How to find distribution of $\Phi \left( x \right)$

Question

Given $X \sim N\left(\mu, \sigma^2 \right)$, I want to find the distribution of $$Y = \Phi \left( X \right),$$ with $\Phi \left(x\right) = \int_{-\infty}^{x} \frac{1}{ \sqrt{2\pi}} e^{-\frac{t^2}{2}}dt$, i.e., CDF of a Standard Normal Distribution.

More specifically, I want to get an expression for $SD \left[Y \right]$, the standard deviation of $Y$.

For the specific case with $\mu = 0, \sigma=1$, I would easily have $Y \sim U\left(0,1 \right)$, but what will be the case for any normal distribution?

Any pointers will be very helpful.

Answer 1

There is no closed form expression for the standard deviation but it may be useful to express it in terms of the bivariate normal cdf as follows. First, note that \begin{align} E(Y)&=E(\Phi(X)) \\&=P(Z\le X) && \text{where $Z\sim N(0,1)$} \\&=P(Z-X\le 0) \\&=\Phi\Big(\frac{\mu}{\sqrt{1+\sigma^2}}\Big). \end{align} Similarly, \begin{align} E(Y^2)&=E(\Phi(X)^2) \\&=P(Z_1 \le X \cap Z_2\le X) &&\text{where $Z_1,Z_2$ are iid $N(0,1)$} \\&=P(Z_1-X\le 0 \cap Z_2-X\le 0) \\&=\Phi_2\Big(\frac{\mu}{\sqrt{1+\sigma^2}},\frac{\mu}{\sqrt{1+\sigma^2}}\Big) \end{align} where $\Phi_2$ is the cdf of a bivariate standardnormal distribution with correlation $$ \rho=\frac{\sigma^2}{1+\sigma^2} $$ available as the function mvtnorm::pmvnorm in R. So from this you can compute the variance and standard deviation of $Y$.