Wald Test for Sharpe Ratio Equality Under Serial Correlation

Worked Solution

How to Think About It: Comparing Sharpe ratios sounds simple -- just take the difference and check if it is zero. The catch is that each Sharpe ratio is a nonlinear function of two estimated quantities (mean and standard deviation), and the returns themselves may be autocorrelated. So even under the null, the sampling distribution of $\hat{S}^A - \hat{S}^B$ has a variance that depends on serial correlations and cross-correlations of both return series. The delta method converts the problem from "nonlinear function of correlated estimators" into a linear approximation whose variance you can estimate with HAC methods.

Key Insight: The Sharpe ratio $S = \mu / \sigma$ is a function of two moments: $\mu = E[r_t]$ and $\sigma^2 = \text{Var}(r_t)$. Stack all the moments you need, apply the delta method to the function $g(\mu_A, \sigma_A^2, \mu_B, \sigma_B^2) = \mu_A/\sigma_A - \mu_B/\sigma_B$, and then estimate the covariance matrix of the moment vector using a HAC estimator.

Formal Derivation:

*Step 1: Define the moment vector.* Let $\theta = (\mu_A, \sigma_A^2, \mu_B, \sigma_B^2)'$ be the population moments. The sample analogue is $\hat{\theta} = (\bar{r}_A, \hat{\sigma}_A^2, \bar{r}_B, \hat{\sigma}_B^2)'$. By the CLT for stationary processes:

$\sqrt{T}(\hat{\theta} - \theta) \xrightarrow{d} N(0, \Omega)$

where $\Omega$ is the long-run covariance matrix of the moment conditions, incorporating all autocovariances.

*Step 2: Apply the delta method.* Define $g(\theta) = \mu_A / \sigma_A - \mu_B / \sigma_B$ where $\sigma_k = \sqrt{\sigma_k^2}$. Compute the gradient:

$\nabla g = \left(\frac{1}{\sigma_A},\; -\frac{\mu_A}{2\sigma_A^3},\; -\frac{1}{\sigma_B},\; \frac{\mu_B}{2\sigma_B^3}\right)'$

Then under $H_0$:

$\sqrt{T}(\hat{S}^A - \hat{S}^B) \xrightarrow{d} N(0, \nabla g' \, \Omega \, \nabla g)$

*Step 3: The Wald test statistic.* Let $V = \nabla g' \, \hat{\Omega} \, \nabla g$ where $\hat{\Omega}$ is the HAC estimate of $\Omega$. The Wald statistic is:

$W = \frac{T(\hat{S}^A - \hat{S}^B)^2}{V} \xrightarrow{d} \chi^2_1 \quad \text{under } H_0$

Equivalently, $\sqrt{W}$ is asymptotically standard normal, so you can use a two-sided $z$-test.

*Step 4: Explicit variance formula.* To write out $V = \nabla g' \, \Omega \, \nabla g$ explicitly, define the influence function for each observation. Let $\psi_t = (\psi_t^A, \psi_t^B)'$ where:

$\psi_t^A = \frac{r_t^A - \mu_A}{\sigma_A} - \frac{S^A}{2} \cdot \frac{(r_t^A - \mu_A)^2 - \sigma_A^2}{\sigma_A^2}$

and similarly for $\psi_t^B$. Then:

$\text{Var}(\hat{S}^A - \hat{S}^B) = \frac{1}{T} \sum_{l=-L}^{L} k\left(\frac{l}{L}\right) \hat{\Gamma}(l)$

where $\hat{\Gamma}(l) = \frac{1}{T}\sum_{t=|l|+1}^{T} (\hat{\psi}_t^A - \hat{\psi}_t^B)(\hat{\psi}_{t-|l|}^A - \hat{\psi}_{t-|l|}^B)$ is the sample autocovariance of the difference $\hat{\psi}_t^A - \hat{\psi}_t^B$ at lag $l$, and $k(\cdot)$ is the Newey-West (Bartlett) kernel $k(x) = 1 - |x|$ for $|x| \leq 1$.

Expanding $\hat{\Gamma}(l)$ in terms of the original return series, the cross-terms involve:

• $\text{Cov}(r_t^A, r_{t-l}^A)$ and $\text{Cov}(r_t^B, r_{t-l}^B)$ -- the auto-covariances of each series
• $\text{Cov}(r_t^A, r_{t-l}^B)$ and $\text{Cov}(r_t^B, r_{t-l}^A)$ -- the cross-covariances between series
• Higher-order terms involving $(r_t - \mu)^2$ cross-products, which capture how the variance estimates co-move across time

The lag truncation $L$ should grow with $T$ (typically $L \sim T^{1/3}$) to ensure consistency.

*Step 5: Overlapping returns and kernel choice.* When returns are computed over overlapping windows of length $q$ (e.g., weekly returns from daily data using a rolling 5-day sum), the return series acquires an MA($q-1$) structure even if the underlying daily returns are i.i.d. This has two consequences:

1. Bandwidth must increase. The induced autocorrelation extends to at least lag $q - 1$, so the bandwidth $L$ must satisfy $L \geq q - 1$. Data-driven bandwidth selectors (Andrews 1991, Newey-West 1994) will automatically pick a larger $L$, but if using a fixed rule, you must account for the overlap.

1. Kernel choice matters more. The Bartlett kernel guarantees a positive semi-definite $\hat{\Omega}$ but has slower convergence. With heavily overlapping returns (large $q$), the Quadratic Spectral or Parzen kernels offer better bias-variance tradeoffs because they downweight distant lags more smoothly. In practice, the Bartlett kernel with $L = q - 1$ (or a small multiple thereof) is a reasonable default for moderate overlap.

1. Variance scaling. Under overlap, the effective sample size is closer to $T/q$ than $T$, so confidence intervals widen roughly by a factor of $\sqrt{q}$. Many practitioners forget this and over-reject.

Answer: The Wald test statistic is $W = T(\hat{S}^A - \hat{S}^B)^2 / V$ where $V = \nabla g' \hat{\Omega} \nabla g$, with $\nabla g = (1/\sigma_A, -S^A/(2\sigma_A^2), -1/\sigma_B, S^B/(2\sigma_B^2))'$ and $\hat{\Omega}$ the Newey-West HAC estimator of the long-run covariance of $(\bar{r}_A, \hat{\sigma}_A^2, \bar{r}_B, \hat{\sigma}_B^2)$. Under $H_0$, $W \to \chi^2_1$. The variance of the Sharpe difference is computed via the influence function approach, summing kernel-weighted autocovariances of $\hat{\psi}_t^A - \hat{\psi}_t^B$ up to lag $L$. Overlapping returns inflate the required bandwidth to at least the overlap length and favor smoother kernels.