# Probability density function, expected value and variance of each probability distribution

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I’d like to summarize probability density function (f(x)), expected value (E(X)) and variance (V(X)) of following probability distribution.

 Supergeometric distribution $\displaystyle f(x) = \frac{{}_{M}C_x\cdot{}_{N-M}C_{n-x}}{{}_{N}C_{n}}$ $\displaystyle x = Max(0, n - (N - M)), \cdots, Min(n, M)$ $\displaystyle E(X) = np$ $\displaystyle p = \frac{M}{N}$ $\displaystyle V(X) = np(1 - p)\frac{N - n}{N - 1}$ Binomial distribution $\displaystyle f(x) = {}_{n}C_{x}p^{x}(1 - p)^{n - x}$ $\displaystyle x = 0, 1, \cdots, n$ $\displaystyle E(X) = np$ $\displaystyle 0 < p < 1[/latex] $\displaystyle V(X) = np(1 - p)$ Poisson distribution $\displaystyle f(x) = \frac{\exp(- \lambda)\cdot\lambda^x}{x!}$ $\displaystyle x = 0, 1, 2, \cdots$ $\displaystyle E(X) = \lambda$ $\lambda = np$ $\displaystyle V(X) = \lambda$ Geometric distribution $\displaystyle f(x) = p(1 - p)^{x - 1}$ $\displaystyle x = 1, 2, 3, \cdots$ $\displaystyle E(X) = \frac{1}{p}$ $\displaystyle 0 < p < 1$ $\displaystyle V(X) = \frac{1 - p}{p^2}$ Negative binomial distribution $\displaystyle f(x) = {}_{k + x - 1}C_{x}p^{k}(1-p)^{x}$ $\displaystyle x = 0, 1, 2, \cdots$ $\displaystyle E(X) = \frac{k(1 - p)}{p}$ $\displaystyle V(X) = \frac{k(1 - p)}{p^2}$ Normal distribution $\displaystyle f(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x - m)^2}{2\sigma^2}\right)$ $\displaystyle - \infty < x < \infty$ $\displaystyle E(X) = m$ $\displaystyle V(X) = \sigma^2$ $\displaystyle \sigma > 0$ Exponential distribution $\displaystyle f(x) = \lambda\exp(-\lambda x)$ $\displaystyle x \ge 0$ $\displaystyle f(x) = 0$ $\displaystyle x < 0$ $\displaystyle E(X) = \frac{1}{\lambda}$ $\displaystyle V(X) = \frac{1}{\lambda^2}$ Gamma distribution $\displaystyle f(x) = \frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha - 1}\exp(- \lambda x)$ $\displaystyle x \ge 0$ $\displaystyle f(x) = 0$ $\displaystyle x < 0$ $\displaystyle E(X) = \frac{\alpha}{\lambda}$ $\displaystyle \lambda, \alpha > 0$ $\displaystyle V(X) = \frac{\alpha}{\lambda^2}$ Beta distribution $\displaystyle f(x) = \frac{x^{\alpha - 1}(1 - x)^{\beta - 1}}{B(\alpha, \beta)}$ $\displaystyle 0 < x < 1$ $\displaystyle f(x) = 0$ $\displaystyle x \le 0, 1 \le x$ $\displaystyle E(X) = \frac{\alpha}{\alpha + \beta}$ $\displaystyle V(X) = \frac{\alpha\beta}{(\alpha + \beta)^2(\alpha + \beta + 1)}$ Cauchy distribution $\displaystyle f(x) = \frac{\alpha}{\pi(\alpha^2 + (x - \lambda)^2)}$ $\displaystyle \alpha > 0$ Log-normal distribution $\displaystyle f(x) = \frac{1}{\sqrt{2\pi}\sigma x}\exp\left(-\frac{(\log(x) - m)^2}{2\sigma^2}\right)$ $\displaystyle x > 0$ $\displaystyle f(x) = 0$ $\displaystyle x \le 0$ $\displaystyle E(X) = \exp\left(m + \frac{\sigma^2}{2}\right)$ $\displaystyle \sigma > 0$ $\displaystyle V(X) = \exp[2m + \sigma^2](\exp[\sigma^2] - 1)$ Pareto distribution $\displaystyle f(x) = \frac{x_0^\alpha}{x^{\alpha + 1}}$ $\displaystyle x \ge x_0$ $\displaystyle f(x) = 0$ $\displaystyle x < 0$ $\displaystyle E(X) = \frac{ax_0}{a - 1}$ $\displaystyle a > 1$ $\displaystyle V(X) = \frac{ax_0^2}{a - 2} - \left(\frac{ax_0}{a - 1}\right)^2$ $\displaystyle a > 2$ Weibull distribution $\displaystyle f(x) = \frac{\gamma x^{\gamma - 1}}{\lambda^b}\exp\left[-\left(\frac{x}{y}\right)^\gamma\right]$ $\displaystyle x \ge 0$ $\displaystyle f(x) = 0$ $\displaystyle x < 0$ [latex]\displaystyle E(X) = \lambda\Gamma\left(1 + \frac{1}{\gamma}\right)$ $\displaystyle \lambda > 0$ $\displaystyle V(X) = \lambda^2\left[\Gamma\left(2 + \frac{1}{\gamma}\right) - \left(\Gamma\left(1 + \frac{1}{\gamma}\right)\right)^2\right]$ $\displaystyle \gamma > 0$

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