Expected Value of a Geometric Average Under Brownian Motion
Summary
The document derives the expectation of a geometric average of a stock modeled as geometric Brownian motion. It starts from the lognormal stock process, takes the time average of its log, and separates the deterministic terms from the integral of Brownian motion. That integral is normally distributed with mean zero and variance equal to one third of the time horizon cubed.
Using the moment generating function of a normal variable, the derivation shows that the geometric average is lognormally distributed and gives its expected value as the initial stock price multiplied by an exponential involving one half of the drift term and a volatility adjustment. This is an analytical calculation, not an empirical result. The excerpt does not develop a change-of-measure comparison or clarify the separate expression mentioned in the original question, and it assumes the stated continuous-time model and parameters apply over the averaging interval.
Key ideas
- The logarithm of the geometric average is built from the time integral of the log stock price.
- The integrated Brownian motion has a normal distribution with variance one third of the horizon cubed.
- The geometric average is lognormally distributed under the stated model.
- Its expectation follows by applying the normal exponential moment formula.
- The result depends on the geometric Brownian motion assumptions in the setup.
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# change of measure expectation
# change of measure expectation
How to find expectation of this stochastic process? Also, to show that the expectation of a stochastic process expression [Xt - St] in one measure is equal to expectation of another expression (of the mentioned stochastic process) in another measure?
Given: $S_t=S_0e^{\sigma W_t+(r-\sigma^2/2)t}$
$dS_t = rS_tdt + \sigma S_tdW_t$
## Answer by Kevin (score 1)
https://quant.stackexchange.com/a/46665
Your process for $(S_t)$ is a geoemtric Brownian motion and since $S_t=S_0 e^{\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t}$, we have \begin{align*} \ln(S_t) &= \ln(S_0)+\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t \\ &\sim N\left(\ln(S_0)+\left(r-\frac{1}{2}\sigma^2\right)t,\sigma^2 t\right). \end{align*} Thus, \begin{align*} X_t &= e^{\frac{1}{t}\int_0^t \ln(S_u)\mathrm{d}u} \\ &= e^{\frac{1}{t}\int_0^t \left(\ln(S_0)+\left(r-\frac{1}{2}\sigma^2\right)u\right)\mathrm{d}u}\cdot e^{\frac{1}{t}\sigma\int_0^t W_u\mathrm{d}u} \\ &= S_0\cdot e^{\frac{1}{2}\left(r-\frac{1}{2}\sigma^2\right)t}\cdot e^{\frac{1}{t}\sigma\int_0^t W_u\mathrm{d}u}. \end{align*} Fortunately, the time integral of a Brownian motion is well-known to be normally distributed with mean zero and variance $\frac{1}{3}t^3$, see here. Remember that if $Z\sim N(0,1)$, then $\mathbb{E}\left[e^Z\right]=e^{\frac{1}{2}}$ and thus $\mathbb{E}\left[e^{m+s Z}\right]=e^{m+\frac{1}{2}s^2}$. As a consequence, $X_t$ is log-normally distributed with \begin{align*} \mathbb{E}[X_t] &= S_0\cdot e^{\frac{1}{2}\left(r-\frac{1}{2}\sigma^2\right)t}\cdot \mathbb{E}\left[e^{\frac{1}{t}\sigma\int_0^t W_u\mathrm{d}u}\right] \\ &= S_0\cdot e^{\frac{1}{2}\left(r-\frac{1}{2}\sigma^2\right)t}\cdot \mathbb{E}\left[e^{\frac{1}{t}\sigma \sqrt{\frac{1}{3}t^3}Z} \right] \\ &= S_0\cdot e^{\frac{1}{2}\left(r-\frac{1}{2}\sigma^2\right)t}\cdot \mathbb{E}\left[e^{ \sqrt{\frac{1}{3}t\sigma^2}Z} \right] \\ &= S_0\cdot e^{\frac{1}{2}\left(r-\frac{1}{2}\sigma^2\right)t}\cdot e^{ \frac{1}{2}\frac{1}{3}\sigma^2t} \\ &= S_0\cdot e^{\frac{1}{2}\left(r-\frac{1}{6}\sigma^2\right)t}. \end{align*}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.