Suppose $\mu$ and $\nu$ are two probability measures on $[0,1]$. Let their $n$-th moments be denoted by $\mu_n$ and $\nu_n$, respectively, for $n \in \mathbb{N}$.

If we know that $\mu_n=\nu_n$ for infinitely many $n$, can we conclude that $\mu=\nu$?

One way to resolve this would be to see if the span of $\{ x^n \mid n \in S \}$ with $|S|=\infty$ is dense in the set of continuous functions $C[0,1]$. Is such a set always dense?

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    $\begingroup$ @Davide Giraudo: Thanks for the Muntz-Szacz theorem! $\endgroup$ – Santhosh Kumar Aug 25 '13 at 21:58
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    $\begingroup$ Why does this question have two votes to close as "off-topic?" $\endgroup$ – Douglas Zare Aug 26 '13 at 2:00
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    $\begingroup$ @Davide Giraudo: Interesting. Applying the Hahn-Banach theorem, that answers the question. $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac1n=\infty$. $\endgroup$ – George Lowther Aug 26 '13 at 8:45
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    $\begingroup$ @GeorgeLowther Thank you! Now an interesting problem would be to exhibit, if it exists, an example where $\nu_k=\mu_k$ on a set $S$ for which $\sum_{n\in S}\frac 1n<\infty$, but the measures are not the same. $\endgroup$ – Davide Giraudo Aug 26 '13 at 9:07
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    $\begingroup$ @yemon: the signed measure also vanishes on the constant functions. So, take its positive and negative components and scale them to get probability measures. $\endgroup$ – George Lowther Aug 26 '13 at 16:54

We actually have that $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac 1n$ is divergent. It's a condition which translates the fact that the set of indexed $k$ such that $\mu_k=\nu_k$ has to be large enough.

Recall Müntz-Szász theorem, which states (in particular) the following:

Theorem: If $(n_k,k\geqslant 0)$ is an increasing sequence of integers with $n_0=0$, then the following conditions are equivalent:

  1. the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.

  2. the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent.

If $\sum_{n\in S}\frac 1n$ is divergent, we can conclude by density that $\mu$ and $\nu$ coincide.

If the series is convergent, we can find $F\in (C[0,1])'$ such that $F(x^n)=0$ for all $n\in \{0\}\cup S$, but $F$ is not identically vanishing (this comes from Hahn-Banach theorem). We represent $F$ as a (non-zero) signed measure $m:=m^+-m^-$ (Hahn decomposition). Since $m^+[0,1]=m^-[0,1]$, we can rescale these measures in order to get probability measures. Then with the same notations as in the OP, $m^+_n=m^-_n$ for all $n\in \{0\}\cup S$, but $m^+\neq m^-$.

  • $\begingroup$ In your last sentence, where do $\mu$ and $\nu$ come from? If they are given at the start of the question, and assumed to have the same moments-indexed-by-S, then how do you know $\mu-m^+$ and $\nu+m^{-}$ are prob measures? (This is the part that I wasn't sure about, it is trivial from Hahn-Banach and Hahn-Jordan that you can find two positive measures whose moments agree on S) $\endgroup$ – Yemon Choi Aug 26 '13 at 16:15
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    $\begingroup$ @YemonChoi: I didn't understand that part of the answer either (I think Davide wants $+$ twice). But the following should work: Since $F(1) = 0 = m^+(1) - m^-(1)$, the measures $m^{\pm}$ have the same norm, so we can normalize to find two distinct nonzero probability measures measures whose moments-indexed-by-S agree by the choice of $F$. $\endgroup$ – Martin Aug 26 '13 at 16:54
  • $\begingroup$ @Martin thanks. (George Lowther suggested exactly the same, indepdendently, in comments to the main question) $\endgroup$ – Yemon Choi Aug 26 '13 at 16:59
  • $\begingroup$ @YemonChoi Indeed, my initial argument was not quite accurate. Thank you for point this out. I've edited. $\endgroup$ – Davide Giraudo Aug 26 '13 at 19:35
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    $\begingroup$ Moreover, I don't think the chain of logic in your last paragraph is quite right. I don't understand why you are fixing $\mu$ and $\nu$ beforehand. What you want to prove is: if $\sum_{n\in S} n^{-1}<\infty$ then there exist two distinct probability measures $\mu$ and $\nu$ such that $\mu$ and $\nu$ have the same moments on $S$. As Martin and George have pointed out, this follows from taking the two parts of the Hahn-Jordan decomposition of an annihilating measure. $\endgroup$ – Yemon Choi Aug 26 '13 at 20:08

If you are willing to move one dimension higher, then having infinitely many equal moments becomes rather weak condition. There are examples of uniform measures supported on finite polygons that have all their harmonic (a.k.a. complex) moments equal, yet the supports differ. See e.g. page 2 here.


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