Combining Uncorrelated Strategies with Sharpe Ratios of One
Summary
The discussion asks how to combine two uncorrelated strategies, each with a Sharpe ratio of one. For an equally weighted portfolio, it derives the Sharpe ratio as the sum of the strategies’ volatilities divided by the square root of the sum of their variances. This shows that the result depends on the relative volatilities; equal individual Sharpe ratios alone do not determine one fixed equally weighted portfolio Sharpe ratio.
A second answer derives the maximum attainable Sharpe ratio using expected excess returns and the covariance matrix. With uncorrelated strategies and unrestricted optimal weights, the combined Sharpe ratio is the square root of the sum of squared individual Sharpe ratios, giving approximately 1.414 in this example. The derivation assumes a valid covariance matrix and permits the optimization used; practical constraints, such as weight limits or estimation error, can change the attainable result.
Key ideas
- An equally weighted combination’s Sharpe ratio depends on the strategies’ relative volatilities.
- Uncorrelated returns make portfolio variance equal to the sum of weighted variances.
- With optimal unrestricted weights, the maximum Sharpe ratio is the square root of the sum of squared individual Sharpe ratios.
- For two uncorrelated strategies each with Sharpe ratio one, the optimal combined Sharpe ratio is approximately 1.414.
- Weight constraints and estimated inputs can make practical portfolio results differ from the unconstrained derivation.
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Full text
# What is the Sharpe ratio of two uncorrelated strategies, each with Sharpe ratio equal 1?
# What is the Sharpe ratio of two uncorrelated strategies, each with Sharpe ratio equal 1?
Given two uncorrelated strategies, each with a Sharpe ratio of 1, what is the of Sharpe ratio of the ensemble?
## Answer by Quantifeye (score 11, accepted)
https://quant.stackexchange.com/a/30299
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?
## Answer by Johannes Gerer (score 7)
https://quant.stackexchange.com/a/32716
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.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.