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---
title: "Central limit theorem"
firstLetter: "C"
publishDate: 2021-03-09
categories:
- Statistics
- Mathematics
date: 2021-03-09T20:39:38+01:00
draft: false
markup: pandoc
---
# Central limit theorem
In statistics, the **central limit theorem** states that
the sum of many independent variables tends towards a normal distribution,
even if the individual variables $x_n$ follow different distributions.
For example, by taking $M$ samples of size $N$ from a population,
and calculating $M$ averages $\mu_m$ (which involves summing over $N$),
the resulting means $\mu_m$ are normally distributed
across the $M$ samples if $N$ is sufficiently large.
More formally, for $N$ independent variables $x_n$ with probability distributions $p(x_n)$,
the central limit theorem states the following,
where we define the sum $S$:
$$\begin{aligned}
S = \sum_{n = 1}^N x_n
\qquad
\mu_S = \sum_{n = 1}^N \mu_n
\qquad
\sigma_S^2 = \sum_{n = 1}^N \sigma_n^2
\end{aligned}$$
And crucially, it states that the probability distribution $p_N(S)$ of $S$ for $N$ variables
will become a normal distribution when $N$ goes to infinity:
$$\begin{aligned}
\boxed{
\lim_{N \to \infty} \!\big(p_N(S)\big)
= \frac{1}{\sigma_S \sqrt{2 \pi}} \: \exp\!\Big( -\frac{(\mu_S - S)^2}{2 \sigma_S^2} \Big)
}
\end{aligned}$$
We prove this below,
but first we need to introduce some tools.
Given a probability density $p(x)$, its [Fourier transform](/know/concept/fourier-transform/)
is called the **characteristic function** $\phi(k)$:
$$\begin{aligned}
\phi(k) = \int_{-\infty}^\infty p(x) \exp(i k x) \dd{x}
\end{aligned}$$
Note that $\phi(k)$ can be interpreted as the average of $\exp(i k x)$.
We take its Taylor expansion in two separate ways,
where an overline denotes the mean:
$$\begin{aligned}
\phi(k)
= \sum_{n = 0}^\infty \frac{k^n}{n!} \: \phi^{(n)}(0)
\qquad
\phi(k)
= \overline{\exp(i k x)} = \sum_{n = 0}^\infty \frac{(ik)^n}{n!} \overline{x^n}
\end{aligned}$$
By comparing the coefficients of these two power series,
we get a useful relation:
$$\begin{aligned}
\phi^{(n)}(0) = i^n \: \overline{x^n}
\end{aligned}$$
Next, the **cumulants** $C^{(n)}$ are defined from the Taylor expansion of $\ln\!\big(\phi(k)\big)$:
$$\begin{aligned}
\ln\!\big( \phi(k) \big)
= \sum_{n = 1}^\infty \frac{(ik)^n}{n!} C^{(n)}
\quad \mathrm{where} \quad
C^{(n)} = \frac{1}{i^n} \: \dv[n]{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0}
\end{aligned}$$
The first two cumulants $C^{(1)}$ and $C^{(2)}$ are of particular interest,
since they turn out to be the mean and the variance respectively,
using our earlier relation:
$$\begin{aligned}
C^{(1)}
&= - i \dv{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0}
= - i \frac{\phi'(0)}{\exp(0)}
= \overline{x}
\\
C^{(2)}
&= - \dv[2]{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0}
= \frac{\big(\phi'(0)\big)^2}{\exp(0)^2} - \frac{\phi''(0)}{\exp(0)}
= - \overline{x}^2 + \overline{x^2} = \sigma^2
\end{aligned}$$
Let us now define $S$ as the sum of $N$ independent variables $x_n$, in other words:
$$\begin{aligned}
S = \sum_{n = 1}^N x_n = x_1 + x_2 + ... + x_N
\end{aligned}$$
The probability density of $S$ is then as follows, where $p(x_n)$ are
the densities of all the individual variables and $\delta$ is
the [Dirac delta function](/know/concept/dirac-delta-function/):
$$\begin{aligned}
p(S)
