# A Brownian motion starting from 0, it becomes -1 once reach -1, what is its expectation?

$X_t$ is a Brownian Motion, it reaches becomes -1 forever once it reaches -1.

Mathematically, $T = \inf\{t: W_t = -1\}$ is a stopping time.

When $t<T$, $X_t = W_t$ ,while $t\geq T$, $X_t = -1$.

What is the expectation of $X_1$?

Part of my solution is

$\mathbb{E}[X_1] = \mathbb{E}[X_1|T<1]\mathbb{P}(T<1)+\mathbb{E}[X_1|T\geq1]\mathbb{P}(T\geq1)$

I know

$\mathbb{E}[X_1|T<1]=-1$ and $\mathbb{P}(T<1)=2(1-\Phi(-1))$

$\mathbb{P}(T\geq1)=1-2(1-\Phi(-1))$

but how about $\mathbb{E}[X_1|T\geq1]$ ? Can anyone help one help me on it? Thank you so mush!

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$X_t = W_{t \wedge T}$ is a martingale... –  Nate Eldredge May 19 '12 at 12:22
So do you mean $\mathbb{E}[X_1] = \mathbb{E}[X_0]=0$? –  pidig May 19 '12 at 14:51