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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

        $\displaystyle\mu_g=\Big(\prod_{i=1}^L(1+r_i)\Big)^{\frac{1}{L}}-1$

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

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1 Answer 1

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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 $L \to \infty$.

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