I want to solve the SDE $dX_t=bdt+cX_t dW_t$, $X_0=0$ for $b,c\in\mathbb R$. I start by rewriting this as

$$dX_t=(\mu_1+\mu_2 X_t )dt+(\sigma_1+\sigma_2 X_t )dW_t$$

where $\mu_1=b, \mu_2=0, \sigma_1=0, \sigma_2=c$. And the general solution for linear SDE is known to be$X_t=Y_t Z_t$ where

$$dY_t=\mu_2 Y_t dt+\sigma_2 Y_t dW_t, \quad Y_0=1$$ $$dZ_t = \frac{\mu_1-\sigma_1 \sigma_2}{Y_t } dt + \frac{\sigma_1}{Y_t} dW_t, \quad Z_0=X_0=0$$

So $$dY_t=cY_t dW_t \quad \Rightarrow \quad Y_t=\exp\left(-\frac{c^2}{2}t+cW_t\right)$$ and

$$dZ_t=\frac{b}{Y_t}dt = b \exp\left(\frac{c^2}{2}t - cW_t\right)dt$$

But now I am stuck as I am not sure how to find $Z_t$.


Actually, there is nothing left to do. From

$$dZ_t = b \exp \left( \frac{c^2}{2} t - c W_t \right) \, dt$$

it follows that

$$Z_t = \underbrace{Z_0}_{0} + b \int_0^t \exp \left( \frac{c^2}{2} s - c W_s \right) \, ds.$$


$$X_t = Y_t \cdot Z_t = b \exp \left(- \frac{c^2}{2} t+c W_t \right) \cdot \int_0^t \exp \left( \frac{c^2}{2} s - c W_s \right) \, ds.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.