Correlation Between Estimated Sharpe Ratios and Return Series
Summary
The document asks whether estimated Sharpe ratios inherit the correlation between two underlying return series. Under a simplifying setup, paired observations are independent across time, the two assets have known volatilities, and the risk-free rate is zero. Each estimated Sharpe ratio is therefore a scaled sample mean. The response calculates their covariance by summing the covariance contributions from matching observations; cross-time terms vanish under the independence assumption. Dividing by the estimated Sharpe ratios’ standard deviations gives the result that their correlation equals the return correlation.
The derivation clarifies that this is an asymptotic approximation for sample Sharpe behavior and depends on the stated assumptions. In particular, it treats volatility as known rather than estimated and does not address serial dependence, time-varying volatility, non-normal returns, or adjustments to Sharpe ratios for finite samples. The result is useful for understanding how shared return movements affect uncertainty in performance estimates, but it should not be taken as a general formula for every empirical Sharpe estimator.
Key ideas
- With known volatilities and zero risk-free rate, each estimated Sharpe ratio is a scaled sample mean.
- Independence across time removes covariance contributions from observations at different dates.
- The covariance of the two sample means comes from paired returns observed at the same time.
- Under these assumptions, the estimated Sharpe ratios have the same correlation as the returns.
- Estimated volatility and serial dependence require a different or extended analysis.
Tags
Full text
# If returns are correlated, are Sharpe ratios correlated?
# If returns are correlated, are Sharpe ratios correlated?
Suppose we have two correlated return series: $$a \sim N(\mu_a,\sigma_a^2)$$ $$b \sim N(\mu_b,\sigma_b^2)$$ $$correl(a,b)=\rho$$
The sample Sharpe ratios of the two series, after $t$ samples for $t \to \infty$, are approximately distributed as: $$\zeta_a \sim N(\frac {\mu_a} {\sigma_a}, \frac 1 t)$$ $$\zeta_b \sim N(\frac {\mu_b} {\sigma_b}, \frac 1 t)$$
But are the Sharpe ratios correlated? $$correl(\zeta_a,\zeta_b)=?$$
Empirically, I found they are equally correlated: $$correl(\zeta_a,\zeta_b)≈correl(a,b)$$
But what is the math behind?
## Answer by NN2 (score 4, accepted)
https://quant.stackexchange.com/a/74513
> Remark 1: From the information in your question, I think you assumed that the risk free rate $r_f$ is equal to $0$ and the Sharp ratio is $$\frac{\mathbb{E}(a)-r_f}{\sqrt{\mathbb{V}(a)}} = \frac{\mathbb{E}(a)}{\sqrt{\mathbb{V}(a)}} $$ where $\mathbb{E}(a)$ and $\mathbb{V}(a)$ are the expected value and variance of the return $a$.
Now, return to the question, for simplifying the problem, we assume that the variances of the two series are known (and equal to $\sigma_a^2$ and $\sigma_b^2$). Then $$\sqrt{\mathbb{V}(a)}=\sigma_a$$ $$\sqrt{\mathbb{V}(b)}=\sigma_b$$
From $t$ return samples $(a_i,b_i)_{i=1,..,t}$ of $(a,b)$, we can estimate the expected returns as $$\mathbb{E}(a) = \frac{1}{t}\sum_{i=1}^t a_i$$ $$\mathbb{E}(b) = \frac{1}{t}\sum_{i=1}^t b_i$$
> Remark 2: we note that these $(a_i,b_i)$ and $(a_j,b_j)$ are independent if $i \ne j$ and for any $i$, there is a correlation $\rho$ between $a_i$ and $b_i$.
and so, their Sharpe ratio can be estimated as follows $$\zeta_a=\frac{\frac{1}{t}\sum_{i=1}^t a_i}{\sigma_a}$$ $$\zeta_b=\frac{\frac{1}{t}\sum_{i=1}^t b_i}{\sigma_b}$$
> Remarque 3: according the central limit theorem, $$\sqrt{t}\cdot \zeta_a \xrightarrow{t\to+\infty} \mathcal{N}\left(\frac{\mu_a}{\sigma_a},1 \right)$$ Hence, your formula in the question should be some kind like this one $$\zeta_a \xrightarrow{t\to+\infty} \mathcal{N}\left(\frac{1}{\sqrt{t}}\frac{\mu_a}{\sigma_a},\frac{1}{t} \right)$$
Now, we compute the correlation between $\zeta_a$ and $\zeta_b$. It suffices to compute their covariance (their variance is known and equal to $\frac{1}{t}$ from the remark 3).
$$ \begin{align} Cov(\zeta_a, \zeta_b) &= \frac{1}{t^2 \sigma_a \sigma_b} \sum_{1 \leq i,j \leq t}Cov(a_i,b_j) \\ &= \frac{1}{t^2 \sigma_a \sigma_b} \left( \sum_{1 \leq i \leq t}Cov(a_i,b_i) +\underbrace{\sum_{1 \leq i \ne j \leq t}Cov(a_i,b_j)}_{=0 \text{ because of the independence according to remark } 2} \right) \\ &=\frac{1}{t^2 \sigma_a \sigma_b} \cdot t \cdot Cov(a,b)\\ &=\frac{1}{t^2 \sigma_a \sigma_b} \cdot t \cdot \rho \sigma_a \sigma_b\\ &=\frac{\rho}{t} \end{align} $$
Finally, the correlation between the two Sharpe ratios is
$$\rho(\zeta_a,\zeta_b) = \frac{Cov(\zeta_a, \zeta_b)}{\sqrt{\mathbb{V}(\zeta_a)}\sqrt{\mathbb{V}(\zeta_b)}} =\color{red}{\rho }$$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.