Given two uncorrelated strategies, each with a Sharpe ratio of 1, what is the of Sharpe ratio of the ensemble?


2 Answers 2


If we assume that by ensemble you mean an equally weighted portfolio of the two. We can express that portfolio as $$P = \frac{1}{2}x + \frac{1}{2}y$$ and the sharpe ratio of $P$, $S(P)$, will be $$\frac{\frac{1}{2}\mu_x + \frac{1}{2}\mu_y - r_f}{\sigma_{\frac{1}{2}x + \frac{1}{2}y}}$$ becuase $x$ and $y$ are uncorellated, this reduces to $$\frac{\mu_x + \mu_y - 2r_f}{\sqrt{\sigma_x^2 + \sigma_y^2}}$$ becuase the sharpe ratios $$S(x)=\frac{\mu_x - r_f}{\sigma_x}=S(y)=\frac{\mu_y - r_f}{\sigma_y} = 1$$ we get $$\mu_x - r_f = \sigma_x \\\mu_y - r_f = \sigma_y $$ thus $$\mu_x + \mu_y - 2r_f = \sqrt{\sigma_x^2} + \sqrt{\sigma_y^2} $$ and $$S(P) = \frac{\sqrt{\sigma_x^2} + \sqrt{\sigma_y^2}}{\sqrt{\sigma_x^2 + \sigma_y^2}}$$ What can you say about this ratio? How does it relate to Jensen's inequality? what happens if they are perfectly correlated?


Of course, it depends on the weights of your 'ensemble'. The optimal combination will have the following Sharpe ratio:

$$ S_{opt} = \sqrt{S_1^2+S_2^2} $$

i.e. $S_{opt} = \sqrt{2} \approx 1.414$ in you example

Proof: Let $x$ be the expectation, and $V$ the covariance matrix of a vector of assets. The Sharpe ratio of a portfolio with weights $w$ is defined by $S_w=\frac{x^Tw}{\sqrt{w^TVw}}$.

First, we transform the problem in a simpler one:

It follows that if $w_1$ has an optimal Sharpe ratio $S^*$, which is always positive, then $a \: w_1$ has the same Sharpe ratio for any positive real number $a$. Setting $a=1/x^Tw_1$, shows that there exists a portfolio $w$ with optimal Sharpe ratio and $x^Tw=1$.

Now, we can find $S^*$ by maximizing $S_w$ subject to $x^Tw=1$, i.e. minimize $w^TVw$ subject to $x^Tw=1$. Using one Lagrange multiplyer $\lambda$ gives the following conditions: $$ \nabla_w(w^TVw+\lambda x^Tw)=2 Vw + \lambda x\stackrel{!}{=}0 $$ $$ x^Tw=1$$ The solution is $w=\frac{V^{-1}x}{x^TV^{-1}x}$ and the optimal Sharpe ratio is thus $$ S^*=\sqrt{x^TV^{-1}x}$$

Application to your case: Two uncorrelated assets with volas $\sigma_1$ and $\sigma_2$ i.e. $V^{-1}=\left(\begin{array} c\sigma_1^{-2}& 0\\0&\sigma_2^{-2}\end{array}\right)$, and Sharpe ratios $S_i=x_i/\sigma_i$ gives the above result.

  • $\begingroup$ Can you derive this? I only get square root of 2 when the variances are also equal. $\endgroup$
    – John
    Feb 27, 2017 at 14:56
  • $\begingroup$ Does this help? $\endgroup$ Mar 3, 2017 at 21:35
  • $\begingroup$ It does. I'm not sure I grok it yet, but making the logic clear helps. $\endgroup$
    – John
    Mar 4, 2017 at 2:51
  • $\begingroup$ There is a small typo in your solution for $ w $: the denominator should have $ x^TV^{-1}x $ instead of $w^TV^{-1}w$. Great answer otherwise! $\endgroup$
    – derpy
    Apr 27, 2020 at 16:39
  • $\begingroup$ Corrected, thanks $\endgroup$ Apr 30, 2020 at 21:36

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