Deriving Maximum-Sharpe Portfolio Weights with Covariance

# Proof of weights maximizing sharpe of a portfolio


# Proof of weights maximizing sharpe of a portfolio












Given a portfolio of $n$ assets with mean vector $\mu$ and correlation matrix $\Sigma$, the optimal weights $w$ on the $n$ assets to maximize overall sharpe is found by $$\max_{w:||w||=1}{\dfrac{\mu^T w}{\sqrt{w^T\Sigma w}}}$$ and the entries of $w$ are nonnegative. I have seen in this post (and various other places) that the weights solving this optimization problem are given as

$$w=\Sigma^{-1}\mu$$

However, I have not been able to find a reference as to why this is true.

Question: What is the proof that these are the optimal weights? Bonus: what if $\Sigma$ is not-invertible (strictly positive semidefinite)?

## Answer by THATS MY QUANT MY QUANTITATIVE (score 3, accepted)

https://quant.stackexchange.com/a/79794

If the optimization problem is:

$\max_{w} \frac{w'\boldsymbol{\mu}}{(w'\boldsymbol{\Sigma}w)^{\frac{1}{2}}}$

constrained to $w\textbf{1}=1$, then we just use Lagrange multipliers:

$$L = \frac{w'\boldsymbol{\mu}}{(w'\boldsymbol{\Sigma}w)^{\frac{1}{2}}} + \lambda(w\textbf{1} - 1)$$

Then taking the partials wrt $w$ and $\lambda$, we have:

$$\frac{\partial L}{\partial w} = \boldsymbol{\mu} (w'\boldsymbol{\Sigma}w)^{-\frac{1}{2}} - (w'\boldsymbol{\mu}(w'\boldsymbol{\Sigma}w)^{-\frac{3}{2}}\boldsymbol{\Sigma}w + \lambda\textbf{1}=0$$

$$ \frac{\partial L}{\partial \lambda} = w'\textbf{1} - 1 = 0$$ Then the result follows.

For your 2nd question, A matrix is invertible if and only if it has a non-zero determinant. But from the definition of a correlation matrix, the only way for a correlation matrix to have a non-zero determinant is for the correlations to all equal $1$ or $-1$. Of which case, the optimal portfolio is trivially obvious. If there are less datapoints than dimensions, you can perform PCA analysis, or you can use singular value decomposition to compute the pseudoinverse of a matrix.

## Answer by Aleksandar Milivojević (score 1)

https://quant.stackexchange.com/a/82449

As an alternative to the accepted solution, one can use the Cauchy-Schwarz inequality: denote by $A$ the square root of $\Sigma$, and consider the vectors $A w$ and $A^{-1} \mu$. Then we have $${\langle A w , A^{-1} \mu \rangle}^2 \leq \langle A w, A w \rangle \langle A^{-1} \mu, A^{-1} \mu \rangle.$$ Now, since $\Sigma$ is symmetric, so is $A$, and we have $$\langle Aw, A^{-1} \mu \rangle = \langle (A^{-1})^\tau A w, \mu \rangle = \langle A^{-1} A w, \mu \rangle = \langle w, \mu \rangle$$ and similarly $$\langle Aw, Aw \rangle = \langle A^2 w, w \rangle = \langle \Sigma w, w \rangle,$$ so $$\langle w, \mu \rangle^2 \leq \langle \Sigma w, w \rangle {\left\lVert A^{-1}\mu \right\rVert}^2.$$

Now, the squared Sharpe ratio of our portfolio is $$\frac{(w^\tau \mu)^2}{w^\tau \Sigma w} = \frac{{\langle w, \mu \rangle}^2}{\langle \Sigma w, w \rangle},$$ which we now know is $\leq \left\lVert A^{-1}\mu \right\rVert$, with equality achieved when $Aw = cA^{-1}\mu$ for some constant $c$. Solving for $w$ we get $w = c (A^{-1})^2 \mu = c \Sigma^{-1} \mu$. If you want $w$ to have entries summing up to 1, you can then solve for $c$ from that condition.