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' is obvious, and the conclusion can be strengthened from $\exists$ to $\forall$.
'only if' $X^{2n} \ge 0$. By Markov's inequality, $\forall a > 0$,
$$P(X^{2n} \ge a) \le \frac{E[X^{2n}]}{a} = 0$$
$$ \iff P(X^{2n} < a) = 0 \ \forall a > 0$$
$$ \iff P(X^{2n} = 0) = 1$$
$$ \iff P(X = 0) = 1$$
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?
Advanced probability stuff:
$X$ is a random variable in $\mathcal (\Omega, \mathscr F, \mathbb P)$.
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' is obvious, and the conclusion can be strengthened from $\exists$ to $\forall$.
'only if' $X^{2n} \ge 0$. By Markov's inequality, $\forall a > 0$,
$$P(X^{2n} \ge a) \le \frac{E[X^{2n}]}{a} = 0$$
$$ \iff P(X^{2n} < a) = 0 \ \forall a > 0$$
$$ \iff P(X^{2n} = 0) = 1$$
$$ \iff P(X = 0) = 1$$
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?
Advanced probability stuff:
$X$ is a random variable in $\mathcal (\Omega, \mathscr F, \mathbb P)$.
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