Central Limit Theorem

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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.

$\blacksquare$


Sources

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