# 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?

• Your problem arises because you are proceeding mechanically here. Look at the Lagrange multiplier theorem again. Your final equation applies at a fixed stationary point of the Lagrangian. Just absorb the denominator with the square root into the multiplier and solve as before.
– RRL
Nov 26, 2022 at 16:53

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.

• Perfect! Thank you Nov 27, 2022 at 14:01
• @rubikscube09: You're welcome.
– RRL
Nov 28, 2022 at 2:37