# expectation of brownian motion squared with regard to stopping time

let $T_1$ be the first occurrence of a Poisson process at rate $\lambda$, and $X(t) = \sigma B(t) + \mu t$ be another independent Brownian motion with drift, calculate $E(X(T_1))$ and $\operatorname{Var}(X(T_1))$.

I know $E(T_1) = 1/\lambda$, as well as the following

$E(X(t))=\mu t$
$\operatorname{Var}(X(t)) = \sigma^2 t$

but not sure what's the result when $t$ becomes a random variable itself.

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Is $T_1$ supposed to be independent of the Brownian motion $B$? Try conditioning on $T_1$. – Nate Eldredge Dec 12 '12 at 19:52
Yes, $T_1$ is independent of the Brownian motion $B$. Would you mind explain a bit more how shall I condition on $T_1$? – threemonks Dec 12 '12 at 20:29

Hint: You could apply Wald's identities:

Let $(B_t,\mathcal{F}_t)_t$ be a BM$^1$ and $\tau$ an $\mathcal{F}_t$ stopping time such that $\mathbb{E}\tau<\infty$. Then $B_\tau \in L^2$ and $$\mathbb{E}B_\tau = 0 \qquad \qquad \mathbb{E}B_{\tau}^2 = \mathbb{E}\tau$$

In this case you can define $\mathcal{F}_t$ as $\mathcal{F}_t := \sigma\{(B_s, s \leq t),T_1\}$ (which is an admissible filtration since $T_1$ is independent of the BM) and use $\mathbb{E}T_1 = \frac{1}{\lambda}$.

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Clear overkill. – Did Dec 15 '12 at 9:25

I am going out on a limb here but maybe try the following:

We know that $$E(X) = E[E(X|T)]$$$$\text{Var}(X) = E(\text{Var}(X|T))+\text{Var}(E(X|T))$$ from the law of total variance.

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