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.

  • 1
    $\begingroup$ One can prove the wlln using Chebyshev's inequality, and one doesn't need any nontrivial information about the shape of the distribution. $\endgroup$ Nov 16, 2013 at 17:36
  • $\begingroup$ Yes. But it is asked for how to derive it from the much stronger central limit theorem.. $\endgroup$
    – user109265
    Nov 16, 2013 at 17:44
  • $\begingroup$ I understand the question. What I don't understand is why the CLT should be of much use here. $\endgroup$ Nov 16, 2013 at 17:48

1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy