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I am preparing for an exam and one of the problems is:

Compute $\mathbb{E}[ \sigma B^2_{\sigma}]$ when $\sigma = \inf \{t \geq 0 : |B_t| = \sqrt{2} \}$, where B is Standard Brownian Motion.

I am not sure where to begin because I have seen that $\mathbb{E}[B^2_{\sigma}] \ = \ \mathbb{E}\sigma$ for bounded stopping time, yet I am shamefully clueless about how to solve this problem. Could anyone give me a helping hand please?

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    $B$ is a Brownian motion ? – 2017-01-06
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    Yes, in fact, I should have said Standard Brownian Motion. – 2017-01-06

2 Answers 2

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First of all , $\sigma$ is finite almost surely (law of iterated logarithm)

At time $t=\sigma$, it is trivial to see that $B_{\sigma}^2=|B_{\sigma}|^2=\sqrt{2}^2=2$ a.s.

Furthermore , the process $M_t=B_t^2-t$ is a martingale, because $B$ is a martingale and its quadratic variation function is $t$.

The process $U_t=M_{t\wedge\sigma}$ is also a martingale( a stopped martingale is a martingale)

$$U_0=E(U_t)$$ In particular using the limit, and the dominated convergence theorem

$$0=U_0=lim_{t \to \infty}E(U_t)=E(lim_{t \to \infty}U_t)=E(M_{\sigma})=E(B_{\sigma}^2)-E(\sigma)$$

which the results you are looking for.

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    Thanks for your answer, I agree with all that. However it does not answer my question: Calculate: $\mathbb{E} \sigma B^2_{\sigma}$..... – 2017-01-06
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    It does if you consider the answer that $B_\sigma^2 = 2$ and plug that in to the last equation and your question… – 2017-01-06
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    @Novice My second line gives you the answer. – 2017-01-06
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Because of the continuity of the sample paths of Brownian motion, we have $$|B_{\sigma}| = \sqrt{2};$$ hence $B_{\sigma}^2=2$ which implies

$$\mathbb{E}(\sigma B_{\sigma}^2) = 2 \mathbb{E}(\sigma).$$

This means that everything boils down to calculating $\mathbb{E}(\sigma)$. To this end, recall that $M_t := B_t^2-t$ is a martingale. By the optional stopping theorem, $M_{t \wedge \sigma}$, $t \geq 0$, is a martingale. In particular, $$\mathbb{E}(M_{t \wedge \sigma}) = \mathbb{E}(M_0)=0,$$

i.e.

$$\mathbb{E}(B_{t \wedge \sigma}^2) = \mathbb{E}(\sigma \wedge t). \tag{1}$$

Since $|B_{t \wedge \sigma}| \leq \sqrt{2}$ for all $t \geq 0$, we can apply the dominated convergence theorem to conclude that

$$2 = \mathbb{E}(B_{\sigma}^2) = \lim_{t \to \infty} \mathbb{E}(B_{t \wedge \sigma}^2).$$

On the other hand, the monotone convergence theorem yields

$$\mathbb{E}(\sigma) = \lim_{t \to \infty} \mathbb{E}(\sigma \wedge t).$$

Letting $t \to \infty$ in $(1)$, we find

$$\mathbb{E}(\sigma)=2.$$

Hence, $$\mathbb{E}(\sigma B_{\sigma}^2) = 4.$$