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.



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> **Question.** *Suppose $\mathbb E[X^n] = 0$ for all odd $n$. Is it true that $X$ is symmetric ?*
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This question was solved in [Churchill (1946)](https://doi.org/10.1214/aoms/1177730987). In fact, he proved something much stronger



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> **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 $\epsilon$.*
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Of course this theorem immediately implies a negative answer to the above question.


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



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



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



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> 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.
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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



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> 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.
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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.


