Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let's say I have two random variables, $X$ and $Y$.
I know the relationship between their CDF's: $$F_Y(y) = g(F_X(y))$$

For example, I could know that $F_Y(y) = F_X(y)^2$ or something like that. What then can I say about relating $Y$ to $X$? I am pretty sure I can't say that $Y = g(X)$... but I think there must be a relationship there.

share|cite|improve this question
When in the first equality, or both are $y$ or both $x$ not a mixture of them without restrictions. You should correct that. – Josué Tonelli-Cueto Sep 28 '11 at 19:51
The meaning of the equation is unclear: you could explain the relationship between $x$ and $y$. – Did Sep 28 '11 at 19:53
@Iasafro and Didier - I think I fixed the problem -- if not, please let me know what doesn't make sense. – Angada Sep 28 '11 at 20:39
@Angada Now yes, it is clear the equality. – Josué Tonelli-Cueto Sep 28 '11 at 21:23
up vote 2 down vote accepted

As already explained by others, a functional equation between the CDF of two random variables $X$ and $Y$, such as the equation in this question, implies no almost sure relation whatsoever between $X$ and $Y$. A first reason is that $X$ and $Y$ may be defined on different probability spaces. A second reason is that, even if $X$ and $Y$ are defined on the same probability space, their CDF do not determine $X$ and $Y$, at all.

Basic examples to keep in mind in this setting might be $X$ uniform on $(0,1)$ and $Y=1-X$, or $X$ uniform on $(0,1)$ and $Y$ uniform on $(0,1)$ but independent on $X$. Both these $X$ and $Y$, at the level of their CDF, are indistinguishable from $X$ uniform on $(0,1)$ and $Y=X$.

However, a similar, but different, question might be of interest: consider a CDF $G$ and a function $g$ such that $g\circ G$ is also a CDF (hence $G$ replaces $F_X$ and $g\circ G$ replaces $F_Y$). Since every function which is nondecreasing and continuous on the right with limits $0$ and $1$ at $-\infty$ and $+\infty$ is a CDF, one way to ensure that $g\circ G$ is a CDF is to ask that $g$ is nondecreasing, continuous on the right, with limits $0$ and $1$ at $0$ and $1$. Assume that all this holds and that $X$ has CDF $G$, one can ask if there is a way to build a random variable with CDF $g\circ G$, using only $X$. The problem becomes:

Let $X$ with CDF $G$. Find a function $a$ such that the CDF of $a(X)$ is $g\circ G$.

Here is a fact:

If $U$ is uniformly distributed on $(0,1)$ and $G$ is a CDF, there exists a function, which we denote $G^\ast$, such that the CDF of $G^\ast(U)$ is $G$.

The function $G^\ast$ is a pseudo-inverse of $G$, defined as follows: for every nondecreasing and continuous on the right function $h$ and every real number $u$, $$ h^\ast(u)=\inf\{x\mid G(x)\ge u\}. $$ It is a nice analysis exercise to check that $h^\ast$ may also be defined by the equivalence $$ h^\ast(u)\le x\iff u\le h(x). $$ Consequences are that $h^\ast$ is nondecreasing and continuous on the right and that $(h_1\circ h_2)^\ast=h_2^\ast\circ h_1^\ast$. This answers a slightly easier question than ours:

Choose $U$ uniformly distributed on $(0,1)$ and consider the random variables $X=G^\ast(U)$ and $Y=(G^\ast\circ g^\ast)(U)$. Then the CDF of $X$ is $G$ and the CDF of $Y$ is $g\circ G$.

This also suggests that, to get $Y$ as a function of $X$, one could consider:

                         $Y=a(X)$ with $a=G^\ast\circ g^\ast\circ G.$

With this definition, one sees that $$ [Y\le y]=[g^\ast(G(X))\le G(y)]=[G(X)\le g(G(y))], $$ hence the proof would be complete if one knew that $\mathrm P(G(X)\le z)=z$ for every $z$ in the image of $g\circ G$ (note that to ask this for every $z$ is hopeless in general).

When $g$ is the identity, the problem itself becomes trivial (choose $Y=X$) but our function $a=G^\ast\circ G$ is not always the identity. In general, $G^\ast\circ G(x)\le x$ for every $x$ and it seems that $$ (G^\ast\circ G)(x)=\inf\{z\le x\mid\mathrm P(z<X<x)=0\}, $$ which would imply that $(G^\ast\circ G)(X)=X$ almost surely.

share|cite|improve this answer

There need not be any obvious relationship between $X$ and $Y$: even if $g(w)=w$, all you have are two identically distributed random variables, but they may be equal or independent or have some other relationship.

So all you have is a relationship between the distributions. You can see that $g$ must be weakly increasing and that $g(0)=0$ and $g(1)=1$ (perhaps in some cases as limits). But that is about it beyond repeating the equation in words or looking at specific examples.

Take $X$ having a uniform distribution on $(0,1)$: so $Y$ has CDF $g(y)$ on this support, and for example if $g(w)=w^2$ then $Y$ has a triangular distribution on $(0,1)$ with the mode at $1$. I also looked at $X$ being normally distributed and $g(w)=w^2$, and that made $Y$ have close to but not quite a normal distribution.

share|cite|improve this answer

$X$ and $Y$ could be defined on completely different probability spaces. Even if they are defined on the same probability space, there is no particular relationship between the events $X \le a$ and $Y \le b$: all you know about are their probabilities $F_X(a)$ and $F_Y(b)$. So e.g. if $F_X(a) = 1/2$ and $F_Y(b) = 1/2$, $P(X \le a \text{ and } Y \le b)$ might be anything from $0$ to $1/2$.

share|cite|improve this answer

Obviously this can work only if $g$ is a weakly increasing function from $[0,1]$ to itself and fixes $0$ and $1$.

One extreme case is that $Y$ is actually equal to $F_Y^{-1}(g(F_X(X)))$. In that case, the random variable $X$ completely determines $Y$ and the relationship between the two cdf's is just the one you've written. Qualification: provided $F_Y$ and $g$ are invertible, thus strictly increasing.

The opposite extreme case is that $Y$ is independent of $X$ and has the particular probability distribution that you've given.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.