3
$\begingroup$

Let $X$ be a random variable with all moments equal to $p\in(0,1)$, i.e. $\mathbb{E}[X^{n}] = p$ for all $n\geq 1$. How to show rigorously it must be $\operatorname{Bernoulli} (p)$?

$\endgroup$
5
$\begingroup$

New Answer. Assume that $X$ is a $\mathbb{R}$-valued random variable such that $\mathbb{E}[X^2] = \mathbb{E}[X^3] = \mathbb{E}[X^4]$. Then

$$ \mathbb{E}[(X - X^2)^2] = \mathbb{E}[X^2] - 2\mathbb{E}[X^3] + \mathbb{E}[X^4] = 0 $$

and hence $X = X^2$ with probability $1$. This implies that $\mathbb{P}(X \in \{0,1\}) = 1$ and therefore $X$ is a Beronulli random variable with parameter $p = \mathbb{E}[X] = \mathbb{E}[X^2]$.


Old Answer. It is probably another overkill, but notice first that $X$ is a bounded variable:

$$ \|X\|_{\infty} = \lim_{n\to\infty} \mathbb{E}[X^{2n}]^{1/2n} = 1 $$

and hence we can apply the Fubini's theorem to conclude

$$ \mathbb{E}[e^{itX}] = \mathbb{E}\left[ \sum_{n=0}^{\infty} \frac{(it)^n}{n!}X^n \right] = \sum_{n=0}^{\infty} \frac{(it)^n}{n!} \mathbb{E}[X^n] = (1-p) + pe^{it}. $$

Since the characteristic function determines the distribution, we must have $X \sim \operatorname{Bernoulli}(p)$.

$\endgroup$
2
$\begingroup$

It's probably overkill, but this follows from Carleman's condition: $\sum_{n>1}(EX^{2n})^{-1/2n}=+\infty$ so the measure is unique.

$\endgroup$
1
  • $\begingroup$ Feeling foolish not to have seen the Cauchy-Schwarz argument earlier. $\endgroup$ Sep 12 '17 at 12:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy