5.4 Brownian Motion and Stochastic Calculus

In this section, we briefly go over some problems for stochastic calculus, the counterpart of stochastic processes in continuous space. Since the basic definitions and theorems of Brownian motion and stochastic calculus are directly used as interview problems, we’ll simply integrate them into the problems instead of starting with an overview of definitions and theorems.

Brownian motion

A. Define and enumerate some properties of a Brownian motion?¹

Solution: This is the most basic Brownian motion question. Interestingly, part of the definition, such as $W(0)=0$, and some properties are so obvious that we often fail to recite all the details.

A continuous stochastic process $W(t),\ t\geq0$, is a Brownian motion if

• $W(0)=0$;
• The increments of the process $W(t_1)-W(0),W(t_2)-W(t_1),\ldots,W(t_n)-W(t_{n-1})$, $\forall 0\leq t_1\leq t_2\leq\cdots\leq t_n$ are independent;
• Each of these increments is normally distributed with distribution $W(t_{i+1})-W(t_i)\sim N(0,t_{i+1}-t_i)$.

Some of the important properties of Brownian motion are the following: continuous (no jumps); $E[W(t)]=0$; $E[W(t)^2]=t$; $W(t)\sim N(0,t)$; martingale property $E[W(t+s)\mid W(t)]=W(t)$; $\operatorname{cov}(W(s),W(t))=s,\ \forall 0<s<t$; and Markov property (in continuous space).

There are two other important martingales related to Brownian motion that are valuable tools in many applications.

• $Y(t)=W(t)^2-t$ is a martingale.
• $Z(t)=\exp\{\lambda W(t)-\frac12\lambda^2t\}$, where $\lambda$ is any constant and $W(t)$ is a Brownian motion, is a martingale. (Exponential martingale).

We’ll show a proof of the first martingale using Ito’s lemma in the next section. A sketch for the exponential martingale is the following:²

B. What is the correlation of a Brownian motion and its square?

Solution: The solution to this problem is surprisingly simple. At time $t$, $B_t\sim N(0,t)$, by symmetry, $E[B_t]=0$ and $E[B_t^3]=0$. Applying the equation for covariance $\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]$, we have $\operatorname{Cov}(B_t,B_t^2)=E[B_t^3]-E[B_t]E[B_t^2]=0-0=0$. So the correlation of a Brownian motion and its square is 0, too.

C. Let $B_t$ be a Brownian motion. What is the probability that $B_1>0$ and $B_2<0$?

Solution: A standard solution takes advantage of the fact that $B_1\sim N(0,1)$, and $B_2-B_1$ is independent of $B_1$, which is again a normal distribution: $B_2-B_1\sim N(0,1)$. If $B_1=x>0$, then for $B_2<0$, we must have $B_2-B_1<-x$.

But do we really need the integration step? If we fully take advantage of the facts that $B_1$ and $B_2-B_1$ are two IID $N(0,1)$, the answer is no. Using conditional probability and independence, we can reformulate the equation as

This approach is better demonstrated in Figure 5.10. When we have $B_1>0$ and $B_2-B_1<-B_1$, which accounts for 1/8 of the density volume. (All 8 regions separated by $x=0$, $y=0$, $y=x$, and $y=-x$ have the same density volume by symmetry.)

Stopping time/ first passage time

A. What is the mean of the stopping time for a Brownian motion to reach either -1 or 1?

Solution: As we have discussed, $B_t^2-t$ is martingale. It can be proved by applying Ito’s lemma:

So $d(B_t^2-t)$ has no drift term and is a martingale. Let $T=\min\{t;\ B_t=1\text{ or }-1\}$. At continuous time and space, the following property still applies: A martingale stopped at 132 a stopping time is a martingale! So $B_T^2-T$ is a martingale and $E[B_T^2-T]=B_0^2-0=0$. The probability that $B_t$ hits 1 or -1 is 1, so $B_T^2=1\Rightarrow E[T]=E[B_T^2]=1$.

B. Let $W(t)$ be a standard Wiener process and $\tau_x$ ($x>0$) be the first passage time to level $x$ ($\tau_x=\min\{t;\ W(t)=x\}$). What is the probability density function of $\tau_x$ and the expected value of $\tau_x$?

Solution: This is a textbook problem that is elegantly solved using the reflection principle, so we will simply summarize the explanation. For any Wiener process paths that reach $x$ before $t$ ($\tau_x\leq t$), they have equal probability ending above $x$ or below $x$ at time $t$, $P(\tau_x\leq t,W(t)\geq x)=P(\tau_x\leq t,W(t)\leq x)$. The explanation lies in the reflection principle. As shown in Figure 5.11, for each path that reaches $x$ before $t$ and is at a level $y$ above $x$ at time $t$, we can switch the sign of any move starting from $\tau_x$ and the reflected path will end at $2x-y$ that is below $x$ at time $t$. For a standard Wiener process (Brownian motion), both paths have equal probability.

Let $v=\frac{w}{\sqrt t}$, we have $e^{-w^2/2t}=e^{-v^2/2}$ and $dv=\frac{dw}{\sqrt t}$.

