# Value at risk (VaR) conditioning argument

Let $X_1$ be a random variable and let $\text{VaR}_p(X_1)$ denote the quantile of $X_1$ such that $$P(X_1>\text{VaR}_p(X_1))=1-p.$$ Now, let $X_2$ be another random variable with $\text{VaR}_p(X_2)$ such that $$P(X_2>\text{VaR}_p(X_2))=1-p.$$

Let $Y$ be yet another random variable. Now let $\widetilde{\text{VaR}}_p(X_1)$ be $\text{VaR}_p(X_1)$ conditional on $Y$. That is, $$P(X_1>\widetilde{\text{VaR}}_p(X_1)|Y=y)=1-p.$$ Clearly, it holds that $$\mathbb{E}(P(X_1>\widetilde{\text{VaR}}_p|Y))=1-p=P(X_1>\text{VaR}_p).$$

I am wondering if it also holds that $$\mathbb{E}(P(X_1>\widetilde{\text{VaR}}_p(X_1), X_2>\widetilde{\text{VaR}}_p(X_2) |Y))=P(X_1>\text{VaR}_p(X_1), X_2>\text{VaR}_p(X_2))?$$

• Where did you want to place the tilde? I have placed it over VaR, but now you reverted to what I see as a bit messy. May 17, 2017 at 14:44
• I wanted it on top of the VaR as well. May 17, 2017 at 14:47
• Then you can roll back (undo) the change and you will have it as I did it -- on top of VaR. May 17, 2017 at 15:00
• @RichardHardy do you perhaps also know the answer? May 18, 2017 at 12:50
• Unfortunately, I do not know it. Even if I could work it out, it would probably require some time and effort (although I have not looked at the problem very carefully), and I am having a busy time these few days. May 18, 2017 at 13:20

It is not 100% clear what you are asking, since, as has been remarked by ssdecontrol, you confuse the conditioning somewhat. But I'll try to have a stab at it anyway and I'll interpret the conditioning in the density sense, i.e. conditioning on $Y=y$ means you evaluate the densities of $X_1$ and $X_2$ under the condition that $Y=y$.
Then the answer is NO. Your last equation does not hold in general. The problem is that you are fixing the probability of the margins by introducing new $VaR$ thresholds but you do not control the dependency. And of course conditioning will not only change the margins but also the dependency.
For an example, assume $X_1$, $X_2$ are independent standard normal variables, $Y = X_1 - X_2$ and condition on $Y=0$ or $X_1=X_2$. This will produce two new joint normal variable $(Z_1, Z_2)$ which are 100% correlated. Hence the left side of your last equation will be equal to $1-p$ because if one $Z$ is larger than the $VaR$ the other will be as well. But the right side is $(1 - p)^2$ due to independence.