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Looking at Opdyke, J.D., Comparing Sharpe Ratios: So Where are the P-Values?, page 22 (Appendix A) an application is given for the Propagation of Errors formula on a ratio of two random variables:

$$\operatorname{Var} (x_1/x_2)=\left[ \frac{E[x_1]}{E[x_2]}\right]^2\left[\frac{Var(x_1)}{E[x_1]^2} + \frac{Var(x_2)}{E[x_2]^2} - 2\frac{Cov(x_1,x_2)}{E[x_1]E[x_2]}\right]$$

The formula itself is clear, what I don't understand is how the two expectations for $Var(x_2)$ and $Cov(x_1,x_2)$ are evaluated when $x_1=\mu$ (=mean return) and $x_2=\sigma$ (=standard deviation of returns):

$$\operatorname{Var} (\mu/\sigma)=\left[ \frac{\mu^2}{\sigma^2}\right]\left[\frac{\sigma^2 \ / \ n}{\mu^2} + \frac{(\mu_4-\sigma^4) \ / \ 4n\sigma^2}{\sigma^2} - 2\frac{\mu_3 \ / \ 2n\sigma}{\mu\sigma}\right]$$

This might be elementary but I don't see how you go from $Var(\sigma)=(\mu_4-\sigma^4)\ / \ 4n\sigma^2$ and $cov(\mu,\sigma)=\mu_3 \ / \ 2n\sigma$.

Are there certain distributional assumption made here?

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  • $\begingroup$ These subscripts on the mean don't make any sense to me as well. $\endgroup$
    – SuavestArt
    Dec 23, 2022 at 14:15
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    $\begingroup$ @SuavestArt I believe $\mu_n$ denotes the $n$th central moment as done here: en.wikipedia.org/wiki/Central_moment#Univariate_moments $\endgroup$
    – Bob Jansen
    Dec 23, 2022 at 15:04
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    $\begingroup$ That’s right, the final equation shows the standard error of the SR being proportional to the skewness and kurtosis of returns, which I think is a pretty nice result. $\endgroup$
    – oronimbus
    Dec 24, 2022 at 7:45
  • $\begingroup$ In addition to Mertens' work, Bao derives a more detailed asymptotic approximation in terms of higher order moments. It is largely based on Central Limit Theorem and Delta Method. $\endgroup$
    – shabbychef
    Feb 15, 2023 at 20:52
  • $\begingroup$ Some derivations of this also given in my Short Sharpe Course in section 4.1, generalizing to the case of e.g. autocorrelation, heteroskedasticity, and so on. $\endgroup$
    – shabbychef
    Feb 15, 2023 at 20:54

1 Answer 1

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I've found an answer in a comment on Lo (2002) by Mertens (2002) that I can understand and have listed some additional steps below. The approach is slightly different to what I've listed in the original question but the results are exactly the same. The argument goes as follows: dropping Lo's assumption of normality introduces skewness and excess kurtosis in the asymptotic variance of the moment estimator. To get to this result, we use GMM.

Let $\theta= \left[\mu \ \ \sigma^2 \right]'=\begin{bmatrix}\mathbb E[R_t] \\ \mathbb E[(R_t-\mu)^2] \end{bmatrix}$ and aim to solve the following (using GMM):

$$ H(\theta) = \begin{bmatrix} R_t-\mu \\ (R_t-\mu)^2-\sigma^2 \\ \end{bmatrix} $$

$$f(H(\theta)) = \frac{1}{T}\sum^T_{t=1} H_t(\theta)\stackrel{!}{=}0$$

This is solved over the entire data set $T$ but I've dropped some notation for ease from the original paper. The variance-covariance matrix of $H$ is then:

$$ S=\mathbb E\left[H(\theta)H(\theta)'\right]= \mathbb E \begin{bmatrix} (R_t-\mu)^2 && (R_t-\mu)\left((R_t-\mu)^2-\sigma^2\right) \\ (R_t-\mu)\left((R_t-\mu)^2-\sigma^2 \right) && \left((R_t-\mu)^2-\sigma^2) \right)^2\\ \end{bmatrix} $$

Multiplying out and evaluating the expectations we get:

$$ S= \begin{bmatrix} \sigma^2 && \mathbb E\left[(R_t-\mu)^3-(R_t-\mu)\sigma^2 \right] \\ \mathbb E\left[(R_t-\mu)^3-(R_t-\mu)\sigma^2 \right] && \mathbb E \left[ (R_t-\mu)^4 - 2\underbrace{(R_t-\mu)^2}_{\sigma^2}\sigma^2 + \sigma^4\right ] \\ \end{bmatrix} $$

Some terms can be simplified in expectation, notably in the top right and bottom left we have $\mathbb E[(R_t-\mu)\sigma^2]=0$. We then get:

$$ S = \begin{bmatrix} \sigma^2 && \mathbb E\left[(R_t-\mu)^3\right] \\ \mathbb E\left[(R_t-\mu)^3\right] && \mathbb E\left[(R_t-\mu)^4\right] -\sigma^4\end{bmatrix}$$

Next, the delta method applied to the Sharpe Ratio can be written as:

$$V_{SR}=\frac{\partial g}{\partial \theta'}V_\theta \frac{\partial g}{\partial \theta'}'$$

The partial derivatives of the Sharpe Ratio function $g(\theta)=g(\mu,\sigma^2)$ are easy, we get: $$\frac{\partial g}{\partial \theta'}=\begin{bmatrix} 1/\sigma \\ -\frac{\mu}{2\sigma^3}\\ \end{bmatrix}$$

Finally, we're back to what Opdyke (2005) is doing in his paper and can address the original question. Setting $V_\theta=S$ and multiplying out $V_{SR}$ we get (I've dropped the risk free rate $R_f$):

$$V_{SR} =\frac{\sigma^2}{\sigma^2}-2\frac{\mu}{2\sigma^3}\frac{1}{\sigma} \mathbb E \left[ (R_t-\mu)^3\right] + \frac{\mu^2}{4\sigma^6} \left( \mathbb E \left[(R_t-\mu)^4\right]-\sigma^4\right)$$

Let $\gamma_3=\frac{\mathbb E \left[(R_t -\mu)^3 \right]}{\sigma^3}$ and $\gamma_4=\frac{\mathbb E \left[(R_t -\mu)^4 \right]}{\sigma^4}$ be the third and fourth standardized, central momements then we get:

$$V_{SR}=1-SR\cdot\gamma_3 + \frac{1}{4}SR^2\cdot (\gamma_4-1)$$

Mertens recovers Lo's result by adding and subtracting $\frac{1}{2}SR^2$ which then, finally, gives us: $$\boxed{V_{SR}=1+\frac{1}{2}SR^2-SR\cdot\gamma_3 + SR^2\cdot \frac{\gamma_4-3}{4}}$$

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  • $\begingroup$ Unless anyone has a response that directly answers my original question on evaluating the two expectations, then I'll accept this as an answer. $\endgroup$
    – oronimbus
    Jan 4, 2023 at 13:51
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    $\begingroup$ I upvoted but please give it 2 days before accepting. During the break, some people that might have an even better answer where away from their computer. $\endgroup$
    – Bob Jansen
    Jan 4, 2023 at 15:37

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