It is known that a random variable $X$ which is symmetric about $0$ (i.e $X$ and $-X$ have the same distribution) must have all its odd moments (when they exist!) equal to zero. The converse is a strong temptation which has been around for perhaps a hundred years.

Question. Suppose $\mathbb E[X^n] = 0$ for all odd $n$. Is it true that $X$ is symmetric ?

This question was solved in Churchill (1946). In fact, he proved something much stronger

Theorem. Let $X$ be a random variable and let $(a_m)_{m \in \mathbb N}$ be a sequence of real numbers. Then for every $\epsilon > 0$, the exists a random variable $Y$ such that (1) $\mathbb E[Y^{2m+1}] = a_m$ for all $m \in \mathbb N$, and (2) The Kolmogorov distance between $X$ and $Y$ is at most $\epsilon$.

Of course this theorem immediately implies a negative answer to the above question.

The proof given in the paper is constructive, but somewhat mysterious.

Question. Is there simple / modern way to prove the above theorem using functional-analytic tools ?

After all, the theorem simply says (roughly) that the set of random variables of with odd-moments given by the sequence $(a_m)$ is dense in the space of random variables equipped with Komolgorov distance.

Note. The expected advantage of a general functional-analytic solution is that it would perhaps extend to constraints which are more general than those implied by prescribed odd moments.

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    $\begingroup$ Any reason for the anonymous downvote ? A comment would be much more useful. Thanks! $\endgroup$
    – dohmatob
    Feb 9, 2021 at 16:15
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    $\begingroup$ Seems like a sensible question to me. $\endgroup$ Feb 9, 2021 at 16:26
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    $\begingroup$ And to me too... $\endgroup$
    – Yemon Choi
    Feb 9, 2021 at 16:38
  • $\begingroup$ Here is one: telescoper.wordpress.com/2018/07/09/… $\endgroup$
    – Clement C.
    Feb 9, 2021 at 20:09
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    $\begingroup$ The solution in your link is referring to the same Churchill-Stieltjes construction I discussed in the preface to my question... $\endgroup$
    – dohmatob
    Feb 9, 2021 at 20:13

1 Answer 1


This doesn't really require modern functional analytic tools, but we can prove a statement (due originally to Edelheit, according to Jochen Wengenroth in the comments) like

Let $V$ be a Frechet space, complete with respect to a family of norms $||x||_i$, $i=0,1,\dots$. Let $L_1,L_2,\dots$ be linear forms on $V$, with $L_i$ bounded with respect to $||x||_i$ but unbounded with respect to any linear combination of $||x||_j$ for $j<i$. Then we can choose $x\in W$ with $L_i(x)$ arbitrary and $||x||_0$ arbitrarily small.

We can apply this by taking $V$ to be the completion of the space of smooth, compactly supported functions on $\mathbb R$ with respect to the set of norms $||f||_0 = \int |f(x)| dx$ and $||f||_m = \int |f(x)| |x|^{2m+1} dx$ for $m>0$, and $L_m(f) = \int f x^{2m+1} dx$. linear forms obtained by integrating against $x^{2n+1}$. This gives us a function $f(x)$ with desired odd moments and norm $\epsilon$, which we can make into a nonnegative function with the same moments and integral $\epsilon$ by the trick $g(x) =f^+(x) + f^-(-x)$. Then take $(1-\epsilon)$ of any measure plus $\epsilon$ times $g$ for suitable $g$.

To prove this: We fix a bunch of small constants $\epsilon_{ij}$, for $i,j \in \mathbb N$, to be chosen later.

After rescaling $L_1$, we may assume $|L_1(\cdot)| \leq ||\cdot||_1$.

Choose $x_1$ with $L_1(x_1)=1$ and $||x_1||_0 < \epsilon_{01}$ . Rescale $||\cdot||_2$ and $L_2$ so that $||x_1||_2<\epsilon_{21}$ and $|L_2( \cdot) |\leq ||\cdot||_2$ and then find $x_2$ with $||x_2||_0 < \epsilon_{02}$, $||x_2||_1< \epsilon_{12}$, and $L_2(x_2)=1$. Rescale $||\cdot ||_3$ and $L_3$ so that $||x_1||_3<\epsilon_{31}$, $||x_2||_3<\epsilon_{32}$, and $L_3(\cdot) \leq ||\cdot||_3$, and iterate this process.

