Let $X$ be a random variable having standard normal distribution. Let $\Phi$ denote its distribution function. Find
$ \int_0^\infty \operatorname{Prob} (\Phi(X) \geq u) \; du $
Let $X$ be a random variable having standard normal distribution. Let $\Phi$ denote its distribution function. Find
$ \int_0^\infty \operatorname{Prob} (\Phi(X) \geq u) \; du $
HINT: If $X$ is standard normally distributed, then $\Phi(X)$ is uniformly distributed.
Since $\Phi$ is a strictly increasing function whose range is $(0,1)$, we have for $0, $ \Pr(\Phi(X)\ge u) = \Pr(X\ge \Phi^{-1}(u)) = 1 - \Phi(\Phi^{-1}(u)) = 1-u. $ Here we've used the fact that $\Pr(X\ge a) = 1-\Phi(a)$, for all $a\in\mathbb{R}$.
But if $u>1$ then $\Pr(\Phi(X)\ge u)$ is $0$ since that event is impossible.
That tells you what to integrate.