Common probability distributions

October 01, 2025

Table of contents

• Sum of random variables
• Normal distribution
• $\chi$ distribution
• Student-t distribution
• Multinomial distribution
• Poisson distribution
• Erlang distribution

Probability of finding a sum of random variables

The probability density $\textrm{p}(\overline{x})$ of finding the average $\overline{x} = \frac{\sum_i x_i}{n}$ of $n$ random variables can be written using the Delta function as:

$$\textrm{p}(\overline{x}) = \int \delta\left(\overline{x}-\sum_i \frac{x_i}{n}\right)\textrm{p}_1(x_1) \dots \textrm{p}_n(x_n)\textrm{d}x_1 \dots \textrm{d}x_n = \int \delta\left(\overline{x}-\sum_i\frac{x_i}{n}\right)\prod_i \textrm{p}_i(x_i) \textrm{d}x_i$$

Calculating the Fourier transform $\hat{\textrm{p}}(k)$ yields:

$$\hat{\textrm{p}}(k) = \int e^{-ik\overline{x}}\textrm{p}(x)\textrm{d}\overline{x} = \int e^{-ik \sum_i x_i/n}\prod_i \textrm{p}_i(x_i) \textrm{d}x_i = \prod_i \int e^{-ikx_i/n} \textrm{p}_i(x_i) \textrm{d}x_i = \prod_i \hat{\textrm{p}}_i(k/n)$$

The original $\textrm{p}(\overline{x})$ can now be recovered using the inverse Fourier transform:

$$\textrm{p}(\overline{x}) = \frac{1}{2\pi} \int e^{ik\overline{x}}\hat{\textrm{p}}(k)\textrm{d}k = \frac{1}{2\pi} \int e^{ik\overline{x}}\prod_i \hat{\textrm{p}}_i(k/n)\textrm{d}k = \frac{1}{2\pi} \int e^{ik\overline{x}} \left({\textrm{p}}_i(k/n)\right)^n \textrm{d}k$$

Where the last equality holds for identically distributed random variables.

Normal distribution

Taylor expanding any probability density $\hat{\textrm{p}}_i(k/n)$ around $k = 0$ yields:

$$\hat{\textrm{p}}_i(k/n) = \sum_m \frac{\partial_k^m \hat{\textrm{p}}_i(0)}{m!}\left(\frac{k}{n}\right)^m = \frac{\mathbb{E}[x_i^m]}{m!}\left(\frac{ik}{n}\right)^m = 1 + \frac{ik \mathbb{E}[x_i]}{n} + \frac{i^2k^2\mathbb{E}[x_i^2]}{2n^2} + \dots$$

Comparing up to first order in $\frac{1}{n}$ one gets:

$$(\hat{\textrm{p}}(k/n))^n = \left(1 + \frac{ik \mathbb{E}[x_i]}{n} + \frac{i^2k^2\mathbb{E}[x_i^2]}{2n^2} + \dots \right)^n = e^{ik \mathbb{E}[x_i]}e^{i^2k^2(\mathbb{E}[x_i^2]-\mathbb{E}[x_i]^2)/2n} + O\left(\frac{1}{n^2}\right)$$

Writing $\mu \equiv \mathbb{E}[x_i]$ and $\sigma^2 \equiv \mathbb{E}[x_i^2]-\mathbb{E}[x_i]^2$, the original $\textrm{p}(\overline{x})$ can now be recovered using the inverse Fourier transform:

$$\textrm{p}(\overline{x}) \overset{n\rightarrow\infty}{\rightarrow} \frac{1}{2\pi} \int e^{ik\overline{x}}e^{ik \mu}e^{i^2k^2\sigma^2/2n}\textrm{d}k = \frac{1}{\sqrt{2\pi\sigma^2/n}}e^{-\frac{(\overline{x}-\mu)^2}{2\sigma^2/n}}$$

The average of $n$ identically distributed random variables will for large $n$ become normally distributed with the same mean $\mu$ and lower variance $\sigma^2/n$ compared to the original distribution.

