# Central Limit Theorem

Consider a sequence of independent and identically distributed random variables with finite mean $\mu$ and finite variance $\sigma ^{2}$ .

Consider also the sample mean

$M_{n}={1 \over {n}}\sum _{i=1}^{n}{x_{i}}$ the sum
$S_{n}=x_{1}+x_{2}+...+x_{n}$ and the sequence of variables
$Z_{n}={{M_{n}-\mu } \over {\sigma /{\sqrt {n}}}}={{S_{n}-n\mu } \over {\sigma {\sqrt {n}}}}$ Then $Z_{n}{\xrightarrow {D}}Z$ , where $Z\sim {N}(0;1)$ .

Alternatively:

• $\lim _{n\to \infty }P(Z_{n}\leq {x})=\Phi (x)$ • $\lim _{n\to \infty }P(M_{n}\leq \mu +x{\sigma \over {\sqrt {n}}})=\Phi (x)$ • $\lim _{n\to \infty }P(S_{n}\leq {n}\mu +x{\sigma {\sqrt {n}}})=\Phi (x)$ For n sufficiently large ($n\geq 30)\Longrightarrow {P}(Z_{n}\leq {x})\approx \Phi (x)$ • Standardization: $Z_{n}\approx {N}(0;1)$ $n\leq 30$ • Mean: $M_{n}\approx {N}(\mu ;{\sigma ^{2} \over {n}})$ • Sum sequence: $S_{n}\approx {N}(n\mu ;n\sigma ^{2})$ Corollary:

Consider the sequence

$Z_{n}^{*}={{M_{n}-\mu } \over {\sigma _{n}/{\sqrt {n}}}}={{S_{n}-n\mu } \over {\sigma _{n}{\sqrt {n}}}}$ where $\sigma ^{2}$ is a sample function such that $\sigma _{n}{\xrightarrow {a.s.}}\vartheta$ as $n\rightarrow \infty$ . Then $Z_{n}^{*}{\xrightarrow {D}}Z$ where $Z\sim {N}(0;1)$ .

## Applications of the Central Limit Theorem

### Approximation of binomial distribution

For the approximation of a binomial distribution we use the correction of continuity which is 0.5.

$Y_{n}\sim (n,\theta )$ $P(Z_{n}\leq x)=\Phi (x)$ {\begin{aligned}P(Y_{n}\leq x)&=P{\Bigl (}{Y_{n}-n\theta \over {\sqrt {n\theta (1-\theta )}}}\leq {x-n\theta \over {\sqrt {n\theta (1-\theta )}}}{\Bigr )}\\&=P{\Bigl (}Z_{n}\leq {x-n\theta \over {\sqrt {n\theta )1-\theta )}}}{\Bigr )}\\&\approx \Phi {\Bigl (}{x-n\theta \pm \overbrace {0.5} ^{\text{correction of continuity}} \over {\sqrt {n\theta (1-\theta )}}}{\Bigr )}\end{aligned}} $P(Y_{n}=y)=0$ {\begin{aligned}P(Y_{n}\approx y)&\simeq P(y-0.5 ### Assessment of the error of the sample mean

For assessing the error of the sample mean we consider the strong law of large numbers $M_{n}{\xrightarrow {a.s.}}\mu$ and the CLT${\Bigl (}M_{n}\sim N(\mu ;{\sigma ^{2} \over n}){\Bigr )}$ , as well as :

• Margin of error:
{\begin{aligned}0.9974&=\Phi (3)-\Phi (3)\\&=P{\Bigl (}-3\leq {M_{n}-\mu \over \sigma /{\sqrt {n}}}\leq +3{\Bigr )}\\&=P{\Bigl (}-3{\sigma \over {\sqrt {n}}}\leq M_{n}-\mu \leq +3{\sigma \over {\sqrt {n}}}{\Bigr )}\\&=P{\Bigl (}|M_{n}-\mu |\leq \overbrace {3{\sigma \over {\sqrt {n}}}} ^{\text{margin of error}}{\Bigr )}\\&=P{\Bigl (}M_{n}-3{\sigma \over {\sqrt {n}}}\leq \mu \leq M_{n}+3{\sigma \over {\sqrt {n}}}{\Bigr )}\\\end{aligned}} • Sample variance$(s^{2})$ :

$\sigma _{n}{\xrightarrow {a.s}}\sigma$ {\begin{aligned}{\sigma _{n}}^{2}&={1 \over {n-1}}\sum _{i=1}^{n}(x_{i}-M_{n})^{2}\\&={n \over {n-1}}{\Bigl (}{1 \over n}\sum _{i=1}^{n}(x_{i}-M_{n})^{2}{\Bigr )}\\&={n \over {n-1}}{\Bigl (}{1 \over n}\sum _{i=1}^{n}{x_{i}}^{2}-{M_{n}}^{2}{\Bigr )}\end{aligned}} • independent and identically distributed ${x_{i}}^{2}$ :

$E({x_{i}}^{2})=\sigma ^{2}+\mu ^{2}$ $var(x_{i})=E({x_{i}}^{2})-[E(x_{i})]^{2}$ $\sigma ^{2}=E({x_{i}}^{2})-\mu ^{2}$ $\Rrightarrow E({x_{i}}^{2})=\sigma ^{2}+\mu ^{2}$ ${1 \over n}\sum _{i=1}^{n}{x_{i}}^{2}{\xrightarrow {a.s.}}E({x_{i}}^{2})=\sigma ^{2}+\mu ^{2}$ • ${M_{n}}^{2}{\xrightarrow {a.s.}}\mu ^{2}$ Hence:

• ${\sigma _{n}}^{2}={n \over n-1}{\Bigl (}{1 \over n}\sum _{i=1}^{n}{x_{i}}^{2}-{M_{n}}^{2}{\Bigr )}$ • ${1 \over n}\sum _{i=1}^{n}{x_{i}}^{2}{\xrightarrow {a.s.}}\sigma ^{2}+\mu ^{2}$ • ${M_{n}}^{2}{\xrightarrow {a.s.}}\mu ^{2}$ • ${n \over n-1}\rightarrow 1$ • ${\sigma _{n}}^{2}{\xrightarrow {a.s.}}1(\sigma ^{2}+\mu ^{2}-\mu ^{2})=\sigma ^{2}$ • $\sigma _{n}{\xrightarrow {a.s.}}{\sqrt {\sigma ^{2}}}=\sigma$ Conclusive, to assess the error one can use the following interval

${\Bigl (}M_{n}-3{s \over {\sqrt {n}}};M_{n}+3{s \over {\sqrt {n}}}{\Bigr )}$ where

$s^{2}={1 \over n-1}\sum _{i=1}^{n}(x_{i}-M_{n})^{2}$ 