Good evening; I just have a simple question about Value at Risk and the subadditivity property, and I know that it may sound silly

I got that, in general, VaR is not subadditive. However, if a portfolio contain elliptically distributed risk factors, then VaR will be subadditive.

What I do not understand is why an ellipticall distribution, such as the Normal distribution, implies subadditivity of VaR.

Thanks in advance.


Suppose $X\sim N(\mu_X,\sigma_X^2)$ and $Y\sim N(\mu_Y,\sigma_Y^2)$ are correlated jointly normal random variables. Then, $$X+Y\sim N(\mu_X+\mu_Y,\sigma^2_X+\sigma_Y^2+2\rho\sigma_X\sigma_Y).$$

Suppose $X$ and $Y$ denote the profit of your portfolio returns (so negative values for $X,Y$ mean losses). Then, the 5% value at risk is the 0.05 quantile of the profit distribution (the minimum amount you lose in the worst five percent of cases $\Leftrightarrow$ in 95% of the cases you're sure to lose less than this value-at-risk). Thus, using the inverse function of the normal distribution (see here), \begin{align*} \text{VaR}(X,\alpha)&=-\mu_X+\sigma_X\Phi^{-1}(1-\alpha),\\ \text{VaR}(Y,\alpha)&=-\mu_Y+\sigma_Y\Phi^{-1}(1-\alpha), \\ \text{VaR}(X+Y,\alpha)&=-\mu_X-\mu_Y+\sqrt{\sigma^2_X+\sigma_Y^2+2\rho\sigma_X\sigma_Y}\Phi^{-1}(1-\alpha). \end{align*} Thus, \begin{align*} \text{VaR}(X,\alpha) + \text{VaR}(Y,\alpha) &= -\mu_X-\mu_Y+(\sigma_X+\sigma_Y)\Phi^{-1}(1-\alpha) \\ &\geq -\mu_X-\mu_Y+\sqrt{\sigma^2_X+\sigma_Y^2+2\rho\sigma_X\sigma_Y}\Phi^{-1}(1-\alpha) \\ &= \text{VaR}(X+Y,\alpha), \end{align*} because for $\rho\in(-1,1)$, \begin{align*} \sigma_X+\sigma_Y = \sqrt{\sigma^2_X+\sigma_Y^2+2\sigma_X\sigma_Y} \geq \sqrt{\sigma^2_X+\sigma_Y^2+2\rho\sigma_X\sigma_Y}. \end{align*}

Thus, value-at-risk (just like the standard deviation) is always sub-additive for two normally distributed loss distributions, if we assume $\alpha<0.5$, (``benefit of diversification''): a portfolio containing $X$ and $Y$ is less risky than the sum of the individual risks. If we assume a different distribution for $X$ and $Y$, then this result may no longer hold.

You can, of course, fiddle in some weights and consider $wX+(1-w)Y$.

  • $\begingroup$ Thank you for your answer Kevin, however, I still don't get why do we need the Normality assumption to ensure subadditivity. As far as I understand, if you calculate a VaR with non-normal distribution or non-parametric approach such as Historical/Montecarlo Simulation, you may have and may not have subadditivity in the VaR of your portfolio. However, with normal distribution, this is always the case. $\endgroup$ Mar 12 at 0:14
  • $\begingroup$ @AntonioStanco For the normal distribution, we can prove that VaR is always subadditive (see edit). However, this calculation may be different (no longer hold) if we assume different distributions for $X$ and $Y$. Thus, VaR is not sub-additive all the time. It only happens to be sub-additive in the special case (example) of a normal distribution (and some others). $\endgroup$
    – Kevin
    Mar 12 at 0:44
  • $\begingroup$ @AntonioStanco Is it now any clearer? You can compute VaR for any distribution (parametric or non-parametric by simulations as you suggest). The question is whether VaR is always sub-additive. It turns out it is not. Take some stock data and try it out :) You'll quickly find examples where it is violated. If we however assume portfolio returns are normally distributed, then in this very special case, VaR turns out to be always sub-additive. So normality is just an example for which we know that we get the property we want $\endgroup$
    – Kevin
    Mar 12 at 9:22
  • $\begingroup$ Good morning Kevin, thanks for your help. What I don't completely get is why if the underlying distribution of your assets is Normal, then you have FOR SURE sub-additivity $\endgroup$ Mar 12 at 12:02
  • 1
    $\begingroup$ @Kevin My pleasure. Btw, I love the precision and thoroughness of your answers. [You did take the time to provide the proper link for the statement. I just thought it would be nice to stress the assumptions that the link worked with. :) ] $\endgroup$
    – ir7
    Mar 13 at 17:29

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