Comparing distributions with moments Suppose I have two variables $X$ and $Y$ which have continuous p.d.f.s $f$ and $g$ on the positive real line. I know that the moments $\mathrm{E}[X^n] > \mathrm{E}[Y^n]$ for sufficiently large $n$ (and all moments exist). Can I conclude that there exists a $\xi$ s.t. $f(x) > g(x)$ for all $x > \xi$?
 A: Certainly not. Let g be any probability distribution function satisfying g(0)=0 and let
f(x)=g(x-1) if x>1 and f(x)=0 if x<1. Then
$$\int_0^\infty f(x)x^n\,dx=\int_0^\infty g(x)(x+1)^n\,dx>\int_0^\infty x^ng(x)\,dx$$
for any n. But it is not necessarily true that g(x-1)>g(x) for large x.
A: Not quite, in order to compare the pgfs globally, i.e on $[0,1]$, you need to know something about what is happening at both ends of the interval.  The moments only tell you what is going on at $1$.  Granted if you know a lot there, you can extract a lot of information using Taylor expansions to approximate $f$ and $g$ at $1$, but you can't use that to say much beyond a certain neighborhood of $1$. The information at $0$ is essentially about $\mathbb{P}[X=0]$ for the pgf $f$ for example.  If you know about that and about a couple moments, you could be able to say things about how $f$ and $g$ compare on $[0,1]$.
A: I think I've managed to prove the following:

Let $X_1$ and $X_2$ be non-negative random variables. If $\mathbb{E}\left[X_1^n\right] \ge \mathbb{E}\left[X_2^n\right] $ then for sufficiently large $x$, $\mathbb{P}\left[X_1 > w \right] \ge \mathbb{P}\left[X_2 > w \right]$.

This way of stating it avoids Michael's example above with oscillatory p.d.f.s. The proof is roughly consider the contrapositive. That states (after some re-arrangement) that if the tails were the other way round, there would be a moment $k$ of $X_1$ which is smaller than $X_2$. The "closest approach to a counter-example" if $X_1$ is distributed as a single atom at some $x_0$, and $X_2$ is mostly an atom at zero but otherwise smeared out over $x>x_0$. Then one has to find an argument which says that for a sufficiently large $k$ the moment condition will be true.
This seems straightforward enough that I worry about not seeing a statement of this somewhere. So it's probably still wrong.
