Problem: find the explicit solution of the following SDE
$$dY_t=r dt+\alpha Y_t dB_t$$
with initial condition $Y_0=y_0$ and $B_t$ is a Brownian motion.
Hint: consider $Z_t=\exp(-\alpha B_t+\frac{1}{2}\alpha^2 t)Y_t$
I've tried to apply Ito's formula to the process given in hint:
$$dZ_t=f(t,Y_t)=\frac{1}{2}\alpha^2Z_tdt+\exp(-\alpha B_t+\frac{1}{2}\alpha^2t)dY_t$$
$$=\frac{1}{2}\alpha^2\exp(-\alpha B_t+\frac{1}{2}\alpha^2 t)Y_tdt+r\exp(-\alpha B_t+\frac{1}{2}\alpha^2t)dt+\exp(-\alpha B_t+\frac{1}{2}\alpha^2t)\alpha Y_t dB_t$$
I get stuck on that it doesn't seem we can get a equation without $Y_t$. Usually, we would have an Ito process which we can transform to the Integral form and then we can multiply the inverse of the hint to get the solution $Y_t$.
And the process in the hint is the inverse of the stochastic exponential with argument $\alpha$, I don't know if it can be used to solve the problem.
Does anyone have some hints or indicate what I've done mistakenly so far?
Best Answer
Consider $X_t=\alpha X_tdW_t$, $X_0=1$. Then, $X_t=e^{-(\alpha^2/2)t+\alpha W_t}$. The hint then suggests to consider $Z_t=X_t^{-1}Y_t$. Note that by Ito $$d(X_t^{-1})=-X_t^{-2}dX_t+\alpha^2X_t^{-3}X_t^2dt=\alpha^2(X_t^{-1})dt-\alpha (X_t^{-1})dW_t$$ So $$dZ_t=(X_t^{-1})dY_t+Y_td(X_t^{-1})-\alpha^2Z_tdt=r (X_t^{-1})dt$$ Therefore $Y_t=Z_tX_t=y_0X_t+r \int_0^tX_tX_s^{-1}ds$.