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

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.

• If by $\Phi(X)$ you mean $Y=F(X)$ where $F(x)=\Pr(X \le x)$, your $Y\sim U(0,1)$ applies to all continuous distributions Commented Jun 27, 2023 at 10:05
• @utobi $\Phi \left( x \right)$ is CDF of Standard Normal Distribution. I have updated my original post with this information Commented Jun 27, 2023 at 10:13
• Derive $\mathbb P(\Phi(X)\le y)$ then. Commented Jun 27, 2023 at 10:21
• There may not be a simple solution with your $\Phi$ being the standard normal CDF. Clearly $Y$ is constrained to be on $[0,1]$ and has a symmetric distribution about $0.5$ when $\mu=0$. The upper bound on the standard deviation is $0.5$, approached when $\mu$ is constant and $\sigma$ increases without bound (bimodal in this case). The lower bound on the standard deviation is $0$, approached when $\mu$ is constant and $\sigma$ decreases towards $0$. Commented Jun 27, 2023 at 10:22
• The 0.5 is explained at stats.stackexchange.com/questions/45588/….
– whuber
Commented Jun 27, 2023 at 17:32

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 $$Y$$.