Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The following question is from the book [Quantitative Equity Portfolio Management: Modern Techniques and Applications] Page 47. I guess it could be solved using the Law of large numbers, but I'm not sure how to do it.

Given $L$ periods investment return $r_1,\cdots,r_L$, define geometric average as


Suppose $r_i=\mu+\sigma\varepsilon_i$, where $\varepsilon_i$'s are independent standard normal random variables. Prove that as $L\to\infty,\;\;\; \mu_g\approx\mu-\frac{1}{2}\sigma^2$

share|cite|improve this question
up vote 1 down vote accepted

Some quick and dirty calculation leads me to this.

$$\begin{eqnarray} \mu_g & = &\Big(\prod_{i=1}^L(1+r_i)\Big)^{\frac{1}{L}}-1 \\ & = & \exp\Big[\frac{1}{L}\sum_{i=1}^L\log(1+r_i)\Big] - 1 \\ & = & \exp\Big[\frac{1}{L}\sum_{i=1}^L(\mu_i+\sigma \epsilon_i)-\frac{1}{2L}\sum_{i=1}^L(\mu_i+\sigma \epsilon_i)^2+\ldots\Big] - 1 \\ & = & \exp\Big[\log(1+\mu)+\frac{\sigma}{L}\sum_{i=1}^L \epsilon_i-\frac{2\sigma}{2L}\sum_{i=1}^L\epsilon_i-\frac{\sigma^2}{2L}\sum_{i=1}^L \epsilon_i^2+\ldots\Big] - 1 \\ & = & (1+\mu)\exp\Big[\frac{\sigma}{L}\sum_{i=1}^L \epsilon_i-\frac{2\sigma}{2L}\sum_{i=1}^L\epsilon_i-\frac{\sigma^2}{2L}\sum_{i=1}^L \epsilon_i^2+\ldots\Big] - 1 \end{eqnarray}$$

This should give in the limit $L\to \infty$ the following

$$\begin{eqnarray} \mu_g & \approx &(1+\mu)\exp\Big[-\frac{1}{2}\sigma^2\Big]-1 \\ & \approx &(1+\mu)\Big(1-\frac{1}{2}\sigma^2\Big)-1 \\ & \approx &\mu-\frac{1}{2}\sigma^2 \\ \end{eqnarray}$$

because as you mentioned, the $\epsilon_i$ are independent standard normal variables and according to the law of large numbers $\frac{1}{L}\sum_{i=1}^L \epsilon_i \to 0$ and $\frac{1}{L}\sum_{i=1}^L \epsilon_i^2 \to 1$ as $L \to \infty$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.