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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.

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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

4 Answers 4

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.

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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.

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$X$ 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 $1/2$.

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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.

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