# Finding the transition density for $(W_t, B_{1,t}, B_{2,t})$, where $\mathrm{d} W_t = \frac{B_t}{| B_t |} \cdot \mathrm{d} B_t$

Let $B_t$ be 2-dimensional standard Wiener process. Define $W_t$ as $$W_0 = 0, \quad W_t = \int_0^t \frac{B_s \cdot \mathrm{d}B_s}{\sqrt{B_s \cdot B_s}}$$ It is well known that $W_t$ is a standard 1-dimensional Wiener process. I convinced myself of it by noting that the moment generating function of the increment (here $Z_i$ are i.i.d standard normal random variables) is $$\mathbb{E}\left( \exp\left( u \frac{X_1 Z_1 + X_2 Z_2}{\sqrt{X_1^2 + X_2^2}} \right) \right) = \mathbb{E}\left( \frac{u^2}{2} \left( \left( \frac{X_1}{\sqrt{X_1^2+X_2^2} }\right)^2 + \left( \frac{X_2}{\sqrt{X_1^2+X_2^2}} \right)^2 \right) \right) = \mathrm{e}^{u^2/2}$$ hence the sum of these is also normal.

Let $V_t(\omega) = (W_t(\omega), B_{1,t}(\omega), B_{2,t}(\omega))$. I would like to find the probability density function of $V_t | V_s = v$ for $t>s$.

I initially thought that $V_t$ is a Gaussian process, but it mostly likely is not. It is easily seen that $$\mathbb{E}(W_t B_{k,s}) = \int_0^{\min(t,s)} \mathbb{E}\left( \frac{B_{k,u}}{\sqrt{B_{1,u}^2 + B_{2,u}^2}} \right) \mathrm{d} u = 0 \quad \mathbb{E}(B_{n,t} B_{m,s}) = \min(t,s) \delta_{n,m}$$ Thus if $V_t$ was Gaussian it would be 3-dimensional standard Wiener process, but such a pdf does not verify the forward Kolmogorov equation.

Thus I am seeking suggestions on how to go about determining $V_t|V_s=v$ for $t>s$.

If any of my arguments above do not hold water, I am naturally interested to find out why.

@ Sasha : Hi, just a remark, to see that $W_t$ is a Brownian Motion, you can calculate its quadratic variation, see that it's equal to $t$ and conclude from Lévy's caracterisation theorem. This doesn't help for the problem at hand but simplifies your original argument. Best regards – TheBridge Apr 5 '12 at 10:13
@ Sasha : It might be worth trying to calculte the SDE followed by $W_t$ conditionally to $B_t^1=b_1,B_t^2=b_2$. If you can from this SDE calculate $W_t$'s density, it should have the form $P(W_t|b_1,b_2)$ and so you might have a semi-closed form for the joint density of $V$. But I haven't tried to do the calculations, so I don't know if it is really a good idea. Best regards. – TheBridge Apr 5 '12 at 14:10
@ Sasha: I found that by conditioning at $T=1$,$B_{i,1}=b_i$ for $t<1$, $(B^2_{1,t}+B^2_{2,t})^{1/2}dW_t=B_{1,t}d\beta_{1,t}+B_{2,t}d\beta_{2,t}+\frac{d‌​‌t}{1−t}[(b_1−B_{1,t})B_{1,t}+(b_2−B_{2,t})B_{2,t}]$ where $(\beta_1,\beta_2)$ is a 2-d BM under the new filtration, the quadratic variation of $W_t$ is still equal to $t$, but this is about all I can say. The drift is non trivial and I can't tell if $W_t$ is still a gaussian process here, if this were to be the case then the 2 first moments could be computed and taking the limit as $t$ goes to 1 would finish the job. Best regards. – TheBridge Apr 5 '12 at 16:37