--- title: "Central limit theorem" date: 2021-03-09 categories: - Statistics - Mathematics layout: "concept" --- 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 $\ln\!\big(\phi(k)\big)$: $$\begin{aligned} \ln\!\big( \phi(k) \big) = \sum_{n = 1}^\infty \frac{(ik)^n}{n!} C^{(n)} \quad \mathrm{where} \quad C^{(n)} = \frac{1}{i^n} \: \dvn{n}{}{k} \Big(\ln\!\big(\phi(k)\big)\Big) \Big|_{k = 0} \end{aligned}$$ 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: $$\begin{aligned} C^{(1)} &= - i \dv{}{k} \Big(\ln\!\big(\phi(k)\big)\Big) \Big|_{k = 0} = - i \frac{\phi'(0)}{\exp(0)} = \overline{x} \\ C^{(2)} &= - \dvn{2}{}{k} \Big(\ln\!\big(\phi(k)\big)\Big) \Big|_{k = 0} = \frac{\big(\phi'(0)\big)^2}{\exp(0)^2} - \frac{\phi''(0)}{\exp(0)} = - \overline{x}^2 + \overline{x^2} = \sigma^2 \end{aligned}$$ 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/): $$\begin{aligned} p(S) &= \int\cdots\int_{-\infty}^\infty \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( S - \sum_{n = 1}^N x_n \Big) \dd{x_1} \cdots \dd{x_N} \\ &= \Big( p_1 * \big( p_2 * ( ... * (p_N * \delta))\big)\Big)(S) \end{aligned}$$ 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 $\phi_n$: $$\begin{aligned} \phi_z(k) &= \int\cdots\int \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( z - \frac{1}{\sqrt{N}} \sum_{n = 1}^N x_n \Big) \exp(i k z) \dd{x_1} \cdots \dd{x_N} \\ &= \int\cdots\int \Big( \prod_{n = 1}^N p(x_n) \Big) \exp\!\Big( i \frac{k}{\sqrt{N}} \sum_{n = 1}^N x_n \Big) \dd{x_1} \cdots \dd{x_N} \\ &= \prod_{n = 1}^N \phi_n\Big(\frac{k}{\sqrt{N}}\Big) \end{aligned}$$ 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 $N$: $$\begin{gathered} \ln\!\big( \phi_z(k) \big) = \sum_{m = 1}^\infty \frac{(ik)^m}{m!} C^{(m)} \\ C^{(m)} = \frac{1}{i^m} \dvn{m}{}{k} \sum_{n = 1}^N \ln\!\bigg( \phi_n\Big(\frac{k}{\sqrt{N}}\Big) \bigg) = \frac{1}{i^m N^{m/2}} \dvn{m}{}{\kappa} \sum_{n = 1}^N \ln\!\big( \phi_n(\kappa) \big) \end{gathered}$$ 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.