2
$\begingroup$

Let $X_1,X_2,\dots$ be a sequence of equally distributed random variables.

Suppose that $\forall n\in\mathbb{N}:\mathbb{E}[X_n]=0$ and $\lim\limits_{n\to\infty}\mathbb{E}[X_n^4]=0$.

Prove or Disprove: $X_n\stackrel{\text{a.s.}}\to0$.

I feel that the statement is true and I tried to prove it.

I said that since the random variables are all equally distributed then the random variables $|X_1|,|X_2|,\dots$ are also equally distributed.

So, $\forall n,m\in\mathbb{N}:\mathbb{E}[|X_n|]=\mathbb{E}[|X_m|]$. Let's call this value $S$.

Since $\lim\limits_{n\to\infty}\mathbb{E}[X_n^4]=0$, it must be that $S=0$ (not sure about this).

Therefore, $P[X_n=0]=1$ and $X_n\stackrel{\text{a.s.}}\to0$.

Is it true?

$\endgroup$
  • $\begingroup$ Are you sure you have your hypothesis right? Does "equally distributed" mean "identically distributed"? $\endgroup$ – kimchi lover Dec 22 '17 at 18:51
  • $\begingroup$ I have translated it from another language. I think that they mean "equally distributed". Otherwise the sequence is constant, right? @kimchilover $\endgroup$ – Don Fanucci Dec 22 '17 at 18:53
  • $\begingroup$ The implication "$P(X_n=0)=1$" is false. Note that $X_n=\pm 1/n$ with probability 1/2 works. $\endgroup$ – Martín Vacas Vignolo Dec 22 '17 at 18:55
  • $\begingroup$ What is your definition of "equally distributed"? $\endgroup$ – Robert Israel Dec 22 '17 at 18:57
  • $\begingroup$ I think $\forall t\in\mathbb{R}:F_X(t)=F_Y(t)$. @RobertIsrael $\endgroup$ – Don Fanucci Dec 22 '17 at 18:58
0
$\begingroup$

As mentioned in the comments we have $E[X_n^4]=0$, i.e. \begin{align} \int_\Omega |X_n|^4 \,dP = 0 \end{align} And that means $X_n=0$ a.s.

$\endgroup$

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.