0
$\begingroup$

So I understand that the minimum variance hedge ratio minimizes the second moment of the portfolios. My question is how is it related to the size of the risk capital (which is calculated as the Value at Risk - Expected Value). Is the Risk capital also minimized at the optimal hedge ratio? Since the Value at Risk is not considered in the minimization process this should not be the case correct? Also I was wondering if the U-Shape relationship as it holds for the standard deviation also holds for the risk capital.

$\endgroup$
4
  • $\begingroup$ In distributions fully characterized by the first two moments (e.g. the normal distribution), minimizing variance is equivalent to minimizing value at risk. In other distributions, these are not the same. $\endgroup$ May 6, 2020 at 19:58
  • $\begingroup$ @RichardHardy So does that technically mean that risk capital is minimized at a different hedge ratio ? Is it also U-shaped or can it go up and down too (f.e. 50 for hedge ratio of 0.5, then 60 for hedge ratio 0.6 and then again 55 for hedge ratio 0.7) or is it a clear parabola? I guess this should depend on the induced tradeoff between the expected value and the standard deviation right? As to minimize the standard error I give up some expected return. $\endgroup$
    – macro123
    May 6, 2020 at 21:21
  • $\begingroup$ Yes to your first question. The details and answer to the remaining questions depend on the joint distribution of assets. I should note the distribution I mentioned in my first comment was the multivariate one if you think about the individual assets forming the portfolio. $\endgroup$ May 7, 2020 at 6:20
  • $\begingroup$ I considered the bivariate normal case for the log-returns (hence they are normally distributed), where I correlated them using cholesky. But since the risk capital is evaluated at the level of the prices (which are lognormally distributed) the distribution becomes lognormal through the shift. $\endgroup$
    – macro123
    May 7, 2020 at 6:23

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.