# Distribution of the normal CDF of a normal random variable

Let $X\sim N(\mu,\sigma^2)$. It's straightforward to derive (see, e.g., distribution of the normal cdf and https://mathoverflow.net/questions/225868/variance-of-the-normal-cdf?) $E[\Phi(X)]$ and $Var(\Phi(X))$, where $\Phi(\cdot)$ is the normal CDF.

Is it also straightforward to characterize the full distribution of $\Phi(X)$?

• You should clarify what you mean by "characterize the full distribution". For any given real numbers $a(> 0)$ and $b$, the CDF of $\Phi(aZ+b)$, where $Z$ is standard normal with CDF $\Phi$, is just $$F(t) = P(\Phi(aZ+b) \le t)=P(aZ+b \le \Phi^{-1}(t))=\Phi\left(\frac{\Phi^{-1}(t)-b}{a}\right)\quad(-\infty<t<\infty).$$ Doesn't this function "characterize the full distribution"? Commented Dec 13, 2016 at 3:21
• Thanks; great point. I guess my actual question is: Does the distribution of $\Phi(X)$ coincide with that of (perhaps a function of) another random variable with a known distribution? On further reflection I suspect not, though I'm inspired by the related problem of "characterizing" the distribution of $F_Y(Y+\mu)$ where $Y$ is standard Gumbel and $\mu$ is some constant. In that case $F_Y(Y+\mu)=\exp(-\exp(-Y-\mu))=(\exp(-\exp(-Y)))^{1/\exp(\mu)}\sim Beta(\exp(\mu),1)$. I wasn't sure if this was at all generalizeable. Commented Dec 13, 2016 at 3:50

Even more so: for any continuous distribution with CDF $F$:
If $F(x) = y$, $0 < y < 1$, $$F_{F(X)}(y) = \mathbb P(F(X) \le y) = \mathbb P(X \le x) = F(x) = y$$
That is, $F(X)$ has uniform distribution on $[0,1]$.
• Right so if $\mu=0$ and $\sigma=1$ I'd know $\Phi(X)\sim U(0,1)$. But what about in the general case where i'm applying the "wrong" CDF for $X$ (in this case it's the "right" type of r.v. - normal - but with the "wrong" mean and variance). Does that make sense? Thanks for answering! Commented Dec 13, 2016 at 2:57