Distribution of Normal distribution

Question

suppose $X \sim Normal(\mu, \sigma^{2})$. What is the distribution of $Y := N(X)$? where $N$ denotes Standard Normal Cumulative Distribution Function? e.g. in a special case when $\mu = 0$ and $\sigma = 1$, $Y \sim U[0,1]$.

Answer 1

$X$ is a ($\mu, \sigma^2)$-normal variable, so $\mathsf P(X\leq x) = N_{\mu, \sigma^2}(x)$.

Variable $Y$ represents the probability that the realisation of some standard normal variable (say, $Z$) is at least as great as the realisation of normal variable $X$.

Let there be $Z\sim \mathcal N(0,1)$, so hence we have defined the random variable $Y \mathop{:=}\mathcal P(Z\leq X) = N_{0,1}(X)$.

Thus the support of $Y$ is $(0;1)$.   What is the cumulative distribution of $Y$ ?

$$\begin{align} \mathsf P(Y\leq y) & = \mathsf P(\mathsf P(Z\leq X)\leq y)\;\color{silver}{\mathbf 1_{y\in(0,1)}+\mathbf 1_{y\geq 1}} \\ & = \mathsf P(N_{0,1}(X)\leq y)\;\color{silver}{\mathbf 1_{y\in(0,1)}+\mathbf 1_{y\geq 1}} \\ & = \mathsf P(X\leq N^{-1}(y))\;\color{silver}{\mathbf 1_{y\in(0,1)}+\mathbf 1_{y\geq 1}} \\ & = \color{blue}{N_{\mu,\sigma^2}\circ N_{0,1}^{-1}(y)}\;\color{silver}{\mathbf 1_{y\in(0,1)}+\mathbf 1_{y\geq 1}} \\ & = N_{0,1}\left(\frac{N_{0,1}^{-1}(y)-\mu}{\sigma^2}\right)\;\color{silver}{\mathbf 1_{y\in(0,1)}+\mathbf 1_{y\geq 1}} \end{align}$$

Only in the special case where $\mu=0, \sigma=1$ do we then have a uniform distribution over the support $(0;1)$: $$\begin{align} \mathsf P(Y\leq y) & = N_{0,1}\circ N_{0,1}^{-1}(y)\;\color{silver}{\mathbf 1_{y\in(0,1)}+\mathbf 1_{y\geq 1}} \\ & = y\;\color{silver}{\mathbf 1_{y\in(0,1)}+\mathbf 1_{y\geq 1}} \end{align}$$