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

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


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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Ok, let $B_0$ be deterministic, then the next step would be to recall that $$ \Bbb P(B_t\in Y) = \int_Y p_t(y)\mathrm dy $$ where $p_t(y)$ is the density function of the jointly normal variable $B_t \sim \mathcal N(B_0,t I)$. As a result, $$ \int_0^\infty \Bbb P(B_t\in Y)\mathrm dy = \int_Y\left(\int_0^\infty p_t(y)\mathrm dt\right)\mathrm dy $$ which always hold since $p_t(y)\geq 0$ and is jointly measurable. However, as I mentioned in the comment to Bunder's answer, $$ P(y):=\int_0^\infty p_t(y)\mathrm dt = \infty $$ for all $y$, – Ilya Mar 1 '13 at 16:17
which would mean that $\int_Y P(y)\mathrm dy = \infty$ unless $Y$ is of measure $0$. However, it does not seem to be true - and I can't see where the mistake is. – Ilya Mar 1 '13 at 16:18
up vote 0 down vote accepted

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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What about $$ \int_0^\infty t^{-\frac12}\mathrm e^{-\frac{s^2}{2t}}\mathrm dt = \infty $$ – Ilya Mar 1 '13 at 15:07
@Ilya 's complaint seems valid. And I don't see how to address it. – RClark Mar 1 '13 at 15:59
@RClark: that's why I didn't post the same answer before :) – Ilya Mar 1 '13 at 16:13
Which is a kindly reminder that the BM is transient only for $k \geq 3$. Thanks for pointing it out :) i made the correspoinding corrections. – Bunder Mar 1 '13 at 17:11
Wow! I can literally feel the neurons in my brain connecting! Thanks! – RClark Mar 1 '13 at 17:19

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