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Categories: Mathematics, Statistics.

Central limit theorem

In statistics, the central limit theorem states that the sum of many independent variables tends to 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 $p(x_n)$, we define the following totals of all variables, means and variances:

$$\begin{aligned} t \equiv \sum_{n = 1}^N x_n \qquad \qquad \mu_t \equiv \sum_{n = 1}^N \mu_n \qquad \qquad \sigma_t^2 \equiv \sum_{n = 1}^N \sigma_n^2 \end{aligned}$$

The central limit theorem then states that the probability distribution $p_N(t)$ of $t$ for $N$ variables will become a normal distribution when $N$ goes to infinity:

$$\begin{aligned} \boxed{ \lim_{N \to \infty} \!\big(p_N(t)\big) = \frac{1}{\sigma_t \sqrt{2 \pi}} \exp\!\bigg( -\frac{(t - \mu_t)^2}{2 \sigma_t^2} \bigg) } \end{aligned}$$

We prove this below, but first we need to introduce some tools. Given a probability density $p(x)$, its Fourier transform is called the characteristic function $\phi(k)$:

$$\begin{aligned} \phi(k) \equiv \int_{-\infty}^\infty p(x) \exp(i k x) \dd{x} \end{aligned}$$

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:

$$\begin{aligned} \phi(k) = \sum_{n = 0}^\infty \frac{k^n}{n!} \bigg( \dvn{n}{\phi}{k} \Big|_{k = 0} \bigg) \qquad \qquad \phi(k) = \overline{\exp(i k x)} = \sum_{n = 0}^\infty \frac{(ik)^n}{n!} \overline{x^n} \end{aligned}$$

By comparing the coefficients of these two power series, we get a useful relation:

$$\begin{aligned} \dvn{n}{\phi}{k} \Big|_{k = 0} = i^n \: \overline{x^n} \end{aligned}$$

Next, the cumulants $C^{(n)}$ are defined from the Taylor expansion of $\ln\!\big(\phi(k)\big)$:

$$\begin{aligned} \ln\!\big( \phi(k) \big) = \sum_{n = 1}^\infty \frac{(ik)^n}{n!} C^{(n)} \quad \mathrm{where} \quad C^{(n)} \equiv \frac{1}{i^n} \: \dvn{n}{}{k} \ln\!\big(\phi(k)\big) \Big|_{k = 0} \end{aligned}$$

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:

$$\begin{aligned} C^{(1)} &= - i \dv{}{k} \ln\!\big(\phi(k)\big) \Big|_{k = 0} = - i \frac{\phi'(0)}{\exp(0)} = \overline{x} \\ C^{(2)} &= - \dvn{2}{}{k} \ln\!\big(\phi(k)\big) \Big|_{k = 0} = \frac{\big(\phi'(0)\big)^2}{\exp(0)^2} - \frac{\phi''(0)}{\exp(0)} = - \overline{x}^2 + \overline{x^2} = \sigma^2 \end{aligned}$$

Now that we have introduced these tools, we repeat our definition of $t$ as the sum of $N$ independent variables $x_n$, in other words:

$$\begin{aligned} t \equiv \sum_{n = 1}^N x_n = x_1 + x_2 + ... + x_N \end{aligned}$$

The probability density of $t$ is then as follows, where $p(x_n)$ are the densities of all the individual variables and $\delta$ is the Dirac delta function:

$$\begin{aligned} p(t) &= \int\cdots\int_{-\infty}^\infty \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( t - \sum_{n = 1}^N x_n \Big) \dd{x_1} \cdots \dd{x_N} \\ &= \Big( p_1 * \big( p_2 * ( ... * (p_N * \delta))\big)\Big)(t) \end{aligned}$$

In other words, we integrate over all possible combinations of $x_n$, and use the Dirac delta function to pick out the combinations where the $x_n$ add up to the desired $t$-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 states that it is a product in the Fourier domain:

$$\begin{aligned} \phi_t(k) = \prod_{n = 1}^N \phi_n(k) \end{aligned}$$

By taking the logarithm of both sides, the product becomes a sum, which we further expand:

$$\begin{aligned} \ln\!\big(\phi_t(k)\big) = \sum_{n = 1}^N \ln\!\big(\phi_n(k)\big) = \sum_{n = 1}^N \sum_{m = 1}^{\infty} \frac{(ik)^m}{m!} C_n^{(m)} \end{aligned}$$

Consequently, the cumulants $C^{(m)}$ stack additively for the sum $t$ of independent variables $x_m$, and therefore the means $C^{(1)}$ and variances $C^{(2)}$ do too:

$$\begin{aligned} C_t^{(m)} = \sum_{n = 1}^N C_n^{(m)} = C_1^{(m)} + C_2^{(m)} + ... + C_N^{(m)} \end{aligned}$$

We now introduce the scaled sum $z$ as the new combined variable:

$$\begin{aligned} z \equiv \frac{t}{\sqrt{N}} = \frac{1}{\sqrt{N}} (x_1 + x_2 + ... + x_N) \end{aligned}$$

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:

$$\begin{aligned} \ln\!\big( \phi_z(k) \big) &\approx i k C^{(1)} - \frac{k^2}{2} C^{(2)} = i k \mu_z - \frac{k^2}{2} \sigma_z^2 \\ \implies \quad \phi_z(k) &\approx \exp(i k \mu_z) \exp(- k^2 \sigma_z^2 / 2) \end{aligned}$$

We take its inverse Fourier transform to get the density $p(z)$, which turns out to be a Gaussian normal distribution and is even already normalized:

$$\begin{aligned} p(z) = \hat{\mathcal{F}}^{-1} \{\phi_z(k)\} &= \frac{1}{2 \pi} \int_{-\infty}^\infty \exp\!\big(\!-\! i k (z - \mu_z)\big) \exp(- k^2 \sigma_z^2 / 2) \dd{k} \\ &= \frac{1}{\sqrt{2 \pi \sigma_z^2}} \exp\!\Big(\!-\! \frac{(z - \mu_z)^2}{2 \sigma_z^2} \Big) \end{aligned}$$

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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