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Let $X\sim\mathcal{N}(0,1)$ and denote the standard normal CDF by $F$. It is easy to see that $F(X)\sim\mathcal{U}(0,1)$. Namely, for $u\in(0,1)$ we have \begin{align*} P(F(X)\leq u)=P(X\leq F^{-1}(u)) = F(F^{-1}(u))= u, \end{align*} so the expected value is \begin{align*} E(F(X))=0.5. \end{align*} Now consider the random variable $F(X+a)$ where $a>0$ is a constant. How can we find the expected value \begin{align*} E(F(X+a))=\;? \end{align*} I suspect the answer to be $F(a)$. Additionally, can we find a simple form for the density or distribution function of the random variable $F(X+a)$?

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  • $\begingroup$ The CDF for $X+a$ is $F [F^{-1}(u) - a ]$. $\endgroup$
    – mlc
    Mar 12, 2017 at 11:30

1 Answer 1

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Your conjectured value for $E[F(X+a)]$ is close but not quite correct. $$ E[F(X+a)] = \int_{-\infty}^\infty F(x+a)f(x)\, \mathrm dx = \int_{-\infty}^\infty \left[\int_{-\infty}^{x+a}f(y)\, \mathrm dy\right]f(x)\, \mathrm dx $$ Now suppose that $Y\sim N(0,1)$ is independent of $X$. Then, the integral on the right can be interpreted as calculating $P(Y \leq X+a) = P(Y-X \leq a)$. But, $Y-X \sim N(0,2)$ and so $$E[F(X+a)] = F(a/\sqrt 2)$$ and not $F(a)$ as you have conjectured. See also this answer of mine on stats.SE for a more general version of the argument used here.

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