&= \idotsint_{-\infty}^\infty \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( S - \sum_{n = 1}^N x_n \Big) \dd{x_1} \cdots \dd{x_N}
\\
&= \Big( p_1 * \big( p_2 * ( ... * (p_N * \delta))\big)\Big)(S)
\end{aligned}$$
In other words, the integrals pick out all combinations of $x_n$ which
add up to the desired $S$-value, and multiply the probabilities
$p(x_1) p(x_2) \cdots p(x_N)$ of each such case. This is a convolution,
so the [convolution theorem](/know/concept/convolution-theorem/)
states that it is a product in the Fourier domain:
$$\begin{aligned}
\phi_S(k) = \prod_{n = 1}^N \phi_n(k)
\end{aligned}$$
By taking the logarithm of both sides, the product becomes a sum,
which we further expand:
$$\begin{aligned}
\ln\!\big(\phi_S(k)\big)
= \sum_{n = 1}^N \ln\!\big(\phi_n(k)\big)
= \sum_{n = 1}^N \sum_{m = 1}^{\infty} \frac{(ik)^m}{m!} C_n^{(m)}
\end{aligned}$$
Consequently, the cumulants $C^{(m)}$ stack additively for the sum $S$
of independent variables $x_m$, and therefore
the means $C^{(1)}$ and variances $C^{(2)}$ do too:
$$\begin{aligned}
C_S^{(m)} = \sum_{n = 1}^N C_n^{(m)} = C_1^{(m)} + C_2^{(m)} + ... + C_N^{(m)}
\end{aligned}$$
We now introduce the scaled sum $z$ as the new combined variable:
$$\begin{aligned}
z = \frac{S}{\sqrt{N}} = \frac{1}{\sqrt{N}} (x_1 + x_2 + ... + x_N)
\end{aligned}$$
Its characteristic function $\phi_z(k)$ is then as follows,
with $\sqrt{N}$ appearing in the arguments of $\phi_n$:
$$\begin{aligned}
\phi_z(k)
&= \idotsint
\Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( z - \frac{1}{\sqrt{N}} \sum_{n = 1}^N x_n \Big) \exp(i k z)
\dd{x_1} \cdots \dd{x_N}
\\
&= \idotsint
\Big( \prod_{n = 1}^N p(x_n) \Big) \exp\!\Big( i \frac{k}{\sqrt{N}} \sum_{n = 1}^N x_n \Big)
\dd{x_1} \cdots \dd{x_N}
\\
&= \prod_{n = 1}^N \phi_n\Big(\frac{k}{\sqrt{N}}\Big)
\end{aligned}$$
By expanding $\ln\!\big(\phi_z(k)\big)$ in terms of its cumulants $C^{(m)}$
and introducing $\kappa = k / \sqrt{N}$, we see that the higher-order terms
become smaller for larger $N$:
$$\begin{gathered}
\ln\!\big( \phi_z(k) \big)
= \sum_{m = 1}^\infty \frac{(ik)^m}{m!} C^{(m)}
\\
C^{(m)}
= \frac{1}{i^m} \dv[m]{k} \sum_{n = 1}^N \ln\!\bigg( \phi_n\Big(\frac{k}{\sqrt{N}}\Big) \bigg)
= \frac{1}{i^m N^{m/2}} \dv[m]{\kappa} \sum_{n = 1}^N \ln\!\big( \phi_n(\kappa) \big)
\end{gathered}$$
For sufficiently large $N$, we can therefore approximate it using just the first two terms:
$$\begin{aligned}
\ln\!\big( \phi_z(k) \big)
&\approx i k C^{(1)} - \frac{k^2}{2} C^{(2)}
= i k \overline{z} - \frac{k^2}{2} \sigma_z^2
\\
\phi_z(k)
&\approx \exp(i k \overline{z}) \exp\!(- k^2 \sigma_z^2 / 2)
\end{aligned}$$
We take its inverse Fourier transform to get the density $p(z)$,
which turns out to be a Gaussian normal distribution,
which is even already normalized:
$$\begin{aligned}
p(z)
= \hat{\mathcal{F}}^{-1} \{\phi_z(k)\}
&= \frac{1}{2 \pi} \int_{-\infty}^\infty \exp\!\big(\!-\! i k (z - \overline{z})\big) \exp(- k^2 \sigma_z^2 / 2) \dd{k}
\\
&= \frac{1}{\sqrt{2 \pi \sigma_z^2}} \exp\!\Big(\!-\! \frac{(z - \overline{z})^2}{2 \sigma_z^2} \Big)
\end{aligned}$$
Therefore, the sum of many independent variables tends to a normal distribution,
regardless of the densities of the individual variables.
## References
1. H. Gould, J. Tobochnik,
*Statistical and thermal physics*, 2nd edition,
Princeton.
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