VaR backtesting: counting the number of rejections Let's say you calculate the number of VaR rejections for every $r_{1,t}$, $r_{2,t}$,
should you have the same number of rejections in a model, irrespectively of the weights being different?
$[w,\ 1-w] [r_{1,t} r_{2,t}]^T \leq -[w,\ 1-w][\mu_{1}\ \mu_{2}]^T-\sqrt{[w,\ 1-w] \Sigma  [w,\ 1-w]^T }$
To the left are the actual P\&L and on the right the Value-at-Risk.
 A: No, the value at risk ($\text{VaR}$) of a portfolio is generally not the same as the sum of $\text{VaR}$s of the individual elements. E.g. let $r_1$ and $r_2$ be returns on two assets. Let $r_1$ and $r_2$ follow a bivariate normal distribution with standard normal marginals and a correlation approaching negative one, $\rho\approx-1$. Construct a portfolio with equal weights, $w_1=1$ and $w_2=1$. Denote its return $r_p$:
$$
r_p:=r_1+r_2.
$$
Since $r_p$ is a linear combination of two marginals from a bivariate normal distribution, it is normally distributed itself. Note that
\begin{aligned}
\text{Var}(r) &= \text{Var}(w_1 r_1) + \text{Var}(w_2 r_2) + 2\rho\sqrt{\text{Var}(w_1 r_1)\times\text{Var}(w_2 r_2)} \\
&= \text{Var}(r_1) + \text{Var}(r_2) + 2\rho\sqrt{\text{Var}(r_1)\times\text{Var}(r_2)} \\
&\approx 1+1+2(-1)\sqrt{1\times 1} \\
&=0
\end{aligned}
and
\begin{aligned}
\text{Var}(w_1 r_1) + \text{Var}(w_2 r_2) &= \text{Var}(r_1) + \text{Var}(r_2) \\
&= 1+1 \\
&\approx 2.
\end{aligned}
Thus $r_p$ will be highly concentrated around zero and for any chosen probability $q$, $\text{VaR}_q(r_p)$ will be close to zero, too. Meanwhile, $\text{VaR}_q(w_1 r_1)$ and $\text{VaR}_q(w_2 r_2)$ will not be close to zero for $q$ sufficiently large.
