# “almost surely” convergence of a special sequence of random variables

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?

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

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.