# Even moments = 0 iff $X=0$?

Kind of a dumb question. I wasn't able to find this in probability texts, either elementary (Larsen and Marx) or advanced (David Williams), so it's probably wrong.

Is this false?

Let $X$ be a random variable s.t. $E[X^n] < \infty \ \forall n \in \mathbb N$.

$$(\exists n \in \mathbb N, E[X^{2n}] = 0) \iff X=0 \ \text{a.s.}$$

Pf:

'if' $\leftarrow$

is obvious, and the conclusion can be strengthened from $\exists$ to $\forall$.

'only if' $\rightarrow$

$X^{2n} \ge 0$. By Markov's inequality, $\forall a > 0$,

$$P(X^{2n} \ge a) \le \frac{E[X^{2n}]}{a} = 0$$

$$\iff P(X^{2n} < a) = 1 \ \forall a > 0$$

$$\iff P(X^{2n} = 0) = 1$$

$$\iff P(X = 0) = 1$$

QED

Actually the assumption strengthens itself again from $\exists$ to $\forall$.

Is that wrong?

Also, what probability texts, either elementary or advanced, mention anything like this please?

$X$ is a random variable in $\mathcal (\Omega, \mathscr F, \mathbb P)$.
• Another way to say what you showed: the L2 norm of $X^n$ is zero, so $X^n$ is zero almost surely, so the same is true for X. So the 'well known' fact at the center is the the L^2 norm is a norm (which is more than what you showed). Maybe look at any book about lp spaces? Nice proof though, it's different from the usual one if I recall correctly. – Lorenzo May 12 '18 at 21:56