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Let $(B_1,B_2)$ be a two-dimensional Brownian motion. Let $$ X_t = \int\limits_0^t B_1(s)\mathrm \; dB_2(s). $$ Is there a closed form for $X$ or the integral above is all one can get?

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@ Ilya : Hi I think that the best you can get is the law of $X_t$ (at least if $B_1$ and $B_2$ are independent). – TheBridge Mar 1 '12 at 14:05
@TheBridge: hi, thanks for the comment. They are independent due to the definition of $2$d Brownian motion. I also can prove that there is no $C^2$ function $f$ such that $X_t = f(B_1(t),B_2(t))$ - but I thought that there are maybe other nice forms for $X_t$. – Ilya Mar 1 '12 at 14:28
up vote 4 down vote accepted

In order to determine the law of $X_t$ it suffices to determine it for $t=1$ due to scaling property, $X_t \stackrel{d}{=} t X_1$.

Using the vector Ito process $V_t = (X_t, Y_t, Z_t)$ with $$ \mathbb{d} X_t = Y_t \mathrm{d} B^{(2)}_t \qquad \mathrm{d}Y_t = \mathrm{d} B^{(1)}_t \qquad \mathrm{d} Z_t = \mathrm{d} B^{(2)}_t \qquad V_0 = (0,0,0) $$ and noticing that $\mathcal{L}_0$ operator is nilpotent on polynomials in $X_t$, $Y_t$ and $Z_t$ allows to compute $\mathbb{E}(X_t^n)$ as $$ \mathbb{E}(X_t^n) = \frac{t^n}{n!} \mathcal{L}_0^{\circ n}(X_t^n) = \left(1/2\right)_n \cdot {}_1 F_1 \left(-n ; \frac{1}{2} - n; -\frac{1}{2}\right) $$ where $\mathcal{L}_0 = \frac{1}{2} \left( Z_t \frac{\partial}{\partial X_t} + \frac{\partial}{\partial Y_t}+ \frac{\partial}{\partial Z_t} \right)$.

The (exponential) moment generating function corresponding to this sequence of moments is easy to find: $$ \mathcal{M}_{X_t}(u) = \sum_{n=0}^\infty \frac{u^n}{n!} \mathbb{E}(X_t^n) = \frac{\exp(-t u/2)}{\sqrt{1- t u}} $$ which means that $X_t + \frac{t}{2}$ is equal in distribution to $\Gamma(1/2,t)$ random variable, i.e. $X_t \stackrel{d}{=} \frac{t}{2}\left(Z^2-1\right)$ for a standard normal random variable $Z$.

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thank you for a perfect answer – Ilya Mar 2 '12 at 8:05
Could anyone clarify what is $\mathcal{L}_0$? Thank you. – Tom Artiom Fiodorov Mar 2 '12 at 13:49
@ArtiomFiodorov $\mathcal{L}_0$ is the differential operator from Ito's lemma: $\mathbb{E}(f(V_t)) = \mathbb{E}(f(V_0)) + \int_0^t \mathcal{L}_0(f)(V_s) \mathrm{d} s$ – Sasha Mar 2 '12 at 14:30

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