In a course on statistics, this set of non-compulsory exercises were supplied (in Swedish). I'm stuck on 8.10. My translation of the exercise:
The stochastic variable X has a $N(0,1)$ distribution. Calculate the cumulative distribution function for the stochastic variable $Y = \Phi(X)$, where $\Phi(\cdot)$ is the cumulative distribution function of X. Motivate your answer thoroughly!
The answer given in the key is:
$Y ∈ U(0,1)$
I'm not quite sure in which end to start. I've got something like
$$ X \in N(0,1) \\ F_Y(x) = P(Y < x) = P(\Phi(X) < x) = P(\frac{1}{\sqrt{2\pi}} \int^X_{-\infty}{e^{-t^2/2}}dt < x) $$
What is the solution for this exercise?
It is standard notation to refer to the cdf of $N(0,1)$ using the symbol $\Phi$, not $\phi$, so I shall use that.
First, note that the support for $Y$ is $(0,1)$. Denote $F_Y(\cdot)$ as the cdf of $Y$. Hence $F_Y(u)=0 \text{ for } u \le 0 \text{ and } F_Y(u)=1$ for $u \ge 1$. Consider $u \in (0,1).$ Then
$$ F_Y(u) = P(Y \le u) = P(\Phi(X) \le u) = P(X \le \Phi^{-1}(u)) = \Phi(\Phi^{-1}(u))=u $$
Hence $Y$ has the same cdf as that of a $U(0,1)$ random variable. By the uniqueness of cdfs, you're done.
Note that this works for any random variable, not just Gaussian. So if random variable $X$ has a continuous cdf $F$, then $Y=F(X)$ has a $U(0,1)$ distribution. This is also known as a "probability integral transform".