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 $[0,1]$.


2 Answers 2


There is a somewhat more general proposition regarding this problem.

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.

  • $\begingroup$ 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. $\endgroup$ Dec 14, 2017 at 12:54

For $0\le u\le 1,$ $$ \Pr\left( \frac 1 {\sqrt{2\pi}} \int_{-\infty}^X e^{-x^2/2} \, dx \le u \right) = \Pr\left( \Phi(X) \le u \right) = \Pr(X\le \Phi^{-1}(u) ) = \Phi(\Phi^{-1}(u)) = u. $$


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