t \} | \mathbb{F(s)}] \\ = \ &\exp\{\sigma .W(s) -\frac{1}{2}\sigma^2t\}.E[\exp\{\sigma(W(t)-W(s)) \} | \mathbb{F(s)}] \\ \end{aligned}\]
Given, \(\mathbb{F(s)}\), \(W(s)\) is known, because \(W(t)\) is \(\mathbb{F(t)}\) measurable. This gives us \(\exp\{\sigma .W(s) -\frac{1}{2}\sigma^2t\}\) as constant.
Using the independence theorem we can write- \[\begin{aligned} E[\exp\{\sigma(W(t)-W(s)) \} | \mathbb{F(s)}] = E[\exp\{\sigma(W(t)-W(s)) \}] \end{aligned}\]
Now, \(W(t)-W(s) \sim \mathbb{N}(0, t-s)\). This leads us to- \[\begin{aligned} E[\exp\{\sigma(W(t)-W(s)) \} | \mathbb{F(s)}] &= E[\exp\{\sigma(W(t)-W(s)) \}] \\ &= \text{M.G.F of} \ \mathbb{N}(0, t-s) \\ &= \exp\{\frac{1}{2}\sigma^2(t-s) \} \\ \end{aligned}\]
So we can write- \[\begin{aligned} &E[Z(t) | \mathbb{F(s)}] \\ = \ &\exp\{\sigma .W(s) -\frac{1}{2}\sigma^2t\}.E[\exp\{\sigma(W(t)-W(s)) \} | \mathbb{F(s)}] \\ = \ &\exp\{\sigma .W(s) -\frac{1}{2}\sigma^2t\}. \exp\{\frac{1}{2}\sigma^2(t-s) \} \\ = \ &\exp\{\sigma .W(s) -\frac{1}{2}\sigma^2s\} = \mathbb{Z(s)} \\ \end{aligned}\] This proves that \(\mathbb{Z(t)}\) is a martingale.
Let \(m\) be a real number; similar to symmetric random walk, we are going to define first passage time to level \(m\) by- \[\tau_m = \min\{t \geq 0: W(t) = m \}\]
This is the first time the Brownian motion \(W\) reaches level \(m\) and if the Brownian motion never reaches level \(m\), we write- \(\tau_m = \infty\).
Now, if you remember the Optional Sampling Theorem, it states that- a martingale that is stopped at a stopping time is still a martingale, so it has constant expectation.
Similar to symmetric random walk we can write- \[\begin{aligned} 1 = \mathbb{Z(0)} &= \mathbb{E[Z(t \land \tau_m)]} \\ &= \mathbb{E[\exp\{\sigma .W(t \land \tau_m) -\frac{1}{2}\sigma^2(t \land \tau_m) \}]} \\ \end{aligned}\]
For the next step, we assume- \(\sigma > 0\) and \(m > 0\). In this case, the Brownian motion is always at or below level \(m\) for \(t \leq \tau_m\), so we can write- \[0 \leq \exp\{\sigma .W(t \land \tau_m)\} \leq e^{\sigma m}\]
If \(\tau_m < \infty\), we can write- \[\lim_{t \rightarrow \infty} \exp\{-\frac{1}{2}\sigma^2(t \land \tau_m)\} = \exp\{-\frac{1}{2}\sigma^2\tau_m\}\]
If \(\tau_m = \infty\), we can write- \[\lim_{t \rightarrow \infty} \exp\{-\frac{1}{2}\sigma^2(t \land \tau_m)\} = \lim_{t \rightarrow \infty} \exp\{-\frac{1}{2}\sigma^2t \} = 0\]
Combining both the cases we can write- \[\lim_{t \rightarrow \infty} \exp\{-\frac{1}{2}\sigma^2(t \land \tau_m)\} = \mathbb{I_{(\tau_m < \infty)}} \exp\{-\frac{1}{2}\sigma^2\tau_m\}\]
Hence, for \(t \rightarrow \infty\) we can have- \[\begin{aligned} \lim_{t \rightarrow \infty}\exp\{\sigma .W(t \land \tau_m) -\frac{1}{2}\sigma^2(t \land \tau_m) \} = \mathbb{I_{(\tau_m < \infty)}} \exp\{\sigma m - \frac{1}{2}\sigma^2\tau_m\ \} \end{aligned}\]
So, from the Optional Sampling theorem part, we get- \[\begin{aligned} 1 = \mathbb{Z(0)} &= \mathbb{E[Z(t \land \tau_m)]} \\ &= \mathbb{E[\exp\{\sigma .W(t \land \tau_m) -\frac{1}{2}\sigma^2(t \land \tau_m) \}]} \\ &= \mathbb{E[\mathbb{I_{(\tau_m < \infty)}} \exp\{\sigma m - \frac{1}{2}\sigma^2\tau_m\ \}]} \\ \implies e^{\sigma m} &= \mathbb{E[\mathbb{I_{(\tau_m < \infty)}} \exp\{- \frac{1}{2}\sigma^2\tau_m\ \}]} \\ \end{aligned}\]
