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.
Thanks for reading.