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 $$ E\left[\int^T_0\frac{1}SdS\right] \neq \mu T $$


1 Answer 1


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 $e^{\mu t}$.

  • $\begingroup$ Thank you, that makes sense. Just to clarify my understanding,you can't actually evaluate the rv in the LHS expectation in any closed form way can you? $\endgroup$
    – mpirie
    Aug 12, 2019 at 6:40
  • $\begingroup$ What LHS? The expectation of the integral is $\mu T$ which is pretty closed form? $\endgroup$
    – Kevin
    Aug 12, 2019 at 6:44
  • $\begingroup$ Sorry, I meant the random variable inside it (ignoring the expectations), sort of clearing up the first part of my question $\endgroup$
    – mpirie
    Aug 12, 2019 at 6:45
  • $\begingroup$ 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 $\frac{dS}{S}$. $\endgroup$
    – Kevin
    Aug 12, 2019 at 10:16

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