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

Became Hot Network Question

occurred Feb 9, 2021 at 23:56

added 3 characters in body

Source Link

edited Feb 9, 2021 at 19:46

6.9k
1
18
76

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.

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 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.

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.

added 21 characters in body

Source Link

edited Feb 9, 2021 at 19:27

dohmatob

6.9k
1
18
76

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 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 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 evenconstraints which are more general constraintsthan those implied by prescribed 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 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 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 even more general constraints those prescribed 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 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.

deleted 7 characters in body

Source Link

edited Feb 9, 2021 at 18:00

dohmatob

6.9k
1
18
76

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 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 statementtheorem using functional-analytic tools ?

After all, the above theorytheorem simply says (roughly) that the set of random variables of with odd-moments given by the sequence $(a_m)$ is dense in the space 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 even more general constraints those prescribed 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 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 statement using functional-analytic tools ?

After all, the above theory simply says (roughly) that the set of random variables of with odd-moments given by the sequence $(a_m)$ is dense in the space 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 even more general constraints those prescribed 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 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 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 even more general constraints those prescribed odd moments.