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:

Then its characteristic function is given by:

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.