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Let $Z$ be a standard normal random variable and denote by $\Phi$ its cumulative distribution function. Let $a$ be a real number. Question: what is the expected value of the random variable $\Phi(Z+a)$?

Now, if $a=0$, the answer is simple. Since $\Phi$ is continuous, a theorem states that $\Phi(Z)$ has the standard uniform distribution, hence its expected value is $1/2$.

But what happens when $a\neq 0$? I tried to approach the problem transforming the random variable $\Phi(Z+a)$ in a similar way, possibly with some translation, but I got nowhere. The issue here is that also the random variable $\Phi(Z+a)$ takes its values in the interval $[0,1]$ but it cannot be anymore a uniform random variable, so I find problematic to approach this problem in the same way as in the case $a=0$.

Thank you in advance for your ideas.

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    $\begingroup$ This has been asked on the site before: note that $\Phi(z+a)=P(W<z+a)$ with $W$ standard normal independent of $Z$ hence $$E(\Phi(Z+a))=P(W<Z+a)=P((W-Z)/\sqrt2<a/\sqrt2)=\Phi(a/\sqrt2)$$ since $(W-Z)/\sqrt2$ is standard normal. The case $a=0$ yields $E(\Phi(Z))=\Phi(0)=\frac12$, as you computed already. The same approach yields every $E(\Phi(bZ+a))$. $\endgroup$
    – Did
    Mar 4, 2017 at 8:44
  • $\begingroup$ @Did, I don't understand why you have $E(\Phi(Z+a))=P(W<Z+a)$? What is the justification for this? $\endgroup$
    – RandomGuy
    Mar 4, 2017 at 8:53
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    $\begingroup$ The definitions. For every $z$, $E(\Phi(Z+a)\mid Z=z)=\Phi(z+a)=P(W<z+a)$ hence $E(\Phi(Z+a))=P(W<Z+a)$. $\endgroup$
    – Did
    Mar 4, 2017 at 8:55
  • $\begingroup$ @Did thanks. I find myself at odds with conditional expectations, that's probably why I didn't think of this transformation. Can you please address me to some resource where I can find the properties of c.e. needed here, so I can look into these types of problems more confortably? Thanks again. $\endgroup$
    – RandomGuy
    Mar 4, 2017 at 9:03
  • $\begingroup$ An excellent textbook, mathematical but accessible, is David Williams' Probability with martingales. $\endgroup$
    – Did
    Mar 4, 2017 at 10:34

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