Lognormal Stock Price Moments Under Brownian Motion

Summary

The document asks for the expectation and variance of a stock price modeled as an exponential function of drift and standard Brownian motion, evaluated at twice a given time. Its key lesson is that the exponential moment of a normal random variable must be computed using the normal moment-generating function. For a standard normal variable Z, the expectation of exp(aZ) is exp(a²/2), so the random term contributes a factor determined by the square of its coefficient, not by that coefficient alone.

The proposed derivation mishandles this normal expectation, and the posted answer does not fully correct it: its displayed expression still has an incorrect dependence on the coefficient. Applying the moment-generating function directly gives the stock-price expectation; the second moment can be found by squaring the price expression and applying the same rule, then variance follows as the second moment minus the squared expectation. The discussion provides no completed variance calculation, so readers should derive both moments carefully from the stated model rather than rely on the answer’s final line.

Key ideas

• For a standard normal variable, the expectation of its exponential with coefficient a is exp(a²/2).
• The coefficient of Brownian motion must be squared when applying the normal moment-generating function.
• The second moment follows by squaring the price model and evaluating the resulting normal exponential moment.
• Variance is the second moment minus the squared expectation, but the document does not complete this calculation.

Tags

# Expectation and variance of standard brownian motion


# Expectation and variance of standard brownian motion












Assuming that the price of the stock follows the model

$ S(t) = S(0) exp ( mt − (σ^2/ 2) t + σW(t) ) , $ where W(t) is a standard Brownian motion; σ > 0, S(0) > 0, m are some constants.

What is the expectation and variance of S(2t)?

Expectation:

$E[S(2t)]=E[S(0)exp(2mt-(t\sigma^2)+\sigma W(2t)] = $

$S(0)E[exp(2mt-(t\sigma^2)+\sigma W(2t))] = S(0)exp(2mt-\sigma^2 t)E[exp(\sigma W(2t)]$

using that $W(2t)$ is $N(0,2t)$ I get that $W(2t)=\sqrt{2t} Z$, where $Z$ is $N(0,1)$.

$S(0)exp(2mt-\sigma^2 t)E[exp(\sigma \sqrt{2t} Z)]$ =

$S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})E[e^{Z}] =$

$S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})$

Is this solution correct?

Variance: Assuming that the expectation is correctly solved I could just use that $Var(S(2t))= E[(S(2t))^2] - E[S(2t)]^2$ ?

## Answer by Tomas G. (score 1)

https://quant.stackexchange.com/a/44621

No because $$ E(e^Z)=e^{\frac{1}{2}}\neq1 $$

More generally: $$ N \sim \mathcal N(\mu,\sigma^2)\\ E(e^{Nt})=MGF_{\mathcal N(\mu, \sigma^2)}(t)=e^{\mu t+\frac{1}{2}\sigma^2t^2} $$

The last lines should be: $$ S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})E[e^{Z}] =\\ S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})e^{\frac{1}{2}} $$

For the rest it is correct.

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