# Central Limit Theorem

## Theorem

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

expectation $\expect {X_i} = \mu \in \R$
variance $\var {X_i} = \sigma^2 > 0$

Let:

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

Then:

$\ds \dfrac {S_n - n \mu} {\sqrt {n \sigma^2} } \xrightarrow D \Gaussian 0 1$ as $n \to \infty$

that is, converges in distribution to a standard Gaussian.

## Proof

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

We have that:

$\expect {Y_i} = 0$

and:

$\expect {Y_i^2} = 1$

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

$\map {\phi_{Y_i} } t = 1 - \dfrac {t^2} 2 + \map o {t^2}$

Now let:

 $\ds U_n$ $=$ $\ds \frac {S_n - n \mu} {\sqrt {n \sigma^2} }$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \frac {X_i - \mu} {\sqrt {n \sigma^2} }$ $\ds$ $=$ $\ds \frac 1 {\sqrt n} \sum_{i \mathop = 1}^n \paren {\frac {X_i - \mu} \sigma}$ $\ds$ $=$ $\ds \frac 1 {\sqrt n} \sum_{i \mathop = 1}^n Y_i$

Then its characteristic function is given by:

 $\ds \map {\phi_{U_n} } t$ $=$ $\ds \expect {e^{i t U_n} }$ $\ds$ $=$ $\ds \expect {\map \exp {\frac {i t} {\sqrt n} \sum_{i \mathop = 1}^n Y_i} }$ $\ds$ $=$ $\ds \prod_{i \mathop = 1}^n \expect {\map \exp {\frac{i t} {\sqrt n} Y_i} }$ since $Y_i$ are independent identically distributed $\ds$ $=$ $\ds \prod_{i \mathop = 1}^n \map {\phi_{Y_i} } {\frac t {\sqrt n} }$ $\ds$ $=$ $\ds \paren {\map {\phi_{Y_i} } {\frac t {\sqrt n} } }^n$ since $Y_i$ are independent identically distributed $\ds$ $=$ $\ds \paren {1 - \frac {t^2} {2 n} + \map o {t^2} }^n$

Recall that the Characteristic Function of Gaussian Distribution is given by:

 $\ds \map \phi t$ $=$ $\ds e^{i t \mu - \frac 1 2 t^2 \sigma^2}$ $\ds$ $=$ $\ds e^{i t \paren 0 - \frac 1 2 t^2 \paren 1^2}$ $\ds$ $=$ $\ds e^{- \frac 1 2 t^2}$

Indeed the characteristic equations of the series converges to the Characteristic Function of Gaussian Distribution:

$\paren {1 - \dfrac {t^2} {2 n} + \map o {t^2} }^n \to e^{-\frac 1 2 t^2}$ 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 Gaussian 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 Gaussian random variable.

$\blacksquare$