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