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Let $B(t)=(B_{1}(t),B_{2}(t),B_{3}(t))$ be a standard three dimensional Brownian motion (i.e. it has independent components and starts at the origin). Now let $a=(a_{1},a_{2},a_{3})\neq(0,0,0)$ be a point in $\mathbb{R}^{3}$ and define $X(t)=\frac{1}{||a+B(t)||}$. Now compute the expected value of $X(t)$. Im kind of stuck on this one. I think I should use Girsanov theorem or some Ito calculus but im not getting anywhere. Any ideas?

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Right, some Ito calculus is a good idea. What did you try? – Did Nov 29 '12 at 21:30
up vote 1 down vote accepted

Let $f(x) := \frac{1}{|a+x|}$ where $x \in \mathbb{R}^3 \backslash \{-a\}$. Then

$$\partial_{x_j} f(x) = - \frac{1}{2} \cdot \frac{2 (x_j+a_j)}{|a+x|^3} \qquad \qquad \partial_{x_j}^2 f(x) = \frac{1}{|a+x|^3} \cdot \left(-1 + \frac{3}{2} \frac{2 (x_j+a_j)^2}{|a+x|^2} \right)$$

We can apply Itô's Formula and obtain

$$f(B_t)-f(0) = \sum_{j=1}^3 \int_0^t \frac{-(B_s^j+a_j)}{|a+B_s|^3} \, dB_s^j + \frac{1}{2} \sum_{j=1}^3 \int_0^t \frac{1}{|a+B_s|^3} \cdot \left(-1+ 3 \frac{(B_s^j+a_j)^2}{|a+B_s|^2} \right) \, ds \quad (\ast)$$

Now we have

$$\sum_{j=1}^3 \int_0^t \frac{1}{|a+B_s|^3} \cdot \left(-1+ 3 \frac{(B_s^j a_j)^2}{|a+B_s|^2} \right) \, ds = \int_0^t \frac{1}{|a+B_s|^3} \cdot \bigg(-3+ 3 \underbrace{\sum_{j=1}^3 \frac{(B_s^j+a_j)^2}{|a+B_s|^2}}_{1} \bigg) \, ds=0$$

Since the first term on the righthand side in $(\ast)$ is a martingale (with expectation 0) we conclude

$$\mathbb{E} \left( \frac{1}{|a+B_t|}\right) = \mathbb{E}f(B_t) = \mathbb{E}f(B_0) = \frac{1}{|a|}$$

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