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

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2 Answers 2

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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?

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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.

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  • $\begingroup$ 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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