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.

$\begingroup$ 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. $\endgroup$ – genneth May 7 '12 at 0:35

1$\begingroup$ 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. $\endgroup$ – Brendan McKay May 7 '12 at 0:57

$\begingroup$ @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. $\endgroup$ – genneth May 7 '12 at 2:14


1$\begingroup$ 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. $\endgroup$ – Brendan McKay May 7 '12 at 12:27
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 $[0,1]$.

1$\begingroup$ 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 (visavis extension probability) using their moments. I hope this helps. $\endgroup$ – tipanverella May 7 '12 at 3:39

$\begingroup$ 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 supercritical 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. $\endgroup$ – genneth May 7 '12 at 11:19

$\begingroup$ 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. $\endgroup$ – tipanverella May 7 '12 at 13:33

$\begingroup$ 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. $\endgroup$ – tipanverella May 7 '12 at 13:34

$\begingroup$ 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). $\endgroup$ – tipanverella May 7 '12 at 13:35
I think I've managed to prove the following:
Let $X_1$ and $X_2$ be nonnegative 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 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.

$\begingroup$ (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. $\endgroup$ – Brendan McKay May 16 '12 at 3:33