# CDF raised to a power?

If $F_Z$ is a CDF, it looks like $F_Z(z)^\alpha$ ($\alpha \gt 0$) is a CDF as well.

Q: Is this a standard result?

Q: Is there a good way to find a function $g$ with $X \equiv g(Z)$ s.t. $F_X(x) = F_Z(z)^\alpha$, where $x \equiv g(z)$

Basically, I have another CDF in hand, $F_Z(z)^\alpha$. In some reduced form sense I’d like to characterize the random variable that produces that CDF.

EDIT: I’d be happy if I could get an analytical result for the special case $Z \sim N(0,1)$. Or at least know that such a result is intractable.

I like the other answers, but nobody has mentioned the following yet. The event $\{U \leq t,\ V\leq t \}$ occurs if and only if $\{\mathrm{max}(U,V)\leq t\}$, so if $U$ and $V$ are independent and $W = \mathrm{max}(U,V)$, then $F_{W}(t) = F_{U}(t)*F_{V}(t)$ so for $\alpha$ a positive integer (say, $\alpha = n$) take $X = \mathrm{max}(Z_{1},...Z_{n})$ where the $Z$‘s are i.i.d.

For $\alpha = 1/n$ we can switcheroo to get $F_{Z} = F_{X}^n$, so $X$ would be that random variable such that the max of $n$ independent copies has the same distribution as $Z$ (and this would not be one of our familiar friends, in general).

The case of $\alpha$ a positive rational number (say, $\alpha = m/n$) follows from the previous since

For $\alpha$ an irrational, choose a sequence of positive rationals $a_{k}$ converging to $\alpha$; then the sequence $X_{k}$ (where we can use our above tricks for each $k$) will converge in distribution to the $X$ desired.

This might not be the characterization you are looking for, but it least gives some idea of how to think about $F_{Z}^{\alpha}$ for $\alpha$ suitably nice. On the other hand, I’m not really sure how much nicer it can really get: you already have the CDF, so the chain rule gives you the PDF, and you can calculate moments till the sun sets…? It’s true that most $Z$‘s won’t have an $X$ that’s familiar for $\alpha = \sqrt{2}$, but if I wanted to play around with an example to look for something interesting I might try $Z$ uniformly distributed on the unit interval with $F(z) = z$, $0.

EDIT: I wrote some comments in @JMS answer, and there was a question about my arithmetic, so I'll write out what I meant in the hopes that it's more clear.

@cardinal correctly in the comment to @JMS answer wrote that the problem simplifies to

or more generally when $Z$ is not necessarily $N(0,1)$, we have

My point was that when $F$ has a nice inverse function we can just solve for the function $y = g(x)$ with basic algebra. I wrote in the comment that $g$ should be

Let's take a special case, plug things in, and see how it works. Let $X$ have an Exp(1) distribution, with CDF

and inverse CDF

It is easy to plug everything in to find $g$; after we're done we get

So, in summary, my claim is that if $X \sim \mathrm{Exp}(1)$ and if we define

then $Y$ will have a CDF which looks like

We can prove this directly (look at $P(Y \leq y)$ and use algebra to get the expression, in the next to the last step we need the Probability Integral Transform). Just in the (often repeated) case that I'm crazy, I ran some simulations to double-check that it works, ... and it does. See below. To make the code easier I used two facts:

The plot of the simulation results follows. The R code used to generate the plot (minus labels) is

n <- 10000; alpha <- 0.7
z <- rbeta(n, shape1 = alpha, shape2 = 1)
y <- -log(1 - z)
plot(ecdf(y))
f <- function(x) (pexp(x, rate = 1))^alpha
curve(f, add = TRUE, lty = 2, lwd = 2)


The fit looks pretty good, I think? Maybe I'm not crazy (this time)?