Why $\Phi(\Phi^{-1}(x))-\Phi(\Phi^{-1}(1-y))$ $=\max\{x+y-1,0\}$, where $\Phi$ is the cdf of a standard normal RV?

Question

My question is: why is the L.H.S equal to $\textbf{max}\{x+y-1,0\}$? I just find that it is equal to $x+y-1$.

Or maybe I made a mistake in the previous step:

If $X\sim N(0,1)$ then $\mathbb P[X\le \Phi^{-1}(x),\Phi^{-1}(1-y)\le X]=L.H.S (=\Phi(\Phi^{-1}(x))-\Phi(\Phi^{-1}(1-y)))$

Answer 1

Your first equality is valid only when $\Phi^{-1} (1-y)\leq \Phi^{-1}(x)$ or $1-y \leq x$ or $x+y-1 \geq 0$.

Answer 2

The statement $$\Pr[X \le \Phi^{-1}(x), \Phi^{-1}(1-y) \le X]$$ can actually be written $$\Pr[ 1-y \le \Phi(X) \le x],$$ which illustrates that $1-y \le x$ must be true in order for the probability to be nonzero. Hence $$x + y - 1 \ge 0,$$ otherwise the given probability is zero.