$\chi$ distribution

The probability of finding the root of the sum of squares of $n$ standard-normal distributed random variables is proportional to the surface area of a $n$ dimensional sphere:

$$\textrm{p}(\chi) = \int\delta\left(\chi-\sqrt{\sum_ix_i^2}\right)\prod_i \frac{e^{-x_i^2/2}}{\sqrt{2\pi}}\textrm{d}x_i = \int\textrm{d}A\int\delta(\chi-r)\frac{e^{-r^2/2}}{\sqrt{2\pi}^n}r^{n-1}\textrm{d}r = \frac{e^{-\chi^2/2}\chi^{n-1}}{\sqrt{2^{n-2}}\Gamma\left(\frac{n}{2}\right)}$$

The surface area of the n-dimensional sphere can be calculated from:

$$1=\prod_i \int\frac{e^{-x_i^2/2}}{\sqrt{2\pi}}\textrm{d}x_i = \int \textrm{d}A \int \frac{e^{-r^2/2}r^{n-1}}{\sqrt{2\pi}^n}\textrm{d}r = \int \textrm{d}A \int \frac{e^{-t}t^{\frac{n}{2}-1}}{2\sqrt{\pi^n}}\textrm{d}t \equiv \frac{\Gamma\left(\frac{n}{2}\right)}{2\sqrt{\pi^n}} \int \textrm{d}A$$

Where the gamma function is defined as $\Gamma(x) \equiv \int_0^\infty e^{-t}t^{x-1}\textrm{d}t$.

$\chi^2$ distribution

A change of variables yields the distribution for the sum of squares, called the $\chi^2$ distribution:

$$\textrm{p}(\chi^2) = \textrm{p}(\chi)\frac{\textrm{d} \chi}{\textrm{d} \chi^2} = \frac{e^{-\chi^2/2}\chi^{n-2}}{\sqrt{2^n}\Gamma\left(\frac{n}{2}\right)}$$

Student-t distribution

When estimating the true mean $\mu$ of a normally-distributed random variable $x$ by the sample mean $\overline{x} \equiv \frac{\sum_i x_i}{n}$, the deviation $\overline{x}-\mu$ is again normally-distributed with a standard deviation of $\frac{\sigma}{\sqrt{n}}$, making $\delta \equiv \frac{\overline{x}-\mu}{\sigma/\sqrt{n}}$ standard-normal distributed. By Cochran’s theorem the ratio $\chi^2 \equiv \frac{(n-1)s^2}{\sigma^2}$ between sample variance $s^2 \equiv \frac{\sum_i (x_i-\overline{x})^2}{n-1}$ and true variance $\sigma^2$ follows a $\chi^2$-distribution of degree $n-1$. The deviation $t \equiv \frac{\overline{x}-\mu}{s/\sqrt{n}} = \frac{\delta}{\chi/\sqrt{n-1}}$ will then depend only on the measurable quantities $\overline{x}$ and $s^2$, following a so-called Student-t distribution of degree $n-1$:

$$\begin{aligned} p(t) &= \iint\delta\left(t-\frac{\delta}{\chi/\sqrt{n-1}}\right)\underbrace{\frac{e^{-\delta^2/2}}{\sqrt{2\pi}}}_{p(\delta)}\underbrace{\frac{e^{-\chi^2/2}\chi^{n-3}}{\sqrt{2^{n-1}}\Gamma\left(\frac{n-1}{2}\right)}}_{p(\chi^2)}\textrm{d}\delta\textrm{d}\chi^2 \\ &= \int\frac{e^{-(t\chi)^2/2(n-1)}}{\sqrt{2\pi}}\frac{e^{-\chi^2/2}\chi^{n-3}}{\sqrt{2^{n-1}}\Gamma\left(\frac{n-1}{2}\right)}\frac{\chi}{\sqrt{n-1}}\textrm{d}\chi^2 \\ &= \int\frac{e^{-\chi^2(t^2/(n-1)+1)/2}}{\sqrt{2^n\pi(n-1)}}\frac{(\chi^2)^{\frac{n-2}{2}}}{\Gamma\left(\frac{n-1}{2}\right)}\textrm{d}\chi^2 \\ &= \int\frac{e^{-u}u^{\frac{n}{2}-1}}{\sqrt{2^n\pi(n-1)}}\frac{\left(\frac{2}{t^2/(n-1)+1}\right)^{\frac{n}{2}}}{\Gamma\left(\frac{n-1}{2}\right)}\textrm{d}u \\ &= \frac{\Gamma\left(\frac{n}{2}\right)}{\sqrt{\pi (n-1)}\Gamma\left(\frac{n-1}{2}\right)}\left(1+\frac{t^2}{n-1}\right)^{-\frac{n}{2}} \end{aligned}$$

Multinomial coefficient

Suppose you wants to arrange $k_i$ copys of $i$ different things into $n = \sum_i k_i$ bins. How many distinct ways $\binom{n}{k_1,\dots,k_i}$ are there to do this?