³

Take the derivative with respect to $t$, we have

From part A, it’s easy to show that the expected stopping time to reach either $\alpha$ ($\alpha>0$) or $-\beta$ ($\beta>0$) is again $E[N]=\alpha\beta$. The expected first passage time to level $x$ is

essentially the expected stopping time to reach either $x$ or $-\infty$ and $E[\tau_x]=x\times\infty=\infty$. Although we have $P(\tau_x\leq\infty)=2-2N(x/\sqrt\infty)=1$, the expected value of $\tau_x$ is $\infty$!

C. Suppose that $X$ is a Brownian motion with no drift, i.e. $dX(t)=dW(t)$. If $X$ starts at 0, what is the probability that $X$ hits 3 before hitting -5? What if $X$ has drift $m$, i.e. $dX(t)=m\,dt+dW(t)$?

Solution: A Brownian motion is a martingale. Let $p_3$ be the probability that the Brownian motion hits 3 before -5. Since a martingale stopped at a stopping time is a martingale, we have $3P_3+(-5)(1-P_3)=0\Rightarrow P_3=5/8$. Similar to random walk, if we have stopping boundaries ($\alpha>0$) and $-\beta$ ($\beta>0$), the probability that it stops at $\alpha$ instead of $-\beta$ is $p_\alpha=\beta/(\alpha+\beta)$. The expected stopping time to reach either $\alpha$ or $-\beta$ is again $E[N]=\alpha\beta$.

When $X$ has drift $m$, the process is no longer a martingale. Let $P(t,x)$ be the probability that the process hits 3 before hitting -5 when $X=x$ at time $t$. Although $X$ is no longer a 134 martingale process, it is still a Markov process. So $P(t,x)=P(x)$ is actually independent of $t$. Applying the Feynman-Kac equation⁴, we have

We also have boundary conditions that $P(3)=1$ and $P(-5)=0$.

$mP_x(x)+1/2P_{xx}(x)=0$ is a homogeneous linear differential equation with two real roots: $r_1=0$ and $r_2=-2m$. So the general solution is $P(x)=c_1e^{0x}+c_2e^{-2mx}=c_1+c_2e^{-2mx}$. Applying the boundary conditions, we have

A different and simpler approach takes advantage of the exponential martingale: $Z(t)=\exp\{\lambda W(t)-\frac12\lambda^2t\}$. Since $W(t)=X(t)-mt$, $X(t)-mt$ is a Brownian motion as well. Applying the exponential martingale, we have $E[\exp(\lambda(X-mt)-\frac12\lambda^2t)]=1$ for any constant $\lambda$. To remove the terms including time $t$, we can set $\lambda=-2m$ and the equation becomes $E[\exp(-2mX)]=1$. Since a martingale stopped at a stopping time is a martingale, we have $P_3\exp(-2m\times3)+(1-P_3)\exp(-2m\times-5)=1\Rightarrow\frac{e^{10m}-1}{e^{10m}-e^{-6m}}$.

D. Suppose that $X$ is a generalized Weiner process $dX=dt+dW(t)$, where $W(t)$ is a Brownian motion. What is the probability that $X$ ever reaches -1?

Solution: To solve this problem, we again can use the equation $E[\exp(-2mX)]=1$ from the previous problem with $m=1$. It may not be obvious since we only have one apparent boundary, -1. To apply the stopping time, we also need a corresponding positive boundary. To address this problem, we can simply use $+\infty$ as the positive boundary and the equation becomes

Ito’s lemma

Ito’s lemma is the stochastic counterpart of the chain rule in ordinary calculus. Let $X(t)$ be an Ito process satisfying $dX(t)=\beta(t,X)dt+\gamma(t,X)dW(t)$, and $f(X(t),t)$ be a twice-differentiable function of $X(t)$ and $t$. Then $f(X(t),t)$ is an Ito process satisfying

A. Let $B_t$ be a Brownian motion and $Z_t=\sqrt t B_t$. What is the mean and variance of $Z_t$? Is $Z_t$ a martingale process?

Solution: As a Brownian motion, $B_t\sim N(0,t)$, which is symmetric about 0. Since $\sqrt t$ is a constant at $t$, $Z_t=\sqrt t B_t$ is symmetric about 0 and has mean 0 and variance $t\times\operatorname{var}(B_t)=t^2$. More exactly, $Z_t\sim N(0,t^2)$.

Although $Z_t$ has unconditional expected value 0, it is not a martingale. Applying Ito’s lemma to $Z_t=\sqrt t B_t$, we have $dZ_t=\frac{\partial Z_t}{\partial B_t}dB_t+\frac{\partial Z_t}{\partial t}dt+\frac12\times\frac{\partial^2Z_t}{\partial B_t^2}dt=\frac12t^{-1/2}B_t\,dt+\sqrt t\,dB_t$. For all the cases that $B_t\neq0$, which has probability 1, the drift term $\frac12t^{-1/2}B_t\,dt$ is not zero.⁵ Hence, the process $Z_t=\sqrt t B_t$ is not a martingale process.

B. Let $W(t)$ be a Brownian motion. Is $W(t)^3$ a martingale process?

Solution: Applying Ito’s lemma to $f(W(t),t)=W(t)^3$, we have $\frac{\partial f}{\partial W(t)}=3W(t)^2$, $\frac{\partial f}{\partial t}=0$, $\frac{\partial^2 f}{\partial W(t)^2}=6W(t)$, and $df(W(t),t)=3W(t)dt+3W(t)^2dW(t)$. So again for the cases $W(t)\neq0$, which has probability 1, the drift term is not zero. Hence, $W(t)^3$ is not a martingale process.