# Why does risk aversion use variance instead of standard deviation?

The risk-aversion component of a portfolio utility function is expressed as the variance of the portfolio. Why the variance, instead of standard deviation, is used in here?

I'm asking this question because of the following calculation: suppose I only have a single stock and also a fixed risk-aversion parameter. Then I use mean-variance tradeoff to determine the optimal amount of the stock to hold. If the variance is used in the utility function, then I can get an optimal position because the variance is quadratic of the position. However, if standard deviation is used, then both the expected return and risk are linear in the position, hence we cannot get an optimal position in this setup. On the other hand, if indeed standard deviation is also a reasonable choice of expression of risk aversion, then it seems the "optimal position" obtained using the variance is purely an artifact of the functional form selected.

• could you show the formula of this portfolio utility function, with the risk-aversion component pointed out – develarist Jun 28 '19 at 2:49

Let us start with some underlying math. First, $$\sigma=\sqrt{\sigma^2}$$, but the minimum variance unbiased estimator (MVUE) for standard deviation is not the square root of the MVUE of the variance, $$\hat{\sigma}\ne\sqrt{\hat{\sigma^2}}.$$ Taking the square root of the unbiased sample estimator of the variance introduces bias because it is a non-linear function. See derivation of MVUE of SD If the parameters are known, then it doesn't matter which way you do it.
• Because the variance is the second central moment. Standard deviation is just a transform if it. Kind of like $e=mc^2$. You could create a squared variable, say k, and write it as $e=mk$. It's the same thing, but c has a definite meaning. – Dave Harris Jul 3 '19 at 22:37