Relating Standard-Deviation and Variance Mean-Variance Objectives

# Alternative form of mean-variance optimization that uses standard deviation


# Alternative form of mean-variance optimization that uses standard deviation












I'm curious about an exercise found in Optimization Methods in Finance. Exercise 8.2 (pg 143) explores a variant of the more commonly used form of MVO. When I refer to the more common variant I'm talking about:

$$ \begin{aligned} \operatorname{max}_x \mu^Tx - \frac{\delta}{2}x^T\Sigma x & \\ Ax &= b \\ Cx &\ge d \end{aligned} $$

The variant that directly uses standard deviation by taking the square root of $x^T\Sigma x$ is:

$$ \begin{aligned} \operatorname{max}_x \mu^Tx - \eta \sqrt{x^T\Sigma x} & \\ Ax &= b \\ Cx &\ge d \end{aligned} $$

The exercise at hand is stated in the book as:

> For each $\eta$, let $x^*(\eta)$ denote the optimal solution of the second form. Show that there exists a $\delta > 0$ such that $x^*(n)$ solves the first form for that $\delta$.

I'm interested in making this conversion because of the more intuitive interpretation of subtracting $\eta$ standard deviations from the mean (which are denominated in the same units), versus subtracting the variance (which doesn't have as clear cut of an interpretation for me).

Seems somewhat similar to the currently unanswered question The Viability/Usefulness of Mean Standard Deviation Optimization?.

## Answer by ethor (score 0)

https://quant.stackexchange.com/a/74569

I've been trying to answer my own question, and this is what I have so far. It should be a good starting point for a discussion.

First, fix $\eta$ and take the gradients of each of the objective functions. If we denote the first objective function as $f_\delta(x)$, then

\begin{aligned} \nabla f_\delta(x) &= \mu - \frac{\delta}{2}\left(2\Sigma x\right) = \mu - \delta\Sigma x \end{aligned}

And if we denote the second objective function as $g(x)$, then

\begin{aligned} \nabla g(x) &= \mu - \frac{\eta\Sigma x}{\sqrt{x^T\Sigma x}} \end{aligned}

We know that $x^*(\eta)$ is a critical point, so $\nabla g(x^*(\eta)) = 0$ holds.

\begin{aligned} \nabla g(x^*(\eta)) &= \mu - \frac{\eta\Sigma x^*(\eta)}{\sqrt{x^*(\eta)^T\Sigma x^*(\eta)}} = 0 \\ \frac{\eta\Sigma x^*(\eta)}{\sqrt{x^*(\eta)^T\Sigma x^*(\eta)}} &= \mu \\ x^*(\eta) &= \frac{\Sigma^{-1}\mu\sqrt{x^*(\eta)^T\Sigma x^*(\eta)}}{\eta} \end{aligned}

We also know that a solution for the first problem, call it $y$, must satisfy $\nabla f_\delta(y) = 0$.

\begin{align} \nabla f_\delta(y) &= \mu -\delta\Sigma y = 0 \\ \delta\Sigma y &= \mu \\ y &= \frac{\Sigma^{-1}\mu}{\delta} \end{align}

Now, we want to find some $\delta^*$ so that $y = x^*(\eta)$.

\begin{align} \frac{\Sigma^{-1}\mu}{\delta^*} &= \frac{\Sigma^{-1}\mu\sqrt{x^*(\eta)^T\Sigma x^*(\eta)}}{\eta} \\ \frac{1}{\delta^*} &= \frac{\sqrt{x^*(\eta)^T\Sigma x^*(\eta)}}{\eta} \\ \delta^* &= \frac{\eta}{\sqrt{x^*(\eta)^T\Sigma x^*(\eta)}} \end{align}

Just to be sure, we can double check that $x^*(\eta)$ is a critical point of $f_{\delta^*}(x)$.

\begin{align} \nabla f_{\delta^*}(x^*(\eta)) &= \mu - \frac{\eta\Sigma x^*(\eta)}{\sqrt{x^*(\eta)^T\Sigma x^*(\eta)}} = \nabla g(x^*(\eta)) = 0 \end{align}

My takeaways from $\delta^*$ are:

- $\delta^* \propto \eta$. $\eta$ chooses how many standard deviations below the mean we should maximize, so the bigger $\eta$ is, the more risk-averse we are. Similarly, higher $\delta$ chooses the ratio of extra returns we need to be willing to accept additional risk. These explanations agree with each other.

- $\delta^* \propto \frac{1}{\sqrt{x^*(\eta)^T\Sigma x^*(\eta)}}$. The term in the denominator is the standard deviation for the optimal solution to the alternative form of the MVO problem. The bigger this standard deviation is, the more risk-seeking we are.