# 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.

• Yes, that's a pretty well-known result and is easy to generalize. (How?) You can also find $g$, at least implicitly. It's essentially an application of the inverse probably transform technique commonly used to generate random variates of an arbitrary distribution. – cardinal May 13 '11 at 15:28
• @cardinal Please, answer. The team is later complaining that we are not fighting with low answered ratio. – user88 May 13 '11 at 16:25
• @mbq: Thanks for your comments, which I understand and respect greatly. Please understand that sometimes considerations of time and/or place do not allow me to post an answer, but do permit a quick comment that can get the OP or other participants off to a start. Be assured that, going forward, if I am able to post an answer, I will do so. Hopefully my continued participation through comments will be ok as well. – cardinal May 13 '11 at 23:18
• @cardinal Some of us are also guilty of the same, for the same reasons... – whuber May 14 '11 at 0:16
• @brianjd Yes, this is a well-known result that has been used for industrially producing "generalised" distributions, see. There exist many transformations like this one and people use them for this purpose: they find a parametric transformation, apply it to a distribution and voilá, you have a paper just by just calculating its properties. And of course, the normal is the first 'victim'. – user10525 May 8 '12 at 18:43

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 $$\left(F_{Z}\right)^{m/n} = \left(F_{Z}^{1/n}\right)^{m}.$$

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 $F(z) = z$, $0<z<1$.

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 $$g^{-1}(y) = \Phi^{-1}(\Phi^{\alpha}(y)),$$ or more generally when $Z$ is not necessarily $N(0,1)$, we have $$x = g^{-1}(y) = F^{-1}(F^{\alpha}(y)).$$ 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 $$y = g(x) = F^{-1}(F^{1/\alpha}(x)).$$

Let's take a special case, plug things in, and see how it works. Let $X$ have an Exp(1) distribution, with CDF $$F(x) = (1 - \mathrm{e}^{-x}),\ x > 0,$$ and inverse CDF $$F^{-1}(y) = -\ln(1 - y).$$ It is easy to plug everything in to find $g$; after we're done we get $$y = g(x) = -\ln \left( 1 - (1 - \mathrm{e}^{-x})^{1/\alpha} \right)$$ So, in summary, my claim is that if $X \sim \mathrm{Exp}(1)$ and if we define $$Y = -\ln \left( 1 - (1 - \mathrm{e}^{-X})^{1/\alpha} \right),$$ then $Y$ will have a CDF which looks like $$F_{Y}(y) = \left( 1 - \mathrm{e}^{-y} \right)^{\alpha}.$$ 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: $$\mbox{If X \sim F then U = F(X) \sim \mathrm{Unif}(0,1).}$$ $$\mbox{If U \sim \mathrm{Unif}(0,1) then U^{1/\alpha} \sim \mathrm{Beta}(\alpha,1).}$$

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)?

• See my comment in @JMS answer. For $Z\sim N(0,1)$ it looks like the answer is $g(z) = \Phi^{-1}(\Phi^{1/\alpha}(z))$ which isn't closed form but can be calculated easily enough. And you can make it easier by recognizing that the input to the inverse CDF is a suitably chosen Beta distribution. The answer will be nice in cases where the inverse CDF is nice, and there are some of those running around. – user1108 May 16 '11 at 13:43
• It would be good to double-check your arithmetic. – cardinal May 16 '11 at 14:40
• @cardinal errr... OK, I did,... and it's right? Would you please point out the error? – user1108 May 20 '11 at 18:01
• (+1) Apologies. I'm not sure where my head was when I first looked at this. It's obviously (well, should have been!) correct. – cardinal May 20 '11 at 19:09
• @cardinal, no harm, no foul. I admit, though, you really had me sweating for a minute! :-) – user1108 May 20 '11 at 19:24

## Proof without words

The lower blue curve is $F$, the upper red curve is $F^\alpha$ (typifying the case $\alpha \lt 1$), and the arrows show how to go from $z$ to $x = g(z)$.

• Nice pic! Q: What was that drawn in? TikZ? – lowndrul May 14 '11 at 15:47
• @brianjd: If I recall, @whuber does many of his plots using Mathematica. – cardinal May 14 '11 at 15:52
• @cardinal You're right. Actually, I use whatever is handy and seems like it will do a good job quickly. FWIW, here's the code: Module[ {y, w, a = 0.1, z = 3.24, f = ChiDistribution[7.6], xmin=0, xmax=5}, y = CDF[f,z]; w = InverseCDF[f, y^(1/a)]; Show[ Plot[{CDF[f, x],CDF[f,x]^a} , {x, xmin, xmax}, Filling->{1->{2}}], Graphics[{ Dashed, Arrow[{{z,0}, {z,y}}], Arrow[{{z,y}, {w,y}}], Arrow[{{w,y}, {w,0}}] }] ] ] – whuber May 14 '11 at 18:10

Q1) Yes. It's also useful for generating variables which are stochastically ordered; you can see this from @whuber's pretty picture :). $\alpha>1$ swaps the stochastic order.

That it's a valid cdf is just a matter of verifying the requisite conditions: $F_z(z)^\alpha$ has to be cadlag, nondecreasing and limit to $1$ at infinity and $0$ at negative infinity. $F_z$ has these properties so these are all easy to show.

Q2) Seems like it would be pretty difficult analytically, unless $F_Z$ is special

• @JMS: What about $Z \sim N(0,1)$ ? – lowndrul May 14 '11 at 15:45
• @brianjd: I don't believe so. Let $g$ be a continuous strictly monotonic function (hence, having a well-defined inverse $g^{-1}$) that satisfies your conditions. Then, it must be that $\renewcommand{\Pr}{\mathbb{P}}\Phi^{\alpha}(u) = \Pr(g(Z) \leq u) = \Pr( Z \leq g^{-1}(u)) = \Phi(g^{-1}(u))$ and so $g^{-1}(u) = \Phi^{-1}(\Phi^{\alpha}(u))$. So the inverse is identified fairly explicitly, but not $g$ itself. This is what I meant in my previous comment about $g$ being found implicitly. – cardinal May 14 '11 at 16:17
• @brianjd - What @cardinal said :) I couldn't even think of a special case for $F_Z$ where you'd get a closed form (not to say there isn't one of course). – JMS May 14 '11 at 23:28
• @JMS: $Z \sim \mathcal{U}[0,1]$ would be one positive example. – cardinal May 15 '11 at 2:38
• @cardinal I never would have thought of such a rare distribution... but now that you mention it a $Beta(a, 1)$ should work in general, giving you back a $Beta(a\alpha, 1)$. – JMS May 15 '11 at 3:11