Expected value of $S_t$ where $dS_t=(\mu S_t+a)dt + (\sigma S_t +b)dW_t$

Question

I have the stochastic differential equation: $$dS_t=(\mu S_t+a)dt + (\sigma S_t +b)dW_t$$ where $W_t$ is a Wiener process with $S_0 > 0$ and $\mu, \sigma, a, b \in \mathbb{R}$.

I have found the solution of this equation: $$S_t= S_0\beta^{-1}_t+(a-\sigma b)\int_0^t \frac{ \beta_s} {\beta_t}ds + b\int_0^t \frac{ \beta_s} {\beta_t}dW_s$$ using an integrating factor of $\beta_t = \exp(-(\mu-\frac{1}{2}\sigma^2)t - \sigma W_t)$ and proved that it is a unique solution of this SDE.

I want to now find the expected value of this solution, so $\mathbb{E}(S_t) $ for all $t\geq 0$, with $- \frac{1}{2} \sigma < \mu < 0$. Could someone just point me in the right direction?

Answer 1

Following @KurtG's comment, you may observe that $\beta_t M_t = \int_0^t \beta_s dW_s$, being the Itô integral of an $L^2(\Omega \times [0,t])$ process, is a martingale, so that $E(\beta_t M_t) = 0$.

Now observe that $$E(\beta_s/\beta_t) = \exp \left( (\mu - 0.5 \sigma^2 ) (t-s)\right) E \exp \left( \sigma (W_t - W_s) \right) = \exp (\mu (t-s)) $$

Where we have used $E \exp \left( \sigma (W_t - W_s) \right) = \exp \left( \frac{1}{2} \sigma^2 (t-s) \right)$, as $\sigma (W_t - W_s) \sim N(0, \sigma^2(t-s))$. Now, use Tonelli's theorem to get: $$E(S_t) = S_0 E(\beta_t^{-1})+(a-\sigma b) \int_0^t E(\beta_s / \beta_t) ds$$ Can you take it from here?

Answer 2

Writing the SDE in integral form we have $$ S_t=S_0+\int_0^t \mu\, S_u+a\,du+\int_0^t\sigma\,S_u+b\,dW_u\,. $$ Therefore, $$ \mathbb E[S_t]=S_0+\int_0^t\mu\,\mathbb E[S_u]+a\,du\,. $$ This is an ODE for $\mathbb E[S_t]$ which is identical to the SDE for $S_t$ when $\sigma=0$ and $b=0$. This ODE therefore has the solution $$ \mathbb E[S_t]=S_0\,e^{\mu t}+a\frac{e^{\mu t}-1}{\mu}. $$