# Proving a sequence of random variables satisfies the weak law of large numbers

I'm trying to understand this exercise:

Let $X_1,X_2,...$ be a sequence of random variables such that $\max_{1 \leq k \leq n} \{|X_k|\} \to_{\mathbb{P}} 0$. Show that this sequence satisfies the weak law of large numbers.

If I understood this law correctly, then I need to prove that $\frac{X_1+...+X_n}{n} \to_{\mathbb{P}} \rm{E}(X_n)$. but my teacher solved this problem proving that $\rm{P}(|\frac{X_1+...+X_n}{n}| > \epsilon) \to 0$. How does this solve the problem? Wasn't it supposed to be proved that $\rm{P}(|\frac{X_1+...+X_n}{n} - \rm{E}(X_n)| > \epsilon) \to 0$

• Uh, are you sure you copied the condition correctly? The sequence $\max_{k\leq n} |X_k|$ is monotonically increasing, so unless your random variables are all 0 a.s. this seems to be a strange condition. Also independent from that, your teacher might have simply recentered the random variables. – E-A Jun 14 '18 at 23:14
• do you mean $\max_{k \geq n} |X_k| \to 0$ in prob? – Daniel Xiang Jun 15 '18 at 1:29
• That's how the exercise is written, but I think it's a typo – creepyrodent Jun 15 '18 at 11:17

## 1 Answer

Your teacher's conclusion is right if you replace the condition with "$\max_{k\geq n}|X_k|\overset{P}{\to}0$". We pick $\delta>0$ and $\epsilon>0$ arbitrarily first.

"$\max_{k\geq n}|X_k|\overset{P}{\to}0$" implies that $$P(\bigcup_{k\geq n}\{|X_k|\geq\delta\})\leq P(\{\max_{k\geq n}|X_k|\geq\delta\})\to0,\ n\to\infty.$$

Choose a $N$ such that $P(\bigcup_{k\geq N}\{|X_k|\geq\delta\})\leq\epsilon/2$.

\begin{eqnarray} P\left(|\frac{1}{n}\sum_{k=1}^{n}X_k|\geq\delta\right)&\leq& P\left(|\frac{1}{n}\sum_{k=1}^{N-1}X_k|\geq\delta\right)+P\left(|\frac{1}{n}\sum_{k=N}^{n}X_k|\geq\delta\right)\\ &\leq&P\left(|\frac{1}{n}\sum_{k=1}^{N-1}X_k|\geq\delta\right)+P\left(\bigcup_{k\geq N}\{|X_k|\geq\delta\}\right)\\ &\leq& \epsilon\quad\text{(when $n$ is big enough)}. \end{eqnarray}

The key to understand this is that $\omega\in\Omega$ where $X_k(k\geq n)$ are relatively large will overall squeeze in a small set when $n\to\infty$.

• can you please explain why you didn't do $P(|1/n \sum X_k - E(X_n) | \geq \delta) \leq \epsilon$? My main doubt is what happened to $E(X_n)$ in this exercise, because I don't see the WLLN in this... – creepyrodent Jun 15 '18 at 11:22
• @dude3221 Actually, we don't even know if $E(X_n)$ does exist or not. Think if $X_n=1/x\cdot\boldsymbol{I}_{(0,1/n]}$, when $x\in[0,1]$, then $E(X_n)=+\infty$ for every $n$. However, $X_n\overset{P}{\to}0$ in this example. – XIAODA QU Jun 15 '18 at 11:54
• @dude3221 the weak convergence law usually holds when the expectations are uniformly bounded by something, but in this problem they're not. If you add up a condition to bound the expectations, then you may find $E(X_n)\to0$, which will be the same with the conclusion. – XIAODA QU Jun 15 '18 at 12:00