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]</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

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投稿者: admin

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