Let $M$ be the $\mathbb N \times \mathbb N$ matrix with entries $M_{ij} = L_i (x_j)$. We have $|M_{ij} |\leq \epsilon ij$ if $i \neq j$ and $1$ if $i=j$. Let $N = I + (I-M) + (I-M)^2 + (I-M)^3+ \dots $ be the inverse matrix, choosing $\epsilon_{ij}$ small enough that this sum converges. Taking $\epsilon_{ij}$ as small as we want, we can make the off-diagonal entries of $N$ as small as desired and the diagonal entries as close to $1$ as desired.

Then letting $$x =\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} N_{ij} x_i a_j $$ we have $L_i(x) = a_i$ and $$ || x||_0 \leq \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |N_{ij}| ||x_i||_0 |a_j| \leq \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |N_{ij}| \epsilon_{0i} |a_j|$$ which we can make as small as desired, and for any $k>0$

$$ || x||_k \leq \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |N_{ij}| ||x_i||_k |a_j| \leq \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |N_{ij}| \epsilon_{ki} |a_j| +\sum_{j=1}^{\infty} |N_{kj}| ||x_k||_k |a_j|$$, where the first term can be made as small as desired and the second term can, at least, be made convergent.

For example $\epsilon_{ij} = 2^{-i-j}/ (|a_j|+1)$ for $i \neq j$ and $\epsilon_{0i} = 2^{-i} \cdot \epsilon / (|a_i|+1) $ will ensure $(N- I)_{ij}$ is bounded by $(3/2) 2^{-i -j} / (|a_j|+1)$ and so $||x||_0 = O(\epsilon)$ and $||x||_k = O ( \epsilon \cdot (1 + ||x_k||_k ) (1+ |a_k|) ) . $

The sign trick to make the function nonnegative is not strictly necessary. We can prove a similar statement like

Let $f_1,f_2,\dots$ be functions on a measure space. Assume that $$ \inf \frac{ f_i(x)}{1 + \sum_{j<i} |f_j(x)| } = - \infty $$ and $$ \sup \frac{ f_i(x)}{1 + \sum_{j<i} |f_j(x)| } = \infty. $$ Then there exists a measure $\mu$ with $\int f_i \mu$ arbirarily and $\int \mu$ arbitarily small.

We just have to keep track of two measures $\mu_i^+, \mu_i^-$, with $\int f_i \mu_i^+=1$, $\int f_i \mu_i^- =-1$, and the other integrals arbitrarily small.

  • $\begingroup$ Great piece of mathematics. Thanks (upvoted). Lemme try and parse it to be sure I follow everything in the constructions, then I'll accept it right away! $\endgroup$
    – dohmatob
    Feb 9, 2021 at 21:05
  • $\begingroup$ @ChristianRemling By suitable $g$ I mean one whose $2m+1$st moment is $a_m$ minus $1-\epsilon$ times the $2m+1$st moment of "any measure". $\endgroup$
    – Will Sawin
    Feb 9, 2021 at 23:32
  • $\begingroup$ Small nitpick/request for clarification: when you say unbounded linear form on a Banach space I guess you mean a densely-defined linear map which is a closed operator, or something similar? To get discontinuous everywhere-defiined linear forms on a Banach space one needs something like the axiom of choice $\endgroup$
    – Yemon Choi
    Feb 9, 2021 at 23:40
  • $\begingroup$ @YemonChoi Good point - I was trying to use the convention that an unbounded linear operator need not be defined on the whole Banach space, but I need to assume that the operators are closed and something about the space of definition. I'm not sure if dense is enough because I need them all to be defined at some points. I'll revise... $\endgroup$
    – Will Sawin
    Feb 9, 2021 at 23:51
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    $\begingroup$ This general result about Fréchet spaces is called Eidelheit theorem. It is, e.g., in the book Introduction to Functional Analysis by Meise and Vogt. $\endgroup$ Feb 10, 2021 at 6:32

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