What happens when $X$ equals its own expected value?

Question

Let $X$ denote a random variable with “moments” $M_1:=E(X)$, $M_2:=E(X^2)$, . . . (the first four of which, at least, are assumed to be finite).

Show that $M_4+6M_2(M_1)^2\ge 4M_3(M_1)+3(M_1)^4$

Under what circumstances would you get equality?

I figured out the first part which is just basically $E((X-E(X))^4)\ge 0$. But the second part when $E((X-E(X))^4)=0$ I suppose this makes $X-E(X)=0$ and I got $X=E(X)$ which I'm not sure what this means. Does that mean all possible $X$ are the same with the same probability? The answer might be simple but somehow I am confused.

Doesn't $X=E(X)$ mean that $X$ is a constant? – bof May 13 '16 at 04:34 — bof, May 13 '16 at 04:34
@bof given that $E(X)$ is a scalar, yes, yes it does. – DRF May 13 '16 at 04:40 — DRF, May 13 '16 at 04:40

score 1 · Accepted Answer · answered May 13 '16 at 04:57

I got $X=E(X)$ which I'm not sure what this means. Does that mean all possible $X$ are the same with the same probability? The answer might be simple but somehow I am confused.

Yes, that is what it means. It is indeed that simple. $\mathsf E(X)$ is a constant. $X$ being constant means that it has a deterministic (aka degenerate) distribution — which means: $\mathsf P(X=\textsf{constant})=1$ and $\mathsf P(X\neq \textsf{constant})=0$.

So when and only when $X$ is constant, then $M_4 + 6M_2M_1^2 = 4M_3M_1 + 3M_1^4$ ; and if and only otherwise, then $M_4 + 6M_2M_1^2 \gt 4M_3M_1 + 3M_1^4$.

What happens when $X$ equals its own expected value?

1 Answers1