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Let $X$ be a random variable having standard normal distribution. Let $\Phi$ denote its distribution function. Find

$$ \int_0^\infty \operatorname{Prob} (\Phi(X) \geq u) \; du $$

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interesting question – dato datuashvili Jan 5 '12 at 11:18
distribution function as in cdf right? – NikBels Jan 5 '12 at 11:19
Um. $\Phi(x)=P(X\leq x)$ – kodyv Jan 5 '12 at 11:28

HINT: If $X$ is standard normally distributed, then $\Phi(X)$ is uniformly distributed.

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Sub-hint: for every continuous random variable $Z$ with CDF $F$, the random variable $F(Z)$ is uniformly distributed on $(0,1)$. – Did Jan 5 '12 at 12:55

Since $\Phi$ is a strictly increasing function whose range is $(0,1)$, we have for $0<u<1$, $$ \Pr(\Phi(X)\ge u) = \Pr(X\ge \Phi^{-1}(u)) = 1 - \Phi(\Phi^{-1}(u)) = 1-u. $$ 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.

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