# Weak law of large numbers where $P(X_n = \pm 2^n) = \frac{1}{2^{2n+1}}$

Let $X_1,X_2,...$ be a sequence of random variables such that $P(X_n = \pm 2^n) = \frac{1}{2^{2n+1}}$ and $P(X_n = 0) = 1 - \frac{1}{2^{2n}}$. I'm supposed to show that this sequence satisfies WLLN, but isn't $Var(X_n) = E(X_n^2) = \sum_{n} \frac{2^{2n}}{2^{2n+1}} = \sum_{n} \frac{1}{2} = \infty$, so the WLLN can't be applied?

I already verified $E(X_n)=0$.

• To check whether the WLLN holds, just check the definition of it and it probably is clear that it works (per definition of $X_n$, essentially since $2^{-2n}\to 0$). As I recall, the assumption of finite variance (in or outside of the limit) is not necessary for the WLLN to apply. – Stan Tendijck Jun 20 '18 at 23:11
• that's odd, in my book one of the hypothesis is "variance uniformly limited" – creepyrodent Jun 21 '18 at 0:19
• Two things: (1) the moment conditions for WLLNs generally depend on the strength of other assumptions such as stationarity and independence and (2) when you say WLLN, do you mean $X_n \to^p 0$ or $(X_1 + ... + X_n)/n \to^p 0$? – Galton Jun 21 '18 at 2:18
• (1) and in this case there is stationarity or independence? I'm failing to see... (2) The latter – creepyrodent Jun 21 '18 at 17:20
• (1) there is no stationarity since the variance changes but there is independence (since there is nothing stated that should suggest otherwise) – Stan Tendijck Jun 23 '18 at 9:30

$P\{X_n \neq 0\}=\frac 1 {2^{2n}}$. Hence $\sum_n P\{X_n \neq 0\}<\infty$. This implies that the $\limsup$ of the events $\{X_n \neq 0\}$ (which is set of points that belong to infinitely many of these events) has probability $0$. Hence $X_n \to 0$ almost surely which implies $\frac {X_1+X_2+...+X_n} n \to 0$ almost surely. This proves a much stronger result without using any major theorem.