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

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
$\displaystyle f(x) = {}_{n}C_{x}p^{x}(1 - p)^{n - x}$	$\displaystyle x = 0, 1, \cdots, n$
$\displaystyle E(X) = np$	$\displaystyle 0 < p < 1[/latex]</td> </tr> <tr> <td>[latex]\displaystyle V(X) = np(1 - p)$
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
$\displaystyle f(x) = p(1 - p)^{x - 1}$	$\displaystyle x = 1, 2, 3, \cdots$
$\displaystyle E(X) = \frac{1}{p}$	$\displaystyle 0 < p < 1[/latex]</td> </tr> <tr> <td>[latex]\displaystyle V(X) = \frac{1 - p}{p^2}$
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
$\displaystyle f(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x - m)^2}{2\sigma^2}\right)$	$\displaystyle - \infty < x < \infty[/latex]</td> </tr> <tr> <td>[latex]\displaystyle E(X) = m$
$\displaystyle V(X) = \sigma^2$	$\displaystyle \sigma > 0$
Exponential distribution
$\displaystyle f(x) = \lambda\exp(-\lambda x)$	$\displaystyle x \ge 0$
$\displaystyle f(x) = 0$	$\displaystyle x < 0[/latex]</td> </tr> <tr> <td>[latex]\displaystyle E(X) = \frac{1}{\lambda}$
$\displaystyle V(X) = \frac{1}{\lambda^2}$
Gamma distribution
$\displaystyle f(x) = \frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha - 1}\exp(- \lambda x)$	$\displaystyle x \ge 0$
$\displaystyle f(x) = 0$	$\displaystyle x < 0[/latex]</td> </tr> <tr> <td>[latex]\displaystyle E(X) = \frac{\alpha}{\lambda}$	$\displaystyle \lambda, \alpha > 0$
$\displaystyle V(X) = \frac{\alpha}{\lambda^2}$
Beta distribution
$\displaystyle f(x) = \frac{x^{\alpha - 1}(1 - x)^{\beta - 1}}{B(\alpha, \beta)}$	$\displaystyle 0 < x < 1[/latex]</td> </tr> <tr> <td>[latex]\displaystyle f(x) = 0$	$\displaystyle x \le 0, 1 \le x$
$\displaystyle E(X) = \frac{\alpha}{\alpha + \beta}$
$\displaystyle V(X) = \frac{\alpha\beta}{(\alpha + \beta)^2(\alpha + \beta + 1)}$
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
$\displaystyle f(x) = \frac{x_0^\alpha}{x^{\alpha + 1}}$	$\displaystyle x \ge x_0$
$\displaystyle f(x) = 0$	$\displaystyle x < 0[/latex]</td> </tr> <tr> <td>[latex]\displaystyle E(X) = \frac{ax_0}{a - 1}$	$\displaystyle a > 1$
$\displaystyle V(X) = \frac{ax_0^2}{a - 2} - \left(\frac{ax_0}{a - 1}\right)^2$	$\displaystyle a > 2$
Weibull distribution
$\displaystyle f(x) = \frac{\gamma x^{\gamma - 1}}{\lambda^b}\exp\left[-\left(\frac{x}{y}\right)^\gamma\right]$	$\displaystyle x \ge 0$
$\displaystyle f(x) = 0$	$\displaystyle x < 0[/latex]</td> </tr> <tr> <td>[latex]\displaystyle E(X) = \lambda\Gamma\left(1 + \frac{1}{\gamma}\right)$	$\displaystyle \lambda > 0$
$\displaystyle V(X) = \lambda^2\left[\Gamma\left(2 + \frac{1}{\gamma}\right) - \left(\Gamma\left(1 + \frac{1}{\gamma}\right)\right)^2\right]$	$\displaystyle \gamma > 0$