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Let $(B_t)_{t\ge 0}$ and $(Y_t)_{t\ge 0}$ be two independent Brownian motion, and let $T$ the following stopping time $$ T=\inf\{t\ge 0 : Y_t=1 \}. $$

I should prove that

$$ \mathbb{P}(B_T \le x) = \mathbb{E} \Bigl (F \Bigl ( \frac{x}{\sqrt{T}} \Bigr ) \Bigr ), $$ where $F$ is the cumulative function of a standard Gaussian.

My attempt was writing $\mathbb{P}(B_T \le x) = \mathbb{E}(\mathbb{I}(B_T \le x)) = \mathbb{E}(\mathbb{E}(\mathbb{I}(B_T \le x)|T))$, with the second identity following from the properties of the conditional expectation. Now I would go one by saying something like $$ \mathbb{E}(\mathbb{I}(B_T \le x)|T)= F \Bigl ( \frac{x}{\sqrt{T}} \Bigr ), $$ But I’ve been failing. Can someone please suggest if this strategy is right and help me?

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Your strategy is correct, and indeed the formula

$ \mathbb{P}(B_T \leq x | T) = F\big(\frac{x}{\sqrt{T}}\big)$

is correct because $B$ and $T$ are independent. I will present what I think is the easiest way to prove it, and I hope I will not overcomplicate it.

Note that the function $G(B,T) = \mathbb{1}\{B_T \leq x\} $ is a function of the whole Brownian path $B$ and $T$. Also, by assumption, $B$ and $T$ are independent. Therefore, the joint law of $(B,T)$ the product of Wiener measure $\mathbb{Q}$, (i.e., the law of Brownian motion) and the law of $T$ which we can denote $\mu$. Therefore,

$\mathbb{E}[G(B,T)] = \int_0^{\infty} \int_{C([0,\infty))} G(\omega,t) \mathbb{Q}(d\omega) \mu(dt)$.

For fixed $t$, the inner integral is, by definition of expectation,

$\int_{C([0,\infty))} G(\omega,t) \mathbb{Q}(d\omega) = \mathbb{P}(B_t \leq x) = F\big(\frac{x}{\sqrt{t}}\big).$

Integrating this against the law of $T$ gives us $\mathbb{E}[F(\frac{x}{\sqrt{T}})]$ which is what you wanted to show. Let me know if this makes sense or if you would like anything clarified further.

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  • $\begingroup$ Yes, this is exactly the formality I was struggling for, it just seems to me a bit strange because of the integration over the space $C[0,\infty)$, wich I’m not so familiar with. So thank you very much $\endgroup$ Commented Jan 15 at 7:05

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