Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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.

share|improve this question
    
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
add comment

1 Answer

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.

share|improve this answer
1  
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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.