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 $x > \xi$?
Certainly not. Let g be any probability distribution function satisfying g(0)=0 and let f(x)=g(x1) 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(x1)>g(x) for large x. 


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 $[0,1]$. 


I think I've managed to prove the following:
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 rearrangement) 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 counterexample" 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. 

