Why Minimum Standard Deviation and Variance Portfolios Match

# Minimum Standard Deviation Portfolio vs Minimum Variance Portfolio


# Minimum Standard Deviation Portfolio vs Minimum Variance Portfolio












When solving for the minimum variance portfolio, we have the object:

$$ f(w) = \frac{1}{2} w^T \Sigma w $$ subject to a basic scaling constraint: $$ \sum_{i=1}^N w_i = 1 $$ or in matrix terms, $w^T \mathbf{1} = 1$ where $\mathbf{1}$ is the $n$ vector of all ones.

Forming the Lagrangian, we get: $$ \Sigma w - \lambda \mathbf{1} = 0 $$ From which we have: $$ w = \lambda \Sigma^{-1} \mathbf{1}$$ Using the constraint, we can solve for $\lambda$ which ends up being a normalizing constant.

I tried to solve for the "minimum standard deviation" portfolio in a similar way, subject to the same constraint. It has the objective function: $$ f(w) = \sqrt{w^T \Sigma w} $$ its solution should be the same as the minimum variance portfolio because the objective is simply a monotone transformation of of the minimum variance objective. Forming the lagrangian again, we get:

$$ \frac{\Sigma w}{\sqrt{w^T\Sigma w}} - \lambda I = 0 $$

I am however, unclear where to proceed from here, as I can't just invert $\Sigma$ to get a solution due to there being another function of $w$ in the denominator. Is there something wrong with apply Lagrange multipliers here, due to the non-differentiability of $\sqrt{.}$? Or am I missing something very obvious?

## Answer by RRL (score 5, accepted)

https://quant.stackexchange.com/a/73887

Recall the conclusion of the Lagrange multiplier theorem. If $w^*$ is an optimal solution for the objective function $f(w)$ and constraint $g(w) = 0$, then there is a unique Lagrange multiplier $\lambda^*$ such that $(w^*,\lambda^*)$ is a stationary point of the Lagrangian $\mathcal{L}(w) = f(w) - \lambda g(w)$. That is, in terms of the derivative operators $Df$ and $Dg$,

$$\tag{1}Df(w^*) - \lambda^*Dg(w^*) = 0$$

In this case, we have the objective function and constraint

$$f(w) = \sqrt{w^T\Sigma w}, \quad g(w) = w^T\mathbf{1} - 1$$

Enforcing (1), the stationary point $(w^*, \lambda^*)$ must satisfy

$$\frac{\Sigma w^*}{\sqrt{(w^*)^T\Sigma w^*}} - \lambda^* \mathbf{1} = 0,$$

and it follows that

$$w^* = \lambda^*\sqrt{(w^*)^T\Sigma w^*}\Sigma^{-1}\mathbf{1}$$

Applying the constraint $(w^*)^T\mathbf{1} - 1= 0$, we can solve for the entire (scalar) expression $\lambda^*\sqrt{(w^*)^T\Sigma w^*}$ and obtain the same solution as in the minimum variance problem.