# Comparing two distributions based of the ratio of their moments

I am looking for some metric for distribution with support on the interval $$[0,1-\epsilon]$$, that will be based on the ratio of their moments. That is, if $$X\sim f(x)$$, $$Y\sim g(y)$$, I'm looking for a metric $$d(f,g)$$ such that $$\frac{\lvert\mathbb{E}X^k-\mathbb{E}Y^k\rvert}{\mathbb{E}X^k}$$ is small for all $$0 $$\iff$$ $$d(f,g)$$ is small.

Of course, I could just define the distance to be the sum of these ratios, but I am not sure what this means. Can two distributions be very different but have a very close moment-ratio?

I was able to show that the 1-Wasserstein distance is small if $$\lvert\mathbb{E}X^k-\mathbb{E}Y^k\rvert$$ is small, and vice-versa, but this is not strong enough, I want the ratio of the moments.

Is there some natural metric to look at?

• What is $k$? What do you want to do if $\mathbb E X^k$ or $\mathbb E Y^k$ is $0$? Dec 29, 2021 at 14:51
• @LSpice : Any positive $k$ will do here. Then $EX^k>0$, because $X$ has a pdf $f$ and hence $X>0$ almost surely. Dec 29, 2021 at 18:41
• Sorry, I meant for all moments, i.e., all $k>0$. Edited the question. Thanks Dec 29, 2021 at 19:14

$$\newcommand\ep\epsilon$$There is no such metric, because for any $$\ep\in(0,1)$$ and any real $$k>0$$ there are random variables $$X$$ and $$Y$$ with different pdf's $$f$$ and $$g$$ supported on $$[0,1-\ep]$$ such that $$EX^k=EY^k$$.

However, the pseudometric $$d_k$$ defined by the formula $$d_k(f,g):=|\ln EX^k-\ln EY^k|=\Big|\ln\frac{EY^k}{EX^k}\Big|$$ will have the desired property: $$d_k(f,g)\to0\iff\frac{EY^k}{EX^k}\to1\iff\frac{|EY^k-EX^k|}{EX^k}\to0. \tag{1}$$

The OP has now said that all moments, for all $$k>0$$, were meant to be close.

Letting then $$d(f,g):=\sum_{k=1}^\infty 2^{-k}\frac{d_k(f,g)}{1+d_k(f,g)}, \tag{2}$$ we get a metric $$d$$ such that $$d(f,g)\to0\iff\frac{|EY^k-EX^k|}{EX^k}\to0\text{ for each natural }k. \tag{3}$$

Indeed, if $$d(f,g)\to0$$ then, by (2), for each natural $$k$$ we have $$\dfrac{d_k(f,g)}{1+d_k(f,g)}\to0$$ and hence $$d_k(f,g)\to0$$. Vice versa, if for each natural $$k$$ we have $$d_k(f,g)\to0$$, then $$\dfrac{d_k(f,g)}{1+d_k(f,g)}\to0$$ and hence, by (2) and dominated convergence, $$d(f,g)\to0$$. So, $$d(f,g)\to0$$ if and only if for each natural $$k$$ we have $$d_k(f,g)\to0$$. Thus, (3) follows from (1).

• Thank you. To clarify- I meant that all moments are close, not just the $k$th moment for a single $k$ Dec 29, 2021 at 19:16
• @Student88 : For all moments close, you can have a true metric, as now added to the answer. Dec 29, 2021 at 19:46
• Note that in terms of topology, this just metrises weak convergence of probability measures, except that it 'excises' the Dirac mass at $0$ by putting it at distance $1$ from everything else. Dec 29, 2021 at 20:53
• @MartinHairer : I am not sure if I understand you correctly. E.g., suppose that, for $a\in(0,(1-\epsilon)/4)$, $f_a$ and $g_a$ are the pdf's of random variables $X_a$ and $Y_a:=2X_a$, respectively, where $X_a$ is uniformly distributed on the interval $[a,2a]$. Then $d(f_a,g_a)$ is $>0$ and does not depend on $a$, whereas (say) the Lévy distance between the cdf's of $X_a$ and $Y_a$ goes to $0$ as $a\downarrow0$. Dec 29, 2021 at 21:19
• @IosifPinelis Yes, but these two sequences themselves don't converge to any limit in the $d$ distance while they precisely converge to the Dirac mass at $0$ in the usual sense. Dec 29, 2021 at 21:44