Does the condition $E[X]=E[X^2]=E[X^3]$ determine the distribution of $X$?

This is a question out of pure curiosity, motivated by this posting. Here I checked that if a $\mathbb{R}$-valued random variable $X$ has finite $4$-th moment and $E[X^2]=E[X^3]=E[X^4]$ then $X$ is a Bernoulli random variable. Indeed, this follows by noticing that

$$ E[X^2(1-X)^2] = E[X^4] - 2E[X^3] + E[X^2] = 0 $$

and hence $X(1-X) = 0$ a.s., i.e. $P(X \in \{0, 1\}) = 1$.

So we may ask whether the requirement can be relaxed, by asking the following question.

Question. If a $\mathbb{R}$-valued random variable $X$ satisfies $E[X] = E[X^2] = E[X^3]$, then does it follow that $X$ has a Bernoulli random variable? I.e,. $P(X = 0 \text{ or } 1) = 1$?

• Notice that there are examples of $X$ where $E[X] = E[X^2]$ but $X$ does not have a Bernoulli distribution. (For instance, Let $Z \sim \mathcal{N}(0, 1)$ and set $X = \frac{1}{2}(1+Z)$.) So the 3rd moment condition cannot be dropped.

• If $X$ satisfies the given condition, so does $1-X$.

• If we assume $X \geq 0$, then the answer is yes, since $X(X-1)^2 \geq 0$ and $E[X(X-1)^2] = 0$.

I tried creating a discrete random variable (other than Bernoulli r.v.s) that satisfy the condition, but this approach have not been successful so far. And to be honest I am not sure whether this will be true or not.

Solution 1:

You have figured out an example by yourself. Indeed, if $X = \frac{1}{2}(1+Z)$, where $Z \sim \mathcal{N}(0, 1)$, then $$E[X^3] = \frac{1}{8}(1+3E[Z]+3E[Z^2]+E[Z^3])=\frac{1}{2}=E[X]=E[X^2].$$