Let X be normally distributed with mean $0$ and variance $1$, find the CDF and density of $Y = \Phi(X)$

Question

Define $\Phi(x)$ as: $$ \Phi(x) = \frac 1{\sqrt{2\pi}}\int_{-\infty}^x \exp\left(-\frac{t^2}{2}\right) dt $$ and let the random variable $Y$ be defined as $\Phi(X)$ where $X$ is a standard normal variable (i.e. with mean $0$ and standard deviation $1$).

What is the CDF and density of $Y$?

Progress

My thoughts are that I can substitute $Y$ into the CDF given, which is $\Phi (x) = \text{ CDF of }X = P(X < x)$, thus getting $P(\Phi(x) < y)$. I then figured that I can inverse both sides to get $P(X < \Phi^{-1}(y))$. At this point, I'm unclear how to proceed, as in I don't know how to express the CDF.

Also, I was told to find the density which I believe I simply have to differentiate the function, but I'm also unclear on how to differentiate since there seems to be 3 different variables x, y and t and I'm not sure which to eliminate or replace...

Answer 1

Based on the following observations

1. $Y$ takes values in $[0,1]$,
2. $Φ(\cdot)$ is strictly monotone increasing and therefore invertible. Let $Φ^{-1}$ denote it's inverse,

the CDF $F_Y(y)$ of $Y$ is given by $$F_Y(y)=P(Y\le y)=P(Φ (X)\le y)=P(X\le Φ^{-1}(y))=Φ(Φ^{-1}(y))=y$$ for $y\in[0,1]$. Thus $$f_Y(y)=\frac{\partial }{\partial y}F_Y(y)=(y)'=1$$ for $y \in [0,1]$ and $0$ elsewhere, which means that $Y$ is uniformly distributed in $[0,1]$. In sum $$F_Y(y)=\begin{cases}0, &y<0\\\frac{y-0}{1-0}=y, & 0\le y\le 1\\1, & y>1 \end{cases}$$ and $$f_Y(y)=\begin{cases}\frac{1}{1-0}=1, & 0\le y\le 1\\0, & \text{ else } \end{cases}$$

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The above definition of $Y$ is related to the concept of inverse transform sampling.

Answer 2

Hint: $Y$ is a uniform distribution on $[0,1]$