0
$\begingroup$

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?


Advanced probability stuff:

$X$ is a random variable in $\mathcal (\Omega, \mathscr F, \mathbb P)$.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Lorenzo May 12 '18 at 21:56
  • $\begingroup$ @AreaMan Ah thanks. I think I can find it somewhere in Royden Fitzpatrick. $\endgroup$ – BCLC May 12 '18 at 22:03
-1
$\begingroup$

Another way to say what you showed: the L2 norm of Xn is zero, so Xn 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 at 21:56

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.