91Ó°ÊÓ

\({\bar X_n} = \frac{1}{n}\sum\limits_{i = 1}^n {{X_i}} \)be the sample mean of the first n random variables in the sequence. We need to prove that \({\mathop{\rm P}\nolimits} \left( {\left| {{X_n} - \mu } \right| \le c} \right)\) converges to 1 as n → ∞.

The sample mean \({\bar X_n}\)has a normal distribution with a mean \(\mu \)and standard deviation

\(\frac{\sigma }{{\sqrt n }}\). This implies that \(\left| {\frac{{{{\bar X}_n} - \mu }}{{\frac{\sigma }{{\sqrt n }}}}} \right|\)has a standard normal distribution.

\(\begin{aligned}{}P\left( {\left| {{X_n} - \mu } \right| \le c} \right) &= P\left( {\left| {\frac{{{X_n} - \mu }}{{\frac{\sigma }{{\sqrt n }}}}} \right| \le \frac{c}{{\frac{\sigma }{{\sqrt n }}}}} \right)\\ &= P\left( {\left| {\frac{{{X_n} - \mu }}{{\frac{\sigma }{{\sqrt n }}}}} \right| \le \frac{{c\sqrt n }}{\sigma }} \right)\end{aligned}\)

Now, as n → ∞, also \(\frac{{c\sqrt n }}{\sigma } \to \infty \) .This implies that we are sure that \(\left| {\frac{{{{\bar X}_n} - \mu }}{{\frac{\sigma }{{\sqrt n }}}}} \right|\)\( \le \frac{{c\sqrt n }}{\sigma }\)as n → ∞.

Hence, \({\mathop{\rm P}\nolimits} \left( {\left| {{X_n} - \mu } \right| \le c} \right)\) converges to 1 as n → ∞.