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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.

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

Standard Frequentist stochastic calculus assumes the parameters are known, even though they are not. Nonetheless, there is an advantage, if you believe people are going to create estimates from the formula to using the variance in that it is the second central moment of a distribution. If a moment generating function exists, then the moment generating function uniquely defines the distribution.

If the parameters are known, then minimizing the variance and minimizing the square of the standard deviation are the same thing. It would also be true in a mean-variance tradeoff situation. If the parameters are not known, and partly depending on your estimation assumptions such as Frequentist, Likelihoodist or Bayesian, then your functional form affects calculations. As economics is primarily a Frequentist discipline, mostly by default, the distinction matters, except in the exceedingly rare case where the parameters were actually known.

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  • $\begingroup$ My summary of your answer is: It does not matter if we use variance or standard deviation in the mean-variance portfolio optimization. On the other hand, if we do use standard deviation for the optimization, we cannot simply take the square root of variance to estimate the standard deviation, which should be estimated using its only MVUE. Is that correct? If so, why do I only see variance used in mean-variance portfolio optimization? $\endgroup$ – DiveIntoML Jul 3 '19 at 22:33
  • $\begingroup$ 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. $\endgroup$ – Dave Harris Jul 3 '19 at 22:37

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