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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 $x > \xi$?

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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.

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That's what I thought, but there is an objection --- all moments exist. That puts some quite tight limits on the behaviour of $g$ at large $x$ already. –  genneth May 7 '12 at 0:35
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Yes, but you just need to start with $g$ having rapidly decreasing tails and yet wriggling a little bit with a period of 2. It's easy to make Michael's counterexample work with all moments existing. To have a chance of success you need to add some condition such as monotonicity of the densities for sufficiently large argument. –  Brendan McKay May 7 '12 at 0:57
    
@BrendanMcKay: For the problem at hand, I actually have that $g ~ e^-cx$ for some $c$, and I want to use the knowledge on the moments to bound $f$. That indeed gives the monotonicity that you mentioned... but I'm still stuck on if it actually works. –  genneth May 7 '12 at 2:14
    
That should read $g \sim e^{-cx}$ above. –  genneth May 7 '12 at 2:21
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Then the $n$-th moment of $g$ for large $n$ is determined by $g(x)$ for $x=n/c + O(n^{1/2})$. I suspect you can wriggle the first derivative of $f$ so it is still decreasing but falls on both sides of $g$ infinitely often. It just has to be above $g$ a little more than it is below $g$. I don't have an explicit example though, so I'm not saying it is solved. –  Brendan McKay May 7 '12 at 12:27
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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 $[0,1]$.

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You might want to look at the following papers: - Braun, Henry "Polynomial Bounds for Probability Generating Functions", Journal of Applied Probability, Volume 12 No 3 (1975) - Brook, D. "Bounds for Moment Generating Functions and for Extinction Probabilities", Journal of Applied Probability, Volume 3 No 1 (1966) Long story short, back in the day people who cared about Branching Processes cared a lot about comparing random variables (vis-a-vis extension probability) using their moments. I hope this helps. –  tipanverella May 7 '12 at 3:39
    
I think you misread p.d.f. as p.g.f. Actually, I am working on branching processes, and indeed I'm trying to bound the tails of some limiting distributions for super-critical processes. In terms of the characteristic function, I can find bounds on the derivatives at zero (i.e. the moments) but I would like to convert that to a control of the tails. In principle it feels quite plausible, but I need to supplement it with some sort of result that rules out the oscillating behaviour which Michael pointed to above. –  genneth May 7 '12 at 11:19
    
In my opinion, it might be fruitful to work with some sort of generating function (pgf, or mgf) if, as I suspected and you confirmed , you are working with branching processes. I therefore suggested the use of pgf instead of pdf because mostly of the nice results in the (Braun, 1975) paper I mentioned in the comment. –  tipanverella May 7 '12 at 13:33
    
Furthermore, a good way to get bounds on tails is to look at logarithmic moment generating functions; the (Brook, 1966) paper is useful because it provides a way for producing moment generating function bounds to the moment generating function of a random variable, based on the moments of that random variable. –  tipanverella May 7 '12 at 13:34
    
Finally, maybe you would like to take a look at Spitzer's Comparaison Lemma (in Athreya and Ney "Branching Processes", 1972, chapter 1, section 9). –  tipanverella May 7 '12 at 13:35
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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.

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(A typo: probably $x$ and $w$ are mixed up.) Frankly I'm dubious. You are right that "if the tails were the other way round, there would be a moment $k$ of $X_1$ which is smaller than $X_2$", but that is not the contrapositive. The contrapositive would be that the set of $w$ such that the tails compare the other way around is unbounded. –  Brendan McKay May 16 '12 at 3:33
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