The above equation holds when \(\sigma > 0\) and \(m > 0\). So we can not substitute \(\sigma = 0\) directly. But since it holds for all \(\sigma > 0\), we can take limit on both sides with respect to \(\sigma \downarrow 0\). \[\begin{aligned} &\lim_{\sigma \downarrow 0} \ \mathbb{E[\mathbb{I_{(\tau_m < \infty)}} \exp\{- \frac{1}{2}\sigma^2\tau_m\ \}]} = \lim_{\sigma \downarrow 0} \ e^{\sigma m} \\ \implies & \ \mathbb{E[\mathbb{I_{(\tau_m < \infty)}}]} = \ \mathbb{\Pr(\tau_m < \infty)} = \ 1 \\ \end{aligned}\]
Now, because \(\tau_m\) is finite with probability 1, (i.e. \(\tau_m\) is finite almost surely) we can remove the indicator function, to get- \[\mathbb{E[\exp\{-\frac{1}{2}\sigma^2 \tau_m\}]} = e^{-\sigma m}\]
This was the hard part. Now let us go through the following theorem.
Theorem: For \(m \in \mathbb{R}\), the first passage time of Brownian motion to level \(m\) is finite almost surely, and a Laplace transformation of its distribution is given by- \[\mathbb{E(e^{-\alpha \tau_m})} = \frac{|m|}{\sqrt{\alpha}}e^{-|m|\sqrt{2\alpha}} \qquad \text{for all} \ \alpha > 0\]
Proof: Let us first consider the case \(m > 0\).
Let \(\alpha\) be a positive number and set \(\sigma = \sqrt{2\alpha}\) which implies- \(\frac{1}{2}\sigma^2 = \alpha\). In that case we can write-
\[\begin{aligned} &\mathbb{E[\exp\{-\frac{1}{2}\sigma^2 \tau_m\}]} = e^{-\sigma m} \\ \implies &\mathbb{E[\exp\{-\alpha \tau_m\}]} = e^{-m\sqrt{2\alpha}} \\ \end{aligned}\]
In case \(m\) is negative, because Brownian motion is symmetric, the first passage time \(\tau_m\) and \(\tau_{|m|}\) have identical distribution.
Hence combining we can write- \[\mathbb{E[\exp\{-\alpha \tau_{m}\}]} = e^{-|m|\sqrt{2\alpha}}\]
But why are we suddenly so interest in \(\mathbb{E[e^{-\alpha \tau_{m}}]}\)?
This is because, Moment Generating Function is unique or if two random variables have the same MGF, then they must have the same distribution. So if we can identify the moment generating function of \(\tau_m\), we can also identify the distribution of \(\tau_m\).
Differentiating both sides of the above equation with respect to \(\alpha\), we get-
\[\begin{aligned} \frac{\delta}{\delta \alpha}\mathbb{E[\exp\{-\alpha \tau_{m}\}]} &= \frac{\delta}{\delta \alpha}e^{-|m|\sqrt{2\alpha}} \\ \implies \mathbb{E[\tau_{m}.\exp\{-\alpha \tau_{m}\}]} &= \frac{|m|}{\sqrt{2\alpha}}e^{-|m|\sqrt{2\alpha}} \qquad \text{For all} \ \alpha >0 \\ \end{aligned}\]
If we decrease \(\alpha\) down to zero, we will obtain- \(\mathbb{E(\tau_m)} = \infty\), so long as \(m \neq 0\).
Reflection Principle
Let us fix a positive level \(m\) and positive time \(t\). We want to count the Brownian motion paths that reach level \(m\) at or before time \(t\). i.e. paths having- \(\tau_m \leq t\).
There are two types of such paths-
- The paths that reach level \(m\) prior to time \(t\) but at time \(t\) it’s some level \(w\) above or below level \(m\). i.e.
\[\tau_m < t \quad \text{and} \quad W(t) = m + w \quad \text{or} \quad m-w\]
- The paths that reach level \(m\) exactly at time \(t\). i.e. \(\tau_m = t\).
