What is the distribution of $\Phi(X) = \int_{-\infty}^x {\exp\left(-{z^2}{/2}\right) / \sqrt{2\pi}} \; \mathrm dx$?

Question

Let $X\sim N(0,1)$. What is the distribution of $$\Phi(X) = \int_{-\infty}^x {\exp\left(-{z^2}{/2}\right) \over \sqrt{2\pi}} \: \mathrm dx \quad?$$

Answer 1

The distribution for $ \Phi(X) $, where $ X \sim N(0,1) $, is the uniform distribution over $ [0,1] $.

We can see it directly from the c.d.f. of $ \Phi(X) $:

$$ \mathbb{P}( \Phi(X)<p)=\mathbb{P}( X<\Phi^{-1}(p))=p \quad \text{for} \quad 0\leq p\leq 1 $$

In fact, the identity $ F(X) \sim U[0,1] $, where $ F(x) $ is the c.d.f. of $ X $, holds for any continuous random variable with strict increasing c.d.f. on the interval $ F^{-1}((0,1)) $.

Answer 2

The distribution of the cdf of a support-$\mathbb{R}$ continuous random variable is $U(0,\,1)$, since $P(F\le f)=P(X\le F^{-1}(f))=F(F^{-1}(f))=f$.