# Expected value of time integral of geometric brownian motion

Given that the stochastic process follows,

$$\frac{dS_t}{S_t} = \mu dt + \sigma dW_t$$

How do i calculate the expected value of,

$$\int_0^T S_te^{r(T-t)} dt$$

in terms of T.

What I tried was the below,

\begin{align} \int_0^T S_te^{r(T-t)} dt &= \left[ -\frac{1}{r}S_te^{r(T-t)}\right]_0^T + \frac{1}{r}\int_0^T e^{r(T-t)} dS_t\\ &=\frac{1}{r}\left[ S_Te^{rT}-S_T \right] + \frac{\mu}{r} \int_0^T S_te^{r(T-t)} dt + \frac{\sigma}{r} \int_0^T S_te^{r(T-t)} dW_t\\ \Rightarrow \left( 1-\frac{\mu}{r}\right) \int_0^T S_te^{r(T-t)} dt &= \frac{1}{r}\left[ S_Te^{rT}-S_T \right] + \frac{\sigma}{r} \int_0^T S_te^{r(T-t)} dW_t\\ \int_0^T S_te^{r(T-t)} dt &= \frac{1}{r-\mu}\left[ S_Te^{rT}-S_T \right] + \frac{\sigma}{r-\mu} \int_0^T S_te^{r(T-t)} dW_t\\ \Rightarrow \mathbb{E}\left[\int_0^T S_te^{r(T-t)} dt\right] &= \frac{1}{r-\mu}\left[ e^{rT}-1 \right]\mathbb{E}[S_T] \end{align}

Can anyone verify if what I did was correct?

• When you take the expected value the nonconstant random variables should disappear. $S_t$ and $S_T$ are nonconstant random variables. – Matt Samuel Jul 5 '15 at 20:49
• Also, $t$ was the dummy variable of integration. It should have gone away too. – Matt Samuel Jul 5 '15 at 20:51
• Oh sorry, i left out the bit that i'm supposed to calculate the expectation in terms of T. and I've fixed the typos related to $S_t$, they should be $S_T$ – Danny Jul 5 '15 at 21:08
• In terms of $T$, yes. So you need to find $E[S_T]$ in terms of $T$. It is not equal to $S_T$. – Matt Samuel Jul 5 '15 at 21:10
• nullUser's answer below is simpler, but you can make yours correct by gleaning from it that $E[S_T]=S_0e^{\mu T}$. – Matt Samuel Jul 5 '15 at 21:19

Assuming $T$ is just a number and not a stopping time, $$E\int_0^T S_t e^{r(T-t)}dt$$ is just just a double integral of a positive function, so we can exchange the order of integration $$= \int_0^T E [S_t] e^{r(T-t)} dt$$ $$= \int_0^T S_0 e^{\mu t} e^{r(T-t)}dt$$ and I presume you can finish from here as nothing left is random.