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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 $0<k$ $\iff$ $d(f,g)$ 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?

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  • $\begingroup$ What is $k$? What do you want to do if $\mathbb E X^k$ or $\mathbb E Y^k$ is $0$? $\endgroup$
    – LSpice
    Dec 29, 2021 at 14:51
  • $\begingroup$ @LSpice : Any positive $k$ will do here. Then $EX^k>0$, because $X$ has a pdf $f$ and hence $X>0$ almost surely. $\endgroup$
    – Iosif Pinelis
    Dec 29, 2021 at 18:41
  • $\begingroup$ Sorry, I meant for all moments, i.e., all $k>0$. Edited the question. Thanks $\endgroup$
    – Student88
    Dec 29, 2021 at 19:14

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$\newcommand\ep\epsilon$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).

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  • $\begingroup$ Thank you. To clarify- I meant that all moments are close, not just the $k$th moment for a single $k$ $\endgroup$
    – Student88
    Dec 29, 2021 at 19:16
  • $\begingroup$ @Student88 : For all moments close, you can have a true metric, as now added to the answer. $\endgroup$
    – Iosif Pinelis
    Dec 29, 2021 at 19:46
  • $\begingroup$ 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. $\endgroup$
    – Martin Hairer
    Dec 29, 2021 at 20:53
  • $\begingroup$ @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 $a\downarrow0$. $\endgroup$
    – Iosif Pinelis
    Dec 29, 2021 at 21:19
  • $\begingroup$ @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. $\endgroup$
    – Martin Hairer
    Dec 29, 2021 at 21:44

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