# Functional-analytic proof of the existence of non-symmetric random variables with vanishing odd moments

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.

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

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 $$j. Then we can choose $$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 $$\inf \frac{ f_i(x)}{1 + \sum_{j and $$\sup \frac{ f_i(x)}{1 + \sum_{j 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.

• 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! Feb 9, 2021 at 21:05
• @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". Feb 9, 2021 at 23:32
• 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 Feb 9, 2021 at 23:40
• @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... Feb 9, 2021 at 23:51
• 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. Feb 10, 2021 at 6:32