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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.

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Fraith17
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    $B_t^4 - 6tB_t^2 + 3t^2$ you should use this equation . It is a martingale, therefore its stopped version is also a martingale – Canardini Nov 17 '16 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). – Fraith17 Nov 17 '16 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 – Canardini Nov 17 '16 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 :)) – Fraith17 Nov 17 '16 at 20:15
  • what is u ? t? You also have to suppose the case where it reaches a before b, and the other. – Canardini Nov 17 '16 at 20:17
  • Limit for editing expired. So I have to say here that I forgot 3 in both equations but the result should be correct. So the last step in the hint is to deduce $E(\tau B_\tau^2)$ from that. But I don't really see that right now. – Fraith17 Nov 17 '16 at 20:29
  • We can rewrite you stopping time $\tau$, as $\tau=min(\tau_a,\tau_b)$ where $\tau_a=\inf{t\geq 0:B_t>a}$ and $\tau_b=\inf{t\geq 0:B_t<b}$ $E(\tau B_\tau^2)=E(\tau B_\tau^21_{\tau_a<\tau_b})+E(\tau B_\tau^21_{\tau_a>\tau_b})$. In the first term, $B_\tau=a$, in the other $B_\tau=b$ – Canardini Nov 17 '16 at 20:46
  • based on that i get $E(\tau B_\tau^2)=\frac{ab(a^3-b^3)}{3(b-a)}$. If this is correct than $E(\tau^2)$ has the same result strangely. If you can confirm I try to formulate an answer. – Fraith17 Nov 17 '16 at 21:45
  • I will write a detailed answer – Canardini Nov 18 '16 at 15:20

1 Answers1

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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 $E(\tau^2)=-ab(a^2-3ab+b^2)$.

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