If $F(x)$ is a valid cdf (in that it is increasing from 0 to 1) is $F^{-1}(x)$ also a valid cdf?
Is the inverse of a cdf always also a cdf?
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0Well, the quantile $F^{-1}(x)$ has already a very restricted domain, so I don't see why it'd be a CDF... – 2011-09-23
1 Answers
No. The standard normal cdf increases from 0 to 1 as its argument increases from $-\infty$ to $+\infty$, so its inverse increases from $-\infty$ to $+\infty$ as its argument increases from 0 to 1. A cdf doesn't go from $-\infty$ to $+\infty$; a cdf goes from 0 to 1. So the inverse is not a cdf.
The only time the inverse of a cdf is a cdf is for random variables that remain between 0 and 1 (and then one must quibble about speaking of an "inverse" of a function that remains 0 on the whole interval $(-\infty,0)$ and on $(1,\infty)$. And even for things that remain within $[0,1]$ you've got a problem for certain types of discrete disteributions. Suppose $X=0$ with probability $1/2$ and $X=1$ with probability $1/2$. Then the cdf is $ F(x) = \begin{cases} 0 & \text{if } x<0,\\ 1/2 & \text{if }0\le x < 1, \\ 1 & \text{if }x>1. \end{cases} $ What is the "inverse" of this function, which remains constant on each of three intervals?