# On if $X \sim N(0,1)$ Then $\frac{1}{\sqrt{2 \pi}} \int_{- \infty}^X e^{-x^2/2} \, dx \sim U[0,1]$

Where could I find a proof of the following theorem in probability theory:

Let $X \sim N(0,1)$ (standard normal distributed) Then

$$\frac{1}{\sqrt{2 \pi}} \int_{- \infty}^X e^{-x^2/2} \, dx \sim U[0,1]$$

where $U[0,1]$ stands for the uniform distribution on $[0,1]$.

If $X$ is a random variable with a continuous and strictly increasing distribution function $F$, then $F(X) \sim U(0, 1)$.
Proof: For any real $0 < x < 1$,$$P(F(X) \leqslant x) = P(X \leqslant F^{-1}(x)) = F(F^{-1}(x)) = x,$$ therefore $F(X) \sim U(0, 1)$.
In your question, $X \sim N(0, 1)$, thus $\displaystyle F(x) = \frac{1}{\sqrt{2\mathrm{\pi}}}\int_{-\infty}^x \mathrm{e}^{-x^2 / 2} \,\mathrm{d}x$. The problem is solved by applying the general proposition.
• Instead of saying $F$ is strictly increasing, you should be saying $F$ is continuous. Any discontinuity disturbs the conclusion, so your proposed theorem in its present form is false. On the other hand, it doesn't need to be strictly increasing. Dec 14, 2017 at 12:54
For $0\le u\le 1,$ $$\Pr\left( \frac 1 {\sqrt{2\pi}} \int_{-\infty}^X e^{-x^2/2} \, dx \le u \right) = \Pr\left( \Phi(X) \le u \right) = \Pr(X\le \Phi^{-1}(u) ) = \Phi(\Phi^{-1}(u)) = u.$$