Let $X_i$ be iid random variables and let $\overline{X}_n=(X_1+\cdots+X_n)/n$. If $EX_i=\mu$ and $\operatorname{Var}X_i = \sigma^2$ then the central limit theorem says that with some conditions we have convergence in some sense

$$\frac{\overline{X}_n -\mu}{\sigma/\sqrt{n}}\to N(0,1)$$

On the other hand the law of large numbers says that if $X_i$ are iid random variables with $EX_i=\mu$, then


Can the central limit theorem be used to prove a form of the law of large number? (obviously weakened since it would have the assumption of existence of $\operatorname{Var} X_i$). Since we expect $\frac{\overline{X}_n -\mu}{\sigma/\sqrt{n}}$ to look like $N(0,1)$ for large $n$, shouldn't we expect $\overline{X}_n -\mu$ to look like $N(0,\sigma^2/n)$. So then the variance of $\overline{X}_n -\mu$ being $\sigma^2/n$ means that the variance is becoming very small and $\overline{X}_n -\mu$ is focusing around $0$ and so something like the law of large numbers should hold.

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    $\begingroup$ I think you can deduce the weak law (for finite variance) but not the strong law...? $\endgroup$ – Qiaochu Yuan May 30 '13 at 0:34

This argument works, but in a sense it's overkill. You have a finite variance $\sigma^2$ for each observation, so $\operatorname{var}\left(\overline{X}_n\right)=\sigma^2/n$. Chebyshev's inequality tells you that $$ \Pr\left(\left|\overline{X}_n - \mu\right|>\varepsilon\right) \le \frac{\sigma^2}{\varepsilon^2 n} \to 0\text{ as }n\to\infty. $$ And Chebyshev's inequality follows quickly from Markov's inequality, which is quite easy to prove.

But the proof of the central limit theorem takes a lot more work than that.


Here is an elementary argument that shows that the central limit theorem (CLT) - actually something weaker stated below - implies the associated weak law of large numbers.

Assume that the following holds

$W_{n}:= \sqrt{n} (\frac{1}{n}\sum_{i=1}^{n} X_{i} - \mu) \Rightarrow W$.

where $\Rightarrow$ denotes convergence in distribution to some distribution (if the CLT is assumed then $W$ is distributed as a normal r.v.). By Slutsky's Theorem the above gives $n^{-1/2} W_{n} \Rightarrow 0$, whenever $W$ is almost surely finite. Of course, convergence in distribution to a constant is known to imply convergence in probability, so that you can retrieve the weak law of large numbers from the CLT.

The idea is that the CLT tells you the scaling factor for "blowing up" the oscillations around the mean. Hope this helps!


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