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
$$
\mathbb{E}(P(X_1>\widetilde{\text{VaR}}_p(X_1), X_2>\widetilde{\text{VaR}}_p(X_2)  |Y))=P(X_1>\text{VaR}_p(X_1), X_2>\text{VaR}_p(X_2))?
$$
 A: 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.
