# Squared hitting time expectation including Brownian motion

I'm refering to this question, which was already answered with a hint.

I repeat it here:
Let $(B_t)_{t\geq 0}$ be a standard Brownian motion and for $a<0<b$ define the stopping time $\tau=\inf\{t\geq 0:B_t\in \{a,b\}\}$. Now I want to calculate $E(\tau^2)$.

But I have trouble using the first hint stated in the link. That is using the fact that $B_t^3-3tB_t$ is a martingale to deduce $E(\tau B_\tau^2)$ and I hope that someone could help at that particular point.

• $B_t^4 - 6tB_t^2 + 3t^2$ you should use this equation . It is a martingale, therefore its stopped version is also a martingale Commented Nov 17, 2016 at 19:29
• Exactly, that is the second hint in the question I linked, but the difficult thing in my opinion is to further calculate $6E(\tau B_\tau^2)$ respectively $E(\tau B_\tau^2)$ (there was a little mistake in my question). Commented Nov 17, 2016 at 19:37
• By continuity you know that $B_{\tau}$ is equal to either a or b. You have to pass to the limit on the stopped martingale Commented Nov 17, 2016 at 19:41
• I thought about your tips a little while and I what I got is, that by looking at $B_{\tau\wedge n}$ I can conclude (since its bounded) that $E(B_\tau^3)=E(\tau B_\tau)$. This gives me if I'm not completly wrong that $E(B_\tau^3|B_\tau =a)=E(\tau B_\tau|B_\tau =a)$ and thus $E(\tau|B_\tau =a)=\frac{1}{3} a^2$ and same with b. (edit: sorry wasn't finished yet and pressed comment :)) Commented Nov 17, 2016 at 20:15
• what is u ? t? You also have to suppose the case where it reaches a before b, and the other. Commented Nov 17, 2016 at 20:17

There are a number of things that need to be calculated before arriving at the final answer.

First we keep in mind that $P(B_{\tau}=a)=\frac{b}{b-a}$ and $P(B_{\tau}=b)=\frac{a}{a-b}$.

Consider the martingale $B_t^2-t$.

\begin{align*} E(B_{\tau}^2-\tau)&=0 \\ E(B_{\tau}^2\mid B_{\tau}=a)P(B_{\tau}=a)+E(B_{\tau}^2\mid B_{\tau}=b)P(B_{\tau}=b)-E(\tau)&=0 \\ a^2\frac{b}{b-a}+b^2\frac{a}{a-b}-E(\tau)&=0 \\ E(\tau)&=\frac{a^2b-b^2a}{b-a} \\ E(\tau)&=-ab. \end{align*}

Use the first hint in the link to get an equation involving $E(\tau\mid B_{\tau}=a)$ and $E(\tau\mid B_{\tau}=b)$.

\begin{align*} E(B_{\tau}^3-3\tau B_{\tau})&=0 \\ E(B_{\tau}^3\mid B_{\tau}=a)P(B_{\tau}=a)+E(B_{\tau}^3\mid B_{\tau}=b)P(B_{\tau}=b) \\ -3E(\tau B_{\tau}\mid B_{\tau}=a)P(B_{\tau}=a)-3E(\tau B_{\tau}\mid B_{\tau}=b)P(B_{\tau}=b)&=0 \\ a^3\frac{b}{b-a}+b^3\frac{a}{a-b}-3aE(\tau\mid B_{\tau}=a)\frac{b}{b-a}-3bE(\tau\mid B_{\tau}=b)\frac{a}{a-b}&=0 \\ a^3b-b^3a-3abE(\tau\mid B_{\tau}=a)+3abE(\tau\mid B_{\tau}=b)&=0 \\ E(\tau\mid B_{\tau}=b)-E(\tau\mid B_{\tau}=a)&=\frac{b^2-a^2}{3}. \end{align*}

We have

\begin{align*} E(\tau\mid B_{\tau}=a)P(B_{\tau}=a)+E(\tau\mid B_{\tau}=b)P(B_{\tau}=b)&=E(\tau) \\ E(\tau\mid B_{\tau}=a)\frac{b}{b-a}+E(\tau\mid B_{\tau}=b)\frac{a}{a-b}&=-ab \\ bE(\tau\mid B_{\tau}=a)-aE(\tau\mid B_{\tau}=b)&=-ab(b-a). \end{align*}

Combine this with the equation earlier to get a system of equations you need to solve.

You will find that $E(\tau\mid B_{\tau}=a)=\frac{a^2-2ab}{3}$ and $E(\tau\mid B_{\tau}=b)=\frac{b^2-2ab}{3}$. Now use the second hint in the link. \begin{align*} E(B_{\tau}^4-6\tau B_{\tau}^2+3\tau^2)&=0 \\ E(B_{\tau}^4\mid B_{\tau}=a)P(B_{\tau}=a)+E(B_{\tau}^4\mid B_{\tau}=b)P(B_{\tau}=b) \\ -6E(\tau B_{\tau}^2\mid B_{\tau}=a)P(B_{\tau}=a)-6E(\tau B_{\tau}^2\mid B_{\tau}=b)P(B_{\tau}=b) \\ +3E(\tau^2)&=0 \\ a^4\frac{b}{b-a}+b^4\frac{a}{a-b}-6a^2\frac{b}{b-a}E(\tau\mid B_{\tau}=a)-6b^2\frac{a}{a-b}E(\tau\mid B_{\tau}=b)+3E(\tau^2)&=0 \\ a^4\frac{b}{b-a}+b^4\frac{a}{a-b}-6a^2\frac{b}{b-a}\frac{a^2-2ab}{3}-6b^2\frac{a}{a-b}\frac{b^2-2ab}{3}+3E(\tau^2)&=0 \\ ab(a^2-3ab+b^2)+E(\tau^2)&=0. \end{align*} Therefore $E(\tau^2)=-ab(a^2-3ab+b^2)$.

• You need to have that the stopping time is bounded in order to apply the Optional Stopping or that the martingale is UI
– asdf
Commented Jun 14, 2018 at 18:50
• I believe you forgot to divide by 3 at the very end. Commented May 2, 2020 at 2:36