Delta-Method Sharpe Ratio Variance with Skewness and Kurtosis
Summary
The document derives the asymptotic variance of an estimated Sharpe ratio using the generalized method of moments and the delta method. It defines the mean return and return variance as moment parameters, constructs their covariance matrix from centered returns, and applies derivatives of the Sharpe ratio function. The resulting expression incorporates the third and fourth central moments, so skewness and kurtosis affect estimation uncertainty.
The derivation presents a form involving standardized third and fourth moments and then gives a rearranged expression that recovers a result associated with Lo through Mertens. The answer says this treatment relaxes a normality assumption, under which higher moments would not contribute in the same way. It drops the risk-free rate in its final calculation and presents an asymptotic argument; the excerpt does not detail additional sampling conditions or finite-sample performance. The result is useful for understanding uncertainty in Sharpe estimates, not a direct trading signal.
Key ideas
- The mean and variance of returns can be estimated jointly as moment conditions in a GMM setup.
- Their covariance matrix depends on the second, third, and fourth centered return moments.
- Applying the delta method transfers parameter uncertainty to the estimated Sharpe ratio.
- Skewness and kurtosis contribute to asymptotic Sharpe ratio variance when normality is relaxed.
- The final derivation omits the risk-free rate and does not assess finite-sample accuracy.
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# Propagation of Errors of Sharpe Ratio
# Propagation of Errors of Sharpe Ratio
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?
## Answer by oronimbus (score 4, accepted)
https://quant.stackexchange.com/a/74258
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: $$\boxed{V_{SR}=1+\frac{1}{2}SR^2-SR\cdot\gamma_3 + SR^2\cdot \frac{\gamma_4-3}{4}}$$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.