Deriving the Second Moment of a Geometric Brownian Motion
Summary
The note derives the expected squared value of a geometric Brownian motion (GBM), starting from its stochastic differential equation. Applying Itô’s lemma to the square of the process shows that the squared process is itself a GBM, with adjusted drift and volatility. The same expression follows by squaring the explicit GBM solution.
Taking the expectation yields a second moment that depends on the initial value, drift, volatility, and time. The derivation illustrates how Itô’s correction affects powers of a stochastic process and notes that the approach extends to other positive powers. It gives an analytical result rather than empirical evidence; the setup assumes constant GBM parameters and does not address model fit or departures from GBM dynamics.
Key ideas
- Applying Itô’s lemma to the square of a GBM produces another GBM with adjusted parameters.
- The explicit GBM solution provides an alternative route to the same squared-process expression.
- The second moment grows at a rate determined by both drift and volatility.
- The power transformation method can be generalized to other positive exponents.
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# What the expectation of S^2 is from GBM?
# What the expectation of S^2 is from GBM?
I was at an interview and was asked to write down the SDE for GBM.
$$ dS = S\mu dt + S\sigma dX $$
Then I was asked how I would compute the expectation of S^2. I didn't know where to start. Any ideas?
$$ {\mathbb{E}} [S_t^2] $$
## Answer by Kevin (score 8)
https://quant.stackexchange.com/a/49241
As Sanjay said, you can apply Itô's Lemma to $f(t,x)=x^2$ and obtain \begin{align*} \mathrm{d} S^2_t=\left(2\mu S_t^2+\sigma^2S_t^2\right)\mathrm{d}t+\left(2\sigma S_t^2\right)\mathrm{d}W_t. \end{align*} Thus, $(S_t^2)$ is again a geometric Brownian motion and hence, for each time point $t$ log-normally distributed with drift $2\mu+\sigma^2$ and volatility $2\sigma$. Then, \begin{align*} S_t^2 &= S_0^2\cdot\exp\left(\left(2\mu+\sigma^2-\frac{1}{2}\cdot 4\sigma^2\right)t+2\sigma W_t\right) \\ &= S_0^2\cdot\exp\left(\left(2\mu-\sigma^2\right)t+2\sigma W_t\right). \end{align*}
Alternatively, you can use that you know $(S_t)$ explicitly and directly obtain \begin{align*} S_t &= S_0\cdot\exp\left(\left(\mu-\frac{1}{2}\sigma^2\right)t+\sigma W_t\right)\\ \implies S_t^2 &= S_0^2 \cdot\exp\left(\left(2\mu-\sigma^2\right)t+2\sigma W_t\right). \end{align*}
In either case, you get that \begin{align*} \mathbb{E}[S_t^2]=S_0^2e^{(2\mu+\sigma^2)t}. \end{align*}
You can of course easily generalise this to the process $(S_t^\gamma)$ for some $\gamma>0$.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.