Maximum Sharpe Weights for Uncorrelated Assets

# Optimise the Sharpe ratio of a portfolio of uncorrelated assets


# Optimise the Sharpe ratio of a portfolio of uncorrelated assets












Given a portfolio of $n$ assets, mean returns vector $\mu$, covariance matrix $K$, one can calculate the portfolio weights $w^*$ that maximise the portfolio Sharpe ratio, by solving:

$$w^*=\text{argmax} \left[\frac {w^T \mu} {\sqrt {w^T K w}} \right]$$

Computing $w^*$ requires building the $K$ matrix and solving a system of quadratic equations, so it becomes computationally expensive when $n$ grows large.

However, if for some reason, we know that the asset returns are uncorrelated, the problem simplifies as: $$ K=\begin{bmatrix} \sigma_1^2 & .. & 0 \\ .. & .. & .. \\ 0 & .. & \sigma_n^2 \\ \end{bmatrix} $$

In this case, I suspect we don't need to bother with the quadratic equations, there should be straightforward formula to compute $w^*$ by just plugging in the individual $\mu_i$ and $\sigma_i$ of each asset:

$$w_i^* = f(\mu_1,\sigma_1,\dots,\mu_n,\sigma_n)$$

I am unable to derive the formula analytically. HELP!

## Answer by Michael Isichenko (score 4)

https://quant.stackexchange.com/a/67876

Formally (and I mean it -- see below), the optimal weights are $w=K^{-1}\mu$. The portfolio pnl then has the mean $Q=\sum_iw_i\mu_i$ and the variance $V=\sum_{ij}w_iw_jK_{ij}$. The Sharpe is $S=Q/\sqrt{V}$. In the uncorrelated case the answer is given by the Pythagorean formula $$ S=\sqrt{\sum_i\frac{\mu_i^2}{\sigma_i^2}}. $$

This answer can be very wrong in any practical context, however. This and related questions are covered in my recent book. A few highlights:

- Two uncorrelated assets with Sharpe ratios $S_1$ and $S_2$. The optimally weighted two-asset book has the Sharpe $\sqrt{S_1^2+S_2^2}$.

- If the asserts are correlated, there is geometric formula involving the circumcircle of a triangle built on the two Sharpes. If the correlation between the two assets is sufficiently positive, the optimal weight of the weaker asset can be negative.

- The case of multiple (say, fewer than 10000) correlated assets is not actually computationally expensive. A LAPACK matrix inversion in C or numpy would probably take under a few seconds or less.

- A more serious issue is the curse of dimensionality (aka as bad sampling of covariance, overfitting, etc) making the optimal combining of multiple assets (or strategies, for that matter) tricky and requiring some regularization. This part is hard to formalize and requires a degree of a prior inductive bias and experience.

## Answer by Aleksandar Milivojević (score 2)

https://quant.stackexchange.com/a/82446

To see quickly how to obtain the weights given by Michael Isichenko, we can proceed as follows: Denote by $s_i = \frac{\mu_i}{\sigma_i}$ the Sharpe ratio of the $i^{\mathrm{th}}$ asset, and consider the vectors $(w_1 \sigma_1, w_2 \sigma_2, \ldots, w_n \sigma_n)$ and $(s_1, s_2, \ldots, s_n)$, where $w_i$ are any weights. Applying the Cauchy-Schwarz inequality, we have $$\left(\sum_i s_i w_i \sigma_i\right)^2 \leq \left(\sum_i s_i^2\right)\left(\sum_i w_i^2 \sigma_i^2\right),$$ and therefore for the squared Sharpe ratio of the portfolio, since $\mu_i = s_i \sigma_i$, we have $$ \frac{(\sum_i w_i \mu_i)^2}{\sum_i w_i^2 \sigma_i^2} = \frac{(\sum_i s_i w_i \sigma_i)^2}{\sum_i w_i^2 \sigma_i^2} \leq \sum_i s_i^2 = \sum_i \frac{\mu_i^2}{\sigma_i^2},$$ with equality achieved when $$(w_1 \sigma_1, w_2 \sigma_2, \ldots, w_n \sigma_n) = c(s_1, s_2, \ldots, s_n)$$ for some constant $c$. We get $w_i = c \frac{\mu_i}{\sigma_i^2}$. Under the restriction $\sum_i w_i = 1$, solving for $c$ by summing up over $i$, we get $$c = \frac{1}{\sum_i \frac{\mu_i}{\sigma_i^2}},$$ and so for $$w_i = \frac{\frac{\mu_i}{\sigma_i^2}}{\sum_i \frac{\mu_i}{\sigma_i^2}}$$ the portfolio achieves the maximum squared Sharpe ratio of $\sum_i \frac{\mu_i^2}{\sigma_i^2}$. Of course, the Sharpe ratio is invariant under positive change of scale for the weights, so we can take $w_i = \frac{\mu_i}{\sigma_i^2}$ as in the other answer.