Solve the SDE $d X_{t}=X_{t} B_{t} d t-X_{t} d B_{t}$

Question

I am not sure about my attempt so far, as I hit a road block:

let $Y_{t}=X_{t} B_{t}$ then $d Y_{t}= B_{t}dX_{t}+X_{t}d B_{t}+ d B_{t} d X_{t}$

Substituting $dX_{t}$ we get:

$dY_{t}=B_{t}\left(X_{t} B_{t} d t-X_{t} d B_{t}\right)+ X_{t} d B_{t}+ d B_{t}\left(X_{t} B_{t} d t-X_{t} d B_{t}\right)$

$d B_{t} d t=0 \quad$ and $\quad d B_{t} d B_{t}=d t$

$d Y_{t}=X_{t} B_{t}^{2} d t-X_{t} B_{t} d B_{t}+X_{t} d B_{t}-X_{t} d t$

$d Y_{t}=X_{t}\left(B_{t}^{2}-1\right) d t-X_{t}\left(B_{t}-1\right) d B_{t}$

I am not sure how to proceed forward now.

2nd attempt

Whilst waiting for someone to reply, I tried a different approach:

$d X_{t}=X_{t} B_{t} d t-X_{t} d B_{t}$

$d X_{t}=X_{t}\left(B_{t} d t-d B_{t}\right)$

Let $d Y_{t} = \left(B_{t} d t-d B_{t}\right)$

$d Y_{t}$ is OU process hence $Y_t$ is $Y_{t}=Y_{0} e^{t} - \int_{0}^{t} e^{(t-s)} d B_{s}$

This makes $d X_{t}= - X_{t} dY_t$ where $X_{t}$ is the stochastic exponential of Y.

Hence: $X_{t} = X_0e^{Y_t}$ and ${Y_t}$ is $Y_{0} e^{t} - \int_{0}^{t} e^{(t-s)} d B_{s}$.

Am I right?

Answer 1

This is similar to a geometric Brownian motion. Try the substitution $Y_t=\ln(X_t)$ (as long as $X_t>0$). Then $$ dY_t=\frac1{X_t}(X_tB_tdt−X_tdB_t)-\frac1{2X_t^2}X_t^2dt=(B_t-\tfrac12)\,dt-dB_t. $$ This integrates to $$ Y_t=Y_0-\frac12t-B_t+\int_0^tB_s\,ds=Y_0-\frac12t+\int_0^t(t-s-1)\,dB_s $$