Expected Itô Integral Under GBM

Article Quant Q&A · Author: mpirie

Summary

The question examines a common integration error for a stock following geometric Brownian motion. Directly treating the differential equation for proportional price changes like ordinary calculus suggests that integrating dS divided by S gives the log price ratio with drift μ. The document points out the concern that stochastic calculus changes the log-price relationship.

The answer evaluates the expectation of the Itô integral directly by substituting the geometric Brownian motion dynamics: the drift component integrates to μT, and the stochastic integral has zero expectation under the stated setup. This does not mean the integral equals the log price ratio. Applying Itô’s lemma to the logarithm adds a variance correction to its drift, so the expected log return differs from the expected accumulated proportional change. The treatment is concise and gives no discussion of assumptions such as integrability or extensions to time-varying parameters.

Key ideas

For geometric Brownian motion, the Itô integral of proportional price changes has expected value μT under the stated model.
The stochastic integral component has zero expectation in the calculation presented.
The integral of dS divided by S is not interchangeable with the logarithm of the price ratio.
Itô’s lemma adds a variance correction to the drift of log price.
The answer does not detail the assumptions needed for the expectation calculation.

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Statistics Equities

Full text

# Integration and expectation of geometric Brownian motion


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

## Answer by Kevin (score 0, accepted)

https://quant.stackexchange.com/a/47078

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 $e^{\mu t}$.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.

The Expected Itô Integral of Proportional Price Changes Under Geometric Brownian Motion

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