Parametric VaR, Normality and Subadditivity

Question

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.

Answer 1

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 $wX+(1-w)Y$.