proof that the sequence satisfies weak law of large numbers

Question

strong textHow to check if this sequence satisfies the weak law of large numbers? $$P(X_n=1)= {{1}\over{3^n}} ,P(X_n=-1)= {{1}\over{3^n}}, P(X_n=0)= {1 -{{2}\over{3^n}}} $$ I calculated that: $$E(X_n)=0 $$ so according to the definition of this law: $${\sum_{n=1}^{N}{(X_n-0)}}\over{N} $$and from this one we can deduce that this expression converges to 0 so the sequence satisfies the weak law of large numbers. Is that correct?

Answer 1

Simply using the definition of WLLN

$$\lim\limits_{n\to +\infty}\mathbb{P}[|X_n|>\epsilon]=\frac{2}{3^n}=0$$

this means that

$$X_n\xrightarrow{\mathcal{P}}0$$

This if $\epsilon$ is chosen $|\epsilon|\leq1$. If it is chosen >1 the probability is exactly zero. Anyway the sequence converges.

Answer 2

$\sum P(X_n \neq 0)=\sum \frac 2 {3^{n}}<\infty$. By Borel Cantelli Lemma $P(\lim \sup (X_n \neq 0))=0$. Hence, with probaility $1$, $X_n=0$ for all $n$ suffciently large. But this implies that the averages $\frac 1n \sum\limits_{k=1}^{n} X_k$ tends to $0$ with probability $1$. Thus, we actually have strong law for this sequence.