# Integration and expectation of geometric Brownian motion

Let the stock price S follows the geometric brownian motion: $$dS=\mu Sdt+\sigma Sdz$$ $$\frac{dS}S=\mu dt+\sigma dz$$

where $$dz$$ is a wiener process.

Naively integrating the second equation above over time $$t$$ gives

$$\int^T_0\frac{1}SdS=\int^T_0\mu dt +\int^T_0\sigma dz$$ $$=ln(S_T/S_0)=\mu(T-0)+\sigma (z_T-z_0)$$

but I recall this being incorrect... $$\frac{dS}S$$ doesn't really have a physical meaning and one needs to apply stochastic calculus rules. But how do I easily explain clearly to someone new to stochastic calculus that this is a wrong statement to make?

Can one then conclude that

$$E\left[\int^T_0\frac{1}SdS\right]=\mu T+ 0$$ ? I am very doubtful, but having some trouble explaining why this doesn't make sense.

Given that from Ito's lemma, the differential of log of $$S$$ is shown to be $$dlog(S_t)=(\mu -\sigma^2/2)dt+\sigma dz$$ I think there must be some correction factor in the above integral such that $$E\left[\int^T_0\frac{1}SdS\right] \neq \mu T$$

Since $$dS_t = \mu Sdt+\sigma Sdz$$, you have by the properties of the Ito Integral,
\begin{align*} E\left[ \int_0^T \frac{1}{S} dS \right] &= E\left[ \int_0^T\frac{1}{S} \mu Sdt\right] + E\left[\int_0^T \frac{1}{S} \sigma S dz\right] \\ &= \int_0^T \mu dt + 0 \\ &= \mu T. \end{align*}
Note that this result does make sense. Integrating $$\frac{dS}{S}$$ is like summing up all the returns of $$S_t$$ whose drift is $$e^{\mu t}$$.
• What LHS? The expectation of the integral is $\mu T$ which is pretty closed form? – Kevin Aug 12 '19 at 6:44
• Yes you can. Using ito calculus. For instance, $\int_0^T \sigma dz = \sigma B_T$ where $(B_t)$ is a Brownian motion which you called $z$. Hence, you can make sense of $\frac{dS}{S}$. – Kevin Aug 12 '19 at 10:16