# Central Limit Theorem

## Theorem

Let $X_1, X_2, \ldots$ be a sequence of independent identically distributed random variables with:

- Mean $E \left[{X_i}\right] = \mu \in \left({-\infty \,.\,.\, \infty}\right)$
- Variance $V \left({X_i}\right) = \sigma^2 > 0$

Let:

- $\displaystyle S_n = \sum_{i \mathop = 1}^n X_i$

Then:

- $\displaystyle \frac {S_n - n \mu} {\sqrt {n \sigma^2} } \xrightarrow {D} N \left({0, 1}\right)$ as $n \to \infty$

that is, converges in distribution to a standard normal.

## Proof

Let $Y_i = \dfrac {X_i - \mu} {\sigma}$.

We have that:

- $E \left[{Y_i}\right] = 0$

and:

- $E \left[{Y_i^2}\right] = 1$

Then by Taylor's Theorem the characteristic function can be written:

- $\phi_{Y_i} = 1 - \dfrac {t^2} 2 + o \left({t^2}\right)$

Now let:

\(\displaystyle U_n\) | \(=\) | \(\displaystyle \frac {S_n - n \mu} {\sqrt {n \sigma^2} }\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^n \frac {X_i - \mu} {\sqrt {n \sigma^2} }\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 {\sqrt n} \sum_{i \mathop = 1}^n \left({\frac {X_i - \mu} {\sigma} }\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 {\sqrt n} \sum_{i \mathop = 1}^n Y_i\) | $\quad$ | $\quad$ |

Then its characteristic function is given by:

\(\displaystyle \phi_{U_n} \left({t}\right)\) | \(=\) | \(\displaystyle E \left[{e^{i t U_n} }\right]\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle E \left[{\exp \left({\frac {i t} {\sqrt n} \sum_n^{i \mathop = 1} Y_i}\right)}\right]\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \prod_n^{i \mathop = 1} E \left[{\exp \left({\frac{i t} {\sqrt n} Y_i}\right)}\right]\) | $\quad$ since $Y_i$ are independent identically distributed | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \prod_n^{i \mathop = 1} \phi_{Y_i} \left({\frac t {\sqrt n} }\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({\phi_{Y_i} \left({\frac t {\sqrt n} }\right)}\right)^n\) | $\quad$ since $Y_i$ are independent identically distributed | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({1 - \frac {t^2} {2 n} + o \left({t^2}\right)}\right)^n\) | $\quad$ | $\quad$ |

Recall that the characteristic equation of a standard normal is given by:

- $e^{-\frac 1 2 t^2}$.

Indeed the characteristic equations of the series converges to the standard normal characteristic equation:

- $\left({1 - \dfrac {t^2} {2 n} + o \left({t^2}\right)}\right)^n \to e^{-\dfrac 1 2 t^2}$ as $n \to \infty$

Then Lévy’s continuity theorem applies.

In particular, the convergence in distribution of the $U_n$ to some random variable with standard normal distribution is equivalent to continuity of the limiting characteristic equation at $t = 0$.

But, $e^{-\frac 1 2 t^2}$ is clearly continuous at $0$.

So we have that $\dfrac{S_n - n \mu} {\sqrt{n \sigma^2} }$ converges in distribution to a standard normal random variable.

$\blacksquare$