Question

Led tin eperaje

Law of Large Number↓

Answer 1

Using Chebyshev's inequality we can get an upper bound for the probability. Here $X_1,X_2,...,X_n$ are IID random variables with $E\left ( X_i \right )=\mu ,Var\left ( X_i \right )=\sigma ^2$

Chebyshev's inequality is $P\left ( \left | X-\mu \right | \geqslant t\right )\leqslant \frac{\sigma ^2}{t^2}$ .

The standard deviation of the mean is $E\left ( \overline{X} \right )=\mu ,Var\left ( \overline{X} \right )=\sigma ^2/n$

This inequality can be written as

$1-P\left ( \left | \frac{S_n-np}{n}\right | \geqslant t\right )\geqslant 1 - \frac{\sigma ^2}{t^2}\\ P\left ( \frac{S_n}{n}-p\leqslant t \right )\geqslant 1-\frac{\sigma ^2}{t^2}\\$

When the $t$ ids small $t<\epsilon$,

$1-P\left ( \left | \frac{S_n-np}{n}\right | \geqslant t\right )\geqslant 1 - \frac{ s^2}{t^2}\\ P\left ( \left |\frac{S_n}{n}-p \right |\leqslant \epsilon \right )\geqslant 1-\frac{\sigma^2/n}{\epsilon^2}\\$

Now as $n\rightarrow \infty$,

$\lim_{n\rightarrow \infty }P\left ( \left |\frac{S_n}{n}-p \right |\leqslant \epsilon \right )\geqslant \lim_{n\rightarrow \infty }1- \lim_{n\rightarrow \infty }\frac{\sigma^2/n}{\epsilon^2}\\ P\lim_{n\rightarrow \infty }P\left ( \left |\frac{S_n}{n}-p \right |\leqslant \epsilon \right )\geqslant \lim_{n\rightarrow \infty }1=1$

But since , the probability is always <= 1, we can write,

$\lim_{n\rightarrow \infty }P\left ( \left |\frac{S_n}{n}-p \right |\leqslant \epsilon \right )=1$

Thus the proof is complete.