Starting with the case that each thing is unique (all $k_i$ = 1), then the first bin can be filled with one of $n$ things, the second with $(n-1)$ remaining things, and so on, yielding $n!$ distinct possibilities:

$$\binom{n}{k_1=1,\dots,k_i=1} = n(n-1)(n-2)(\dots) = n!$$

Grouping together $k_i$ of the previously unique things into a category $i$ reduces the amount by a factor of $k_i!$ since there are now $k_i$ ways to select the first thing out of the category $i$, and $k_i - 1$ ways to select the second, and so on, yielding an overall result of:

$$\binom{n}{k_1,\dots,k_i} = \frac{n!}{k_1!,\dots,k_i!}= \frac{n!}{\prod_i k_i!}$$

What if multiple things can go into one bin? This problem can be reduce to the previous problem by adding $n-1$ separators to the number of bins and treating the separators as an additional category of thing that can be place in a single-occupied bin.

Multinomial distribution

Now what is the probability $\textrm{P}(k_1,\dots,k_i)$ when measuring $n = \sum_i k_i$ outcomes that there will be $k_i$ outcomes of each type $i$? With the probabilities $p_i$ of measuring outcome $i$ in a single observation the total proability $P$ is then the number of possibilities realizing this outcome multiplied by the product all of the $p_i$:

$$\textrm{P}(k_1,\dots,k_i) = \binom{n}{k_1,\dots,k_i} \prod_i p_i^{k_i} = n!\prod_i \frac{p_i^{k_i}}{k_i!}$$

For two possible outcomes with $k_1 + k_2 = n$ and $p_1 + p_2 = 1$ one recovers the binomial distribution:

$$\textrm{P}(n, k) = \frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}$$

Poisson distribution

Now taking the limit $n \rightarrow \infty$ and $p_i \rightarrow 0$ while keeping the expectation values $\lambda_i \equiv n \cdot p_i$ constant yields the Poisson distribution. With the requirement $\sum_i p_i = 1$ we set $p_0 \equiv 1-\sum_i p_i = 1-\sum_i \frac{\lambda_i}{n}$ as the probability of nothing happening during one of the $n$ obervations and $k_0 \equiv n-\sum_i k_i \equiv n-k$ as the number of times this happens. The probability of observing each event ${k_i}$ times is then given by:

$$\begin{aligned} \textrm{P}(k_1,\dots,k_i) &= n!\frac{p_0^{k_0}}{k_0!}\prod_i \frac{p_i^{k_i}}{k_i!} \\ &= \frac{n!}{(n-k)!}p_0^{n-k}\frac{1}{n^k}\prod_i \frac{\lambda_i^{k_i}}{k_i!} \\ &= \underbrace{\frac{n!}{(n-k)!}\frac{1}{n^k}}_{\rightarrow\ 1}\underbrace{\left(1-\frac{\sum_i \lambda_i}{n}\right)^n}_{\rightarrow\ \exp\left(\sum_i \lambda_i\right)} \bigg({\underbrace{1-\frac{\sum_i \lambda_i}{n}}_{\rightarrow\ 1}} \bigg)^{-k}\prod_i \frac{\lambda_i^{k_i}}{k_i!} \\ &\overset{n\rightarrow\infty}{\rightarrow} \prod_i \frac{\lambda_i^{k_i}}{k_i!}e^{\lambda_i} \end{aligned}$$

Geometric distribution

The probability of having the first success in a binomial setup after exactly $k$ trials is simply given by:

$$\textrm{P}(k) = (1-p)^k p$$

Exponential distribution

Writing $x = \frac{k}{n}$, $p = \frac{\lambda}{n}$, $\frac{1}{n} = \textrm{d}x$ and taking the limit $n \rightarrow \infty$ yields the waiting time distribution in a Poisson process:

$$\textrm{P}(x) = \left(1-\frac{\lambda}{n}\right)^{xn} \frac{\lambda}{n} \overset{n\rightarrow\infty}{\rightarrow} e^{-\lambda x}\lambda\textrm{d}x$$

Erlang distribution

The cumulative waiting time for $n$ events is the sum of $n$ exponential distributions. Since the exponential distribution is a special case of a $\chi^2$ distribution with $k=2$ we can get the sum of $n$ exponential distributions simply as $\chi^2$ distribution with $k=2n$:

$$\textrm{p}(\chi^2=2\lambda x, k=2n) = \frac{(2\lambda x)^{n-1}}{2^n\Gamma(n)}e^{-\lambda x}\frac{\textrm{d}\chi^2}{\textrm{d}x} = \frac{\lambda^n x^{n-1}}{(n-1)!}e^{-\lambda x}$$