CDF raised to a power?

If FZ is a CDF, it looks like FZ(z)α (α>0) is a CDF as well.

Q: Is this a standard result?

Q: Is there a good way to find a function g with Xg(Z) s.t. FX(x)=FZ(z)α, where xg(z)

Basically, I have another CDF in hand, FZ(z)α. 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 ZN(0,1). Or at least know that such a result is intractable.

Answer

I like the other answers, but nobody has mentioned the following yet. The event {Ut, Vt} occurs if and only if {max(U,V)t}, so if U and V are independent and W=max(U,V), then FW(t)=FU(t)FV(t) so for α a positive integer (say, α=n) take X=max(Z1,...Zn) where the Z‘s are i.i.d.

For α=1/n we can switcheroo to get FZ=FnX, 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 α a positive rational number (say, α=m/n) follows from the previous since
(FZ)m/n=(F1/nZ)m.

For α an irrational, choose a sequence of positive rationals ak converging to α; then the sequence Xk (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 FαZ for α 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 α=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<z<1.


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
g1(y)=Φ1(Φα(y)),
or more generally when Z is not necessarily N(0,1), we have
x=g1(y)=F1(Fα(y)).
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
y=g(x)=F1(F1/α(x)).

Let's take a special case, plug things in, and see how it works. Let X have an Exp(1) distribution, with CDF
F(x)=(1ex), x>0,
and inverse CDF
F1(y)=ln(1y).
It is easy to plug everything in to find g; after we're done we get
y=g(x)=ln(1(1ex)1/α)
So, in summary, my claim is that if XExp(1) and if we define
Y=ln(1(1eX)1/α),
then Y will have a CDF which looks like
FY(y)=(1ey)α.
We can prove this directly (look at P(Yy) 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:
If XF then U=F(X)Unif(0,1).
If UUnif(0,1) then U1/αBeta(α,1).

The plot of the simulation results follows.

ECDF and F to the alpha

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)?

Attribution
Source : Link , Question Author : lowndrul , Answer Author : Community

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