But since Brownian motion has continuous path; the probability that the paths that reach level \(m\) exactly at time \(t\) is zero, i.e. \(\mathbb{\Pr(\tau_m = t)} = 0\)
The above figure illustrates for each Brownian motion path that reaches level \(m\) before time \(t\), there is also a reflected path that is at level \(2m-w\) at time \(t\). The reflected path is constructed by switching up and down movements after time \(\tau_m\).
Of course, the probability that a Brownian motion path ends at \(m\) or at \(2m-w\) is exactly zero. In order to have non-zero probabilities, we consider the paths that reach level \(m\), prior to time \(t\) and at or below level \(w\) at time \(t\) and we consider a reflection, which are at or above \(2m-w\) at time \(t\). This leads to the reflection equality- \[\mathbb{\Pr(\tau_m \leq t, \mathbb{W(t)} \leq w)} = \mathbb{P(\mathbb{W(t)} \geq 2m-w)} \quad w \leq m, \ m>0\]
First Passage Time distribution
We are going to see the distribution of first passage time, \(\tau_m\).
Theorem: For all \(m \neq 0\), the random variable \(\tau_m\) has cumulative distribution function- \[\mathbb{P(\tau_m \leq t)} = \frac{2}{\sqrt{2\pi}} \int_{\frac{|m|}{\sqrt{t}}}^{\infty} e^{-\frac{y^2}{2}} dy \qquad \text{For all} \ t \geq 0\]
and the density- \[f_{\tau_m}(t) = \frac{d}{dt}\mathbb{P(\tau_m \leq t)} = \frac{|m|}{t\sqrt{2\pi t}} e^{-\frac{m^2}{2t}}, \qquad \text{For} \ t \geq 0\]
Proof: First consider the case \(m > 0\). Substitute \(w=m\) in the reflection formula to obtain- \[\mathbb{\Pr(\tau_m \leq t, \mathbb{W(t)} \leq m)} = \mathbb{P(\mathbb{W(t)} \geq m)} \quad m>0\]
On the other hand, if we have \(\mathbb{W(t)} \geq m\), we must have- \(\tau_m \leq t\); which implies-
\[\mathbb{\Pr(\tau_m \leq t, \mathbb{W(t)} \geq m)} = \mathbb{P(\mathbb{W(t)} \geq m)} \quad m>0\]
So, using the Total probability theorem we can write the Cumulative Distribution Function or CDF of \(\tau_m\), as follows- \[\begin{aligned} &\mathbb{\Pr(\tau_m \leq t)} \\ = &\mathbb{\Pr( \{\tau_m \leq t\} \cap \{\mathbb{W(t)} \geq m \})} \ + \ \mathbb{\Pr( \{\tau_m \leq t\} \cap \{\mathbb{W(t)} \leq m \})} \\ = & \ 2 \ \mathbb{\Pr(\mathbb{W(t)} \geq m)} \\ = &\frac{2}{\sqrt{2 \pi t}} \int_{m}^{\infty} e^{-\frac{x^2}{2t}} \ dx \end{aligned}\]
We can make change of variable- \(y = \frac{x}{\sqrt{t}}\) in the above integral to get-
\[\begin{aligned} \mathbb{\Pr(\tau_m \leq t)} &= \frac{2}{\sqrt{2 \pi t}} \int_{m}^{\infty} e^{-\frac{x^2}{2t}} \ dx \\ &= \frac{2}{\sqrt{2\pi}} \int_{\frac{m}{\sqrt{t}}}^{\infty} e^{-\frac{y^2}{2}} dy \qquad \text{For} \quad t \geq 0 \\ \end{aligned}\] This one is proved for \(m > 0\). Now if \(m < 0\), since \(\tau_{m}\) and \(\tau_{|m|}\) have distributions we can write- \[\begin{aligned} \mathbb{\Pr(\tau_m \leq t)} &= \frac{2}{\sqrt{2\pi}} \int_{\frac{|m|}{\sqrt{t}}}^{\infty} e^{-\frac{y^2}{2}} dy \qquad \text{For} \quad t \geq 0 \\ \end{aligned}\]
Differentiating both sides with respect to \(t\) we get- \[f_{\tau_m}(t) = \frac{d}{dt}\mathbb{P(\tau_m \leq t)} = \frac{|m|}{t\sqrt{2\pi t}} e^{-\frac{m^2}{2t}}, \qquad \text{For} \ t \geq 0 \qquad \text{Q.E.D.}\]
With the distribution of first passage time, \(\tau_m\) our discussion on Brownian motion ends. Now we will move to Stochastic Calculus, which is little different from ordinary calculus. In Stochastic calculus we will learn about Ito integral, Ito-Doeblin formula, Black-Scholes-Metron Equation etc.