Distribution of $\Phi(X)$ where $\Phi$ is standard normal CDF and $X$ is normally distributed

Question

Consider $\theta \sim N(\mu,\sigma^2)$. And $\Phi(\cdot)$ denote the CDF of standard normal distribution. Then what is the distribution of $\Phi(\theta)$? In particular what is the variance of $\Phi(\theta)$?

Answer 1

Let us look for the cumulative distribution function (CDF) $F_{\Phi(\theta)}$ of $\Phi(\theta)$ where $\theta\sim \mathcal{N}(\mu,\sigma^2)$ and $\Phi$ is the CDF of $\mathcal{N}(0,1)$. One has \begin{align} F_{\Phi(\theta)}(t) = P(\Phi(\theta)\leq t) &= \left\lbrace \begin{aligned} &0 &&\text{if } t<0\\ &P\left(\frac{\theta - \mu}{\sigma}\leq \frac{\Phi^{-1}(t)-\mu}{\sigma}\right) = &&\text{if } 0\leq t<1\\ &1 &&\text{if } 1<t \end{aligned} \right. \\ &= \Phi\left(\frac{\Phi^{-1}(t)-\mu}{\sigma}\right) \boldsymbol{1}_{[0,1[} (t) + \boldsymbol{1}_{[1,+\infty[} (t) \, . \end{align} Therefore, the probability density function (PDF) $f_{\Phi(\theta)}(t)$ is $$ f_{\Phi(\theta)}(t) = \frac{f\left((\Phi^{-1}(t)-\mu)/\sigma\right)}{\sigma\,f\left(\Phi^{-1}(t)\right)} \boldsymbol{1}_{[0,1[} (t) \, , $$ where $f$ is the PDF of $\mathcal{N}(0,1)$. One can note that $\Phi(\theta)$ has a uniform distribution over $[0,1[$ if $\theta\sim \mathcal{N}(0,1)$. In this case, the variance of $\Phi(\theta)$ is $1/12$. In the general case, the variance is \begin{aligned} V(\Phi(\theta)) &= E(\Phi(\theta)^2) - E(\Phi(\theta))^2\\ &= \int_{-\infty}^{+\infty} \Phi(t)^2 f_\theta(t) \, dt - \left(\int_{-\infty}^{+\infty} \Phi(t) f_\theta(t) \, dt\right)^2 \, , \end{aligned} where $f_\theta$ denotes the PDF of $\theta\sim \mathcal{N}(\mu,\sigma^2)$. The variance may be not easy to compute in the general case where $\mu\neq0$ and $\sigma\neq 1$.