# Weak Law of Large Numbers

## Theorem

Let $P$ be a population.

Let $P$ have mean $\mu$ and finite variance.

Let $\sequence {X_n}_{n \mathop \ge 1}$ be a sequence of random variables forming a random sample from $P$.

Let:

$\ds {\overline X}_n = \frac 1 n \sum_{i \mathop = 1}^n X_i$

Then:

${\overline X}_n \xrightarrow p \mu$

where $\xrightarrow p$ denotes convergence in probability.

## Proof

Let $\sigma$ be the standard deviation of $P$.

By the definition of convergence in probability, we aim to show that:

$\ds \lim_{n \mathop \to \infty} \map \Pr {\size { {\overline X}_n - \mu} < \epsilon} = 1$

for all real $\epsilon > 0$.

Let $\epsilon > 0$ be a real number.

$\var {{\overline X}_n} = \dfrac {\sigma^2} n$

By the Bienaymé-Chebyshev Inequality, we have for real $k > 0$:

$\map \Pr {\size { {\overline X}_n - \mu} \ge \dfrac {k \sigma} {\sqrt n}} \le \dfrac 1 {k^2}$

As $\sigma > 0$ and $n > 0$, we can set:

$k = \dfrac {\sqrt n} {\sigma} \epsilon$

This gives:

$\map \Pr {\size {{\overline X}_n - \mu} \ge \epsilon} \le \dfrac {\sigma^2} {n \epsilon^2}$

We therefore have:

 $\ds \map \Pr {\size { {\overline X}_n - \mu} < \epsilon}$ $=$ $\ds 1 - \map \Pr {\size { {\overline X}_n - \mu} \ge \epsilon}$ $\ds$ $\ge$ $\ds 1 - \frac {\sigma^2} {n \epsilon^2}$

So:

$1 - \dfrac {\sigma^2} {n \epsilon^2} \le \map \Pr {\size { {\overline X}_n - \mu} < \epsilon} \le 1$

We have:

$\ds \lim_{n \mathop \to \infty} \paren {1 - \dfrac {\sigma^2} {n \epsilon^2} } = 1$

and:

$\ds \lim_{n \mathop \to \infty} 1 = 1$

So by the Squeeze Theorem:

$\ds \lim_{n \mathop \to \infty} \map \Pr {\size { {\overline X}_n - \mu} < \epsilon} = 1$

for all real $\epsilon > 0$.

$\blacksquare$