Central Limit Theorem (CLT)

CLT workd for i.i.d sample from “most” distribution.

Let $x_1,x_2…x_n$ be i.i.d from any distribution, where $M_x(t)$ exists. Denote $E(x_i) = \mu\ and\ Var(x_i) = \sigma^2$. Let $\bar x_n = \frac{1}{n}\sum_{i=1}^{n}x_i\ and\ U = \frac{\bar x_n - \mu}{\sigma/\sqrt{n}}$.
Then $U \Rightarrow N(0,1)\ as\ n\rightarrow \infty$. This means the PDF of U will approximate standard normal distribution when n is efficiently large.

Proof:
Let $y_i = \frac{x_i-\mu}{\sigma},\ then\ E(y_i) = 0,\ Var(y_i) = E(y_i^2) = 1,\ and\ U = \frac{1}{\sqrt{n}}\sum_{i=1}{n}y_i$
$\Rightarrow$ MGF of U is then $M_u(t) = E(e^{tU}) = E[e^{ty_1/\sqrt{n}+ty_2/\sqrt{n}…+ty_n/\sqrt{n}}]$
$=E[e^{ty_1/\sqrt{n}}]E[e^{ty_2/\sqrt{n}}]…E[e^{ty_n/\sqrt{n}}]$
$=M_{y_1}(t/\sqrt{n})M_{y_2}(t/\sqrt{n})…M_{y_n}(t/\sqrt{n})$
$=[M_y(t/\sqrt{n})]^n$

(from Taylor expand around t=0)
$= 1 +E(y_i)\frac{t}{\sqrt{n}}+E(y_i^2)\frac{t^2}{2n}+R$
$= 1 + 0 + \frac{t^2}{2n}+R$
$= M_u(t) = [1+\frac{t^2}{2}\cdot \frac{1}{n}+R]^n$
$\Rightarrow M_y(t/ {\sqrt{n}}) = M_y(0)\frac {(t/ \sqrt{n})^0}{0!}+M_y’(0)\frac{(t/ \sqrt{n})^1}{1!}+M_y’’(0)\frac{(t/ \sqrt{n})^2}{2!}+R$

Recall
$$\lim \limits_{n \rightarrow \infty}(1+\frac{x}{n})^2 = e^x$$

$\Rightarrow \lim \limits_{n \rightarrow \infty}M_u(t) = \lim \limits_{n \rightarrow \infty}[1+\frac{t^2}{2}\frac{1}{n}+R]^n = e^{t^2/2}$
$\Rightarrow$ this is the MGF of $N(0,1)$.

$\Rightarrow$ MGF of U $\rightarrow MGF\ of\ N(0,1)$, since MGF uniquely determines the distribution, $U \rightarrow N(0,1)$.

Ex. Let $x_1,x_2…x_n \overset {\text{i.i.d}}{\sim} Bernouli(p)$, whihc is equal $Binomial(1,p)$.
$E(x_i) = p,\ Var(x_i)=p(1-p)$
By CLT, when n is large
$\Rightarrow \frac{\bar x - p}{\sqrt{p(1-p)}/ \sqrt{n}} \overset{\text{arrpox}}{\sim}N(0,1)$
or $\bar x \overset{\text{arrpox}}{\sim} N(p, \frac{p(1-p)}{n})$
or $\sum_{i=1}^{n}x_i\overset{\text{arrpox}}{\sim}N(np, np(1-p))$
and $\sum_{i=1}^{n}x_i \sim Binomial(n,p)$ which is the exact distribution.

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