I am super confused with the example below, are we given that "the $95\%$ quantile, $\operatorname{VaR}_{0.95}(Z)$, is $90$ because there is a $5\%$ chance of exceeding $90$", or did we find that out somehow. If someone can explain this example in full detail with how they got CDFs and everything else that would be great especially this paragraph:
Similarly, the CDF for risk $Y$ satisfies $F_Y (0) = 0.96$, indicating that there is a $96\%$ chance of no loss. Therefore, the $95\%$ quantile cannot exceed zero, and so $\operatorname{VaR}_{0.95}(Y) \le 0$. Consequently, the sum of the $95\%$ quantiles for $X$ and $Y$ is less than $\operatorname{VaR}_0.95(Z)$, which violates subadditivity.
The value at risk is a quantile. From Wikipedia:
So if $p = 0.05$, the value at risk is an amount for which the probability of loss exceeding such an amount is at most $0.05$, and the probability of a loss less than this amount is at least $0.95$. Your text defines it the other way around; i.e., it writes $\operatorname{VaR}_{0.95}(Z)$ to mean that the chance of a loss exceeding $\operatorname{VaR}_{0.95}(Z)$ is not more than $5\%$.
Since $Z$ is the loss random variable, and we are told that $F_Z(90) = 0.95$, this literally means $$\Pr[Z \le 90] = 0.95.$$ This in turn means the probability the loss is no more than $90$ is $95\%$. Equivalently, $$\Pr[Z > 90] = 1 - \Pr[Z \le 90] = 1 - 0.95 = 0.05.$$
The way to think about the second part is to think of $X$ and $Y$ as labels for $Z$ depending on whether $Z$ is greater than $100$. In other words, if $Z \le 100$, then we give that loss the name $X$. Otherwise, we give it the name $Y$. As a result, we never have a loss greater than $100$ that is called $X$; equivalently, $$\Pr[X > 100] = 0,$$ which of course implies $$F_X(100) = 1 - \Pr[X > 100] = 1.$$ How do we reason for the other quantiles? For example, how did the author arrive at $F_X(1) = 0.95$? We note that in order for $X \le 1$, either $Z > 100$, or $Z \le 1$, since in the first case, if $Z > 100$, then $X = 0$; this happens with probability $1 - F_Z(100) = 0.04$. And in the second case, $X = Z \le 1$, andthis has probability $0.91$. So $\Pr[X \le 1] = 0.04 + 0.91 = 0.95$. We can compute $F_X(90)$ similarly.
For the calculation of the quantiles of $Y$, we note that $$F_Y(0) = \Pr[Y = 0] = \Pr[Z \le 100] = F_Z(100) = 0.96.$$ The first equality is true because $Y$ is never negative; it is either $Z$ if $Z > 100$, or it is $0$ if $Z \le 100$. The second equality is true by definition.
The next statement, "therefore, the $95\%$ quantile cannot exceed zero," arises because $F_Y(0) = 0.96$ means the probability of no loss from $Y$ is $0.96$, so zero is already no more than $4\%$ of the value at risk for $Y$. Since VaR is a quantile, it is monotone decreasing. The larger the VaR, the smaller the chance of exceeding it.
Then the last part says that $$\operatorname{VaR}_{0.95}(X) + \operatorname{VaR}_{0.95}(Y) < \operatorname{Var}_{0.95}(Z),$$ because we calculated $$\operatorname{VaR}_{0.95}(X) = 1, \quad \operatorname{VaR}_{0.95}(Y) \le 0,$$ and we were given $$\operatorname{VaR}_{0.95}(Z) = 90.$$ Subadditivity would require $$\operatorname{VaR}_{0.95}(Z) = \operatorname{VaR}_{0.95}(X+Y) \le \operatorname{VaR}_{0.95}(X) + \operatorname{VaR}_{0.95}(Y);$$ that is to say, the inequality is the wrong way.
Why is this important? The idea is that value at risk is a measure of the relationship between an amount of loss and the probability of realizing that loss. The principle of diversifying one's investments is that it presumes to reduce our overall portfolio risk. But this statement is not actually true in all cases; it depends on the probability distributions of the individual losses. If subadditivity of VaR were always true, it would imply that the value at risk of a portfolio of investments would not exceed the sum of the VaRs of each individual investment (for the same probability threshold), meaning diversification always helps reduce exposure to loss. But as shown by the simplistic example in the text you referenced, this is not true.