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Let $B_t$ $(t \geq 0)$ be a Brownian motion on $\mathbb{R}^3$. That is, $B_t = (B_{t}^{(1)},B_{t}^{(2)},B_{t}^{(3)})$, where each $B_{t}^{(i)}$ is a Brownian motion on $\mathbb{R}$. Let $Y$ be a Borel subset of $\mathbb{R}^3$.

I am being asked to show that

$$

\mathbb{E} \left( \int_{0}^{\infty} I({\{t:B_t \in Y\}})(t)dt \right) = c\int_{Y}\frac{dy}{|B_0 – y|}.

$$

for some constant $c$. Here $I(A)$ denotes the indicator function of a set $A$.

Using Fubini’s Theorem on the left-hand side, I reduced the equation to

$$

\int_{0}^{\infty} \mathbb{P}(B_t \in Y) dt = c\int_{Y}\frac{dy}{|B_0 – y|}.

$$

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Unfortunately, I’m not sure what to do now. I would appreciate any help.

Thanks.

EDIT: As some people have pointed out, the expectation and probability on the left-sides of both equation should probably be conditioned on $B_0$. The professor has been a bit sloppy about this with Brownian motion.

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I suspect that it is either the expectation of the right hand-side or the conditional expectation with respect to $B_0$ on the left side. But the equation as it is cannot be since

the right side is stochastic and left side deterministic. Assuming that you were asked

$$\mathbb{E} \left( \int_{0}^{\infty} I({\{t:B_t \in Y\}})(t)dt | B_0 \right) = c\int_{Y}\frac{dy}{|B_0 – y|}.$$

You can still use Fubini, so the question reduces to calculate:

$$P(B_t \in Y | B_0) = P(B_t – B_0 \in Y – B_0 | B_0)$$

(recall that $P(A | Y) = E( 1_A | Y )$). Since the increment $B_t – B_0$ is

independent of $B_0$ and follows a multivariate Normal$(0,diag(t,t,t))$ the conditional probability is equal to

$$ \int_{Y – B_0} \frac{1}{(2\pi t)^{3/2}} e^{-\frac{s_1^2 + s_2^2 + s_3^2}{2t}} ds_1 \, ds_2 \, ds_3$$

Then you do the change $u = s-B_0$, Fubini again, integrating with respect to $t$ and you should get the result.

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