I want to understand why this holds: $argmax_w ( \frac{\mu^T w}{\sqrt{w^T\Sigma w}})=\Sigma^{-1}\mu $

I just found this post: Derivation of the tangency (maximum Sharpe Ratio) portfolio in Markowitz Portfolio Theory?

In the process you exchanged the optimization problem for the optimal tangency portfolio with the optimization problem for the mean-variance portfolio: $argmax_w (w^T\mu-\frac{1}{2}w^T\Sigma w )$

I want to understand why these optimization problems come to the same conclusion.



1 Answer 1


Your first argmax is actually defined up to a constant multiplier: $argmax\left(\frac{\mu^Tw}{\sqrt{w^T\Sigma w}}\right)=\lambda\Sigma^{-1}\mu$, where $\lambda$ is an arbitrary portfolio size scale. In general, maximizing a scale-invariant ratio of the form $f(w)/g(w)$ can be done in conditional terms: $max(f)$ subject to $g=const$, or, using a Lagrange multiplier $\lambda$, $max(f-\lambda g)$. Formally, the optimality condition $\nabla(f/g)=0$ results in a collinearity of the gradients of $f$ and $g$ -- the same as in the conditional maximum with a Lagrange multiplier. The arbitrariness of the $\lambda$ scale disappears if there are constraints on the portfolio weights $w$ and/or transaction costs incurred when rebalancing the portfolio from its previous state. This involves a more complicated, but still convex optimization problem, c.f. this book.

  • $\begingroup$ Thank you very much. My initial question was, which conditions must apply such that $argmax_w ( \frac{\mu^T w}{\sqrt{w^T\Sigma w}})=\Sigma^{-1}\mu $ $=$ $argmax_w (w^T\mu-\frac{1}{2}w^T\Sigma w )$ holds? In my case there is no budget contrain. $\endgroup$
    – Valentin
    Sep 14, 2021 at 7:49
  • $\begingroup$ The condition for $argmax(\mu^Tw/\sqrt{w^T\Sigma w})=\Sigma^{-1}\mu$ is the normalization of the weights in the form $w^T\Sigma w=\mu^T\Sigma^{-1}\mu$, which is easily verified but has no deep meaning. The solution of your original $argmax$ is defined up to a constant multiplier. $\endgroup$ Sep 14, 2021 at 13:09

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