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?

  • $\begingroup$ I don't know the answer to your question but you are better off writing $ \sqrt{t} \eta \sim N(\frac{\mu}{\sigma},1)$ for the sample sharpe ratios. If you want to keep the $t$ in the standard deviation then it should be $\frac{1}{\sqrt{t}}$ but that still goes to zero as $n \rightarrow \infty$ so its not a good idea to do that. $\endgroup$
    – mark leeds
    Commented Jan 28, 2023 at 11:56
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    $\begingroup$ should be $t \rightarrow \infty$ in above. $\endgroup$
    – mark leeds
    Commented Jan 29, 2023 at 4:40
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    $\begingroup$ Don't think there's a simple expression for it. Perhaps this question on the covariance of two ratios of RV's might help. $\endgroup$
    – oronimbus
    Commented Jan 30, 2023 at 8:29
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    $\begingroup$ An approximate normal form for the vector of Sharpe ratios of correlated assets is given in section 4.2 of my Short Sharpe Course. See also eqn. (4.29) for the expansion when returns are elliptically distributed. $\endgroup$
    – shabbychef
    Commented Feb 15, 2023 at 19:54

1 Answer 1


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

$$\rho(\zeta_a,\zeta_b) = \frac{Cov(\zeta_a, \zeta_b)}{\sqrt{\mathbb{V}(\zeta_a)}\sqrt{\mathbb{V}(\zeta_b)}} =\color{red}{\rho }$$


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