2
$\begingroup$

I was reading this paper where it is mentioned

Let F(·) be the CDF of a continuous random variable X and $\phi^{−1}(·)$ be the inverse of the CDF of N(0,1). Consider the transformation from X to Z by Z = $\phi^{−1}(F(X))$. Then it is easy to see that Z is standard normal regardless of F

I didn't get how it is true. What does $\phi^{−1}(F(X))$ denote here. I am a bit confused. Explanations guys help me out?

$\endgroup$
3
  • $\begingroup$ Think about this question: What is the distribution of $F(X)$ ? $\endgroup$
    – soakley
    May 27 '13 at 0:15
  • $\begingroup$ The CDF $F(\cdot)$ is a real function $F: \mathbb R \to (0,1)$ with the property that for each $\alpha \in (-\infty,\infty)$, $F(\alpha) = P\{X \leq \alpha\} = \beta$. Now, $F(X)$ means we apply the real function $F(\cdot)$ to $X$ to get a new random variable $Y$ that takes on values in $(0,1)$. Also, $\{Y \leq \beta\}$ if and only if the event $\{X \leq \alpha\}$ occurred, and so $$F_Y(\beta) = P\{Y \leq \beta\} = P\{X \leq \alpha\} = \beta.$$ Now see if you can answer the question in soakley's comment. $\endgroup$ May 27 '13 at 3:13
  • $\begingroup$ See stats.stackexchange.com/questions/116221/… $\endgroup$ Nov 6 '18 at 10:18