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Deriving the Mean of a Geometric Brownian Motion Price

Article Quant Q&A · Author: JohnOD25
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Summary

The document derives the expected asset price under geometric Brownian motion. Starting from the exponential solution of the price process, it factors out the deterministic terms and evaluates the expectation of the exponential of a normally distributed Brownian value. Completing the square in the normal density shows that this expectation contributes an exponential volatility adjustment, which cancels the variance correction in the solution’s exponent. The resulting mean is the initial price multiplied by exponential growth at the drift rate.

This provides a direct density-based derivation and connects the calculation to the normal distribution’s moment generating function. It assumes constant drift and volatility in the stated geometric Brownian motion model and does not address estimation of those parameters or whether the model describes observed prices well. The result is the model’s expected price under its probability measure; it should not by itself be interpreted as a forecast or a claim about realized returns.

Key ideas

  • Brownian motion at a fixed time has a normal distribution with variance equal to elapsed time.
  • The exponential moment of a normal variable can be evaluated by completing the square in its density.
  • The volatility adjustment in the geometric Brownian motion solution cancels in the expectation calculation.
  • The expected price grows exponentially at the model’s drift rate.

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# Moments of discrete Asset Price Model


# Moments of discrete Asset Price Model












Say if B is standard Brownian motion then:

$S(t) = S0e^{((𝜇- σ^2)/2)t+σB(t)}$

The mean of this SDE would be

$𝐄[𝑆(𝑡)]=𝑆_0𝑒^{𝜇𝑡}$

I know to do this you use the density function and integrate by parts twice but I keep making a mistake in the integration. Can someone show me how it should look? So I can fix my mistake

## Answer by ab94 (score 2, accepted)

https://quant.stackexchange.com/a/48963

If $S$ is the solution to geometric brownian motion SDE: \begin{equation} dS=\mu S dt + \sigma S dW(t) \end{equation} then \begin{equation} S=S_0e^{(\mu - \sigma^2/2)t + \sigma W(t)} \end{equation} Then if you take expectation \begin{equation} \mathbb{E}[S(t)]=S_0e^{(\mu - \sigma^2/2)t}\mathbb{E}[e^{\sigma W(t)}] \end{equation} Now since $W$ is a wiener process: \begin{equation} W(t) \sim \mathcal{N}(0,t) \end{equation} the expected value is the moment generating function of the normal distribution. This means \begin{equation} \mathbb{E}[e^{\sigma W(t)}]=\int_{-\infty}^{+\infty}e^{\sigma x}\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}dx=\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi t}}e^{-\frac{1}{2t}(x^2 - 2t\sigma x)}dx=\int_{-\infty}^{+\infty}e^{\frac12\sigma^2 t}\frac{1}{\sqrt{2\pi t}}e^{-\frac{1}{2t}(x^2 - 2t\sigma x + \sigma^2t^2)}dx=\int_{-\infty}^{+\infty}e^{\frac12\sigma^2 t}\frac{1}{\sqrt{2\pi t}}e^{-\frac{1}{2t}(x-\sigma t)^2}dx=e^{\frac12\sigma^2 t}\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac12y^2}dy=e^{\frac12\sigma^2 t} \end{equation} if you substitute \begin{equation} \mathbb{E}[S(t)]=S_0e^{\mu t} \end{equation}

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