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

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?

• If you use the fact that the expectation of a well behaved Ito integral is zero you can easily derive an ODE for $\mathbb E[S_t]$ ( which will be a special case of the SDE that you have already solved). Aug 3, 2021 at 17:29

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?
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 $$\mathbb E[S_t]=S_0\,e^{\mu t}+a\frac{e^{\mu t}-1}{\mu}.$$
• +1 Very nice! As a technical aside, do you know of an easy way to argue that $\int_0^t \sigma S_t + b dW_t$ is a true martingale? Aug 4, 2021 at 15:29
• It is enough to know when the local martingale $\int_0^tS_u\,dW_u$ is a martingale. The answer to this question shows that this is the case if and only if $\mathbb E[\int_0^tS_u^2\,du]$ for all $t$. I have not checked if this holds but we know the explicit form of $S_u$ in OP's question. Let me know how it goes. Aug 4, 2021 at 20:18