I read a paper from Ledoit and Wolf that proposes a method to compare two Sharpe ratios and a paper from White that proposes a method to compare $n$ trading rules.
My question is: Can we use White's method to compare two Sharpe ratios? I prefer this method because it's computationally simpler.
As James has pointed out in the comments, White's Reality Check is specifically designed to control the family-wise error rate given $k > 2$ statistics. The theory does not depend on $k$ asymptotics, so there is nothing invalid about using White's Reality Check for $2$ statistics, but in practice there would be little point to doing this. Further, as stevo`america points out above, the Reality check had a patent on it until two years ago - whether it would be enforced in a court case though is another question entirely...
In particular, for $k=2$, it is fairly straightforward to construct a simple statistical test for the difference in two Sharpe ratios. Presumably there is some insight in Ledoit and Wolf's paper that makes their statistic superior to what I am about to suggest. Also, see stevo`americas comments on the question for references to some other sophisticated testing measures. But if what you're after is simplicity, then the following is still perfectly valid:
Let $R_{1,t}$, and $R_{2,t}$ denote returns on the two assets of interest. In this framework, I define the Sharpe ratio: \begin{equation} S_1 = \frac{\mathbb{E} R_{1,t}}{\sqrt{\mathbb{V} R_{1,t}}} \end{equation} For any random variable $X_t$ the sample mean is defined: \begin{equation} \bar{X} = \frac{1}{T} \sum_{t=1}^T X_t \end{equation} Let: \begin{equation} \bar{\sigma}_1 = \sqrt{\frac{1}{T} \sum_{t=1}^T (R_{1,t} - \bar{R}_1)^2} \end{equation} A natural estimator for $S_1$ is: \begin{equation} \hat{S}_1 = \frac{\bar{R}_1}{\bar{\sigma}_1} \end{equation} I assume suitable regularity conditions on $R_{1,t}$ such that $\bar{R}_1 \overset{\mathbb{P}}{\rightarrow} \mathbb{E} R_{1,t}$, $\sqrt{T} \bar{R}_1 \overset{d}{\rightarrow} \mathcal{N}$, and $\bar{\sigma}_1 \overset{\mathbb{P}}{\rightarrow} \sqrt{\mathbb{V}(R_{1,t})}$ (e.g. weak dependence and suitably bounded moments). By Slutsky's theorem, these conditions are sufficient for: \begin{equation} \hat{S}_1 \overset{\mathbb{P}}{\rightarrow} S_1 \end{equation} and note that by Cramer's theorem: \begin{equation} \sqrt{T} \bar{S}_1 = \frac{\sqrt{T} \bar{R}_1}{\bar{\sigma}_1} \overset{d}{\rightarrow} \mathcal N \end{equation} since the numerator is converging in distribution to a Normal, and the denominator is converging in probability to a constant strictly greater than $0$.
So we have a CLT for our statistic. For the purposes testing a difference in two statistics, it is easier if our statistic can be phrased as a single sample mean. This is straightforward. Let: \begin{equation} Y_{1,t} = (\bar{\sigma}_1)^{-1} R_{1,t} , \end{equation} where it is worth emphasizing that it immediately follows that: \begin{equation} \hat{S}_1 = \bar{Y}_1 . \end{equation} Incorporating the second asset, we now define: \begin{equation} d_t = Y_{1,t} - Y_{2,t} . \end{equation} The theory thus far is sufficient to show that under: \begin{equation} H_0 : S_1 = S_2 , \end{equation} we have: \begin{equation} \bar{d} \overset{d}{\rightarrow} \mathcal{N}(0, \alpha) . \end{equation} So we've literally transformed the problem into testing whether a sample mean is equal to zero, with a CLT existing for the sample mean. If you think $d_t$ exhibits time-series dependence, then you will need to estimate $\alpha$ using a HAC estimator, or else you could just bootstrap the statistic. Both are likely to give you similar outcomes. If you aren't worried about time-series dependence then just estimate $\alpha$ using the sample standard deviation of $d_t$ over $\sqrt{T}$.
