10

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.

Sangchul Lee
  • 181,930
  • 2
    Here is a systematic way to find such an example. Consider a normal random variable $Z$ with mean $a$ and variance $\sigma^2$. In order to satisfy your constraints, you want $a=\sigma^2+a^2=3a\sigma^2+a^3$. Solving these equations basically amounts to solving a quadratic equation. The unique answer is given by $a=\sigma=1/2$. – shalin Feb 08 '18 at 02:43
  • 1
    @Shalop, Thank you for the comment. It really teaches me that I should play with well-known examples first. – Sangchul Lee Feb 08 '18 at 02:44

1 Answers1

5

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

Math Lover
  • 15,483
  • Oh yeah, I somehow suspected I was missing a simple counter-example, but I never expected this to be that simple :( Thank you! – Sangchul Lee Feb 08 '18 at 02:37
  • This raises another question: can we classify the family of all such distributions? I guess a trivial comment is that it is closed under mixtures, i.e. convex, so we could ask about extreme points. – Nate Eldredge Feb 08 '18 at 02:45
  • 1
    @NateEldredge I suspect not, because you can take an arbitrary symmetric rv $X$ with mean zero, variance $1$, and finite third moment; then $\frac{1}{2}(1+X)$ will satisfy this. Are there others though? Idk. – shalin Feb 08 '18 at 02:48
  • @NateEldredge What might be interesting is the generalization: for which values of $n$ does $E[X^n] = E[X^{n+1}] = E[X^{n+2}]$ uniquely determine the distribution? – Cameron L. Williams Feb 08 '18 at 02:49
  • @CameronWilliams When $n=2k$ is even this is always true since $E[X^{2k} (1-X)^2]=0$. When it is odd I suspect it is false, perhaps a Gaussian might again provide a counterexample. But idk. – shalin Feb 08 '18 at 02:53
  • @Shalop Definitely on even $n$. It would be interesting to see if there is a simple approach for odd $n$. – Cameron L. Williams Feb 08 '18 at 02:56