Calculating $\mathbb{E}(|B_t|^{-2})$ for 3-dim. Brownian motion

Question

Ok so I'm given a standard 3-dimensional Brownian motion $B(t) = (B_{1}(t),B_{2}(t),B_{3}(t))$, function $f(x,y,z) = \frac{1}{\sqrt{x^2+y^2+z^2}}$ and the process $A(t) = f(B(t))$ and $t \in [1;\infty)$

I need to calculate $E[A^2(t)]$.

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What I did:

I used Ito's formula for the function $f(t,x,y,z) = \frac{1}{\sqrt{x^2+y^2+z^2}}$ and ended up with

$d(f(t,x,y,z)) = -x(x^2+y^2+z^2)^{-\frac{3}{2}}dx - y(x^2+y^2+z^2)^{-\frac{3}{2}}dy - z(x^2+y^2+z^2)^{-\frac{3}{2}}dz$

At this point I'm lost.

$f(0)$ is infinity, so i'm not sure what is going on. Can I even go further and claim, that $A(t)$ = sum of 3 integrals?

Now, if I just forget abut this issue above and go directly for the square, I get:

$E[A^2(t)] = \int_{1}^{t}E[\frac{1}{(x^2+y^2+z^2)^2}]ds$ using the ito isometry and the fubini-tonelli theorem.

And I don't know what to do. Apparently the answer is $\frac{1}{t}$ but for some reason I can't comprehend what am I supposed to do there. Can I go into triple integral and change to spherical coordinates? Also, this constraint on $t$, am I right to assume, that the integration limits will therefore be from 1 and not from 0?

I'd appreciate any advice.

Answer 1

I don't see that Itô's formula is of much use for this problem (in particular, since there is the sinularity at zero which means that you would need to work with local times). It's much easier to use the fact that we know the transition density of $(B_t)$.

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By the very definition of $A(t)$, we need to compute

$$\mathbb{E}(A^2(t)) = \mathbb{E} \left( \frac{1}{|B_t|^2} \right).$$

Plugging in the transition density of $B_t$ we find that

$$\mathbb{E} \left( \frac{1}{|B_t|^2} \right) = \frac{1}{\sqrt{2\pi t}^3} \int_{\mathbb{R}^3} \frac{1}{x^2+y^2+z^2} \exp \left( - \frac{x^2+y^2+z^2}{2t} \right) \, d(x,y,z).$$

Introducing polar coordinates we get

$$\begin{align*} \mathbb{E} \left( \frac{1}{|B_t|^2} \right) &= \frac{4\pi}{\sqrt{2\pi t}^3} \int_{(0,\infty)} \frac{1}{r^2} \exp \left(- \frac{r^2}{2t} \right) \, (r^2 \, dr) \\ &= \sqrt{\frac{2}{\pi t^3}} \int_{(0,\infty)} \exp \left(- \frac{r^2}{2t} \right) \, dr. \end{align*}$$

Hence,

$$\begin{align*} \mathbb{E} \left( \frac{1}{|B_t|^2} \right) &=\frac{1}{t} \underbrace{\frac{1}{\sqrt{2\pi t}} \int_{\mathbb{R}} \exp \left(- \frac{r^2}{2t} \right) \, dr}_{=1} = \frac{1}{t}. \end{align*}$$

Answer 2

Here's a different approach.

1) $$E[e^{-s|B_t|^2}] = E[e^{-ts N(0,1)^2}]^3.$$

2) \begin{align*} E[e^{-u N(0,1)^2}] & = \frac{1}{\sqrt{2\pi}}\int e^{-x^2 ( u + 1/2) }dx\\ & = (2u+1)^{-1/2}. \end{align*}

3) $$x^{-1} = \int_0^\infty e^{-s x} ds.$$

Putting it all together:

\begin{align*} E [|B_t|^{-2} ] & = E [\int_0^\infty [e^{-s|B_t|^2} ]ds \\ & = \int_0^\infty E [ e^{-s|B_t|^2}] ds \\ & = \int_0^\infty (2st+1)^{-3/2} ds \\ & = -\frac{2}{(2t)}(2st+1)^{-1/2} \left.\right|_{s=0}^{s=\infty}\\ & = \frac{1}{t} \end{align*}