The weak law of large numbers from the central limit theorem.

Question

I could not derive the weak law of large numbers from the central limit theorem for i.i.d. random variables with $0 < \operatorname{Var}(X) < \infty$.

The central limit theorem gives $$\frac{\sum X_i - n\cdot E(X_1)}{\sqrt{\operatorname{Var}(X_i)n})} \to N(0,1)$$

Now I want to show

$$\lim P\left(\middle|\frac{\sum X_i - E(X_1)}{n}\middle| \geq \epsilon\right) = 0 \quad\forall \epsilon>0$$

I know that the cdf converges pointwise. It should be really easy as it is intuitively clear that it should hold, but I can't find the correct way to write and pin it down.

Answer 1

For every positive $t$, consider the events $A_n^t=[|S_n-nE[X]|\geqslant t\sqrt{n\mathrm{var}(X)}]$ and $B_n^t=[|S_n-nE[X]|\geqslant nt]$. Then:

1. The CLT says that $P[A_n^t]\to P[|Z|\geqslant t]$ when $n\to\infty$, for every positive $t$, where $Z$ is standard normal.
2. The WLLN says that $P[B_n^t]\to0$ when $n\to\infty$, for every positive $t$.
3. For every positive $t$ and $s$, there exists some finite $N$ such that $B_n^t\subseteq A_n^s$ for every $n$ large enough (can you show this?).

With these elementary facts in mind, assume that 1. holds, that is, that the CLT is valid and consider some positive $t$ and $\varepsilon$. Then:

• There exists some finite $s$ such that $P[|Z|\geqslant s]\leqslant\varepsilon$.
• By 1., this implies that $P[A_n^s]\leqslant2\varepsilon$ for every $n$ large enough, say $n\geqslant N_1$.
• By 3., $P[B_n^t]\leqslant P[A_n^s]$ for every $n$ large enough, say $n\geqslant N_2$.

Thus, for every $n\geqslant\max(N_1,N_2)$, $P[B_n^t]\leqslant2\varepsilon$. Since $\varepsilon$ is arbitrary, this is the WLLN.