Table of Probability Density Functions

	probability density function	argument and parameter(s)	Moment-generating function $M_{X}\left( t \right) = E(e^{\text{tX}})$	Moments $\mu = E\left( X \right) = \int_{- \infty}^{\infty}{x \cdot f\left( x \right)\text{dx}}$<$\sigma^{2} = \text{Var}\left( X \right) = \int_{- \infty}^{\infty}{\left( x - \mu \right)^{2} \cdot f\left( x \right)\text{dx}}$	Random variable generator	Remark	Corresponding of discrete distributions	R
uniform distribution	$u\left( x;\ \alpha,\ \beta \right) = \left\{ \begin{matrix} \frac{1}{\beta - \alpha} & \mathrm{\text{for }}\alpha < x < \beta \\ 0 & \mathrm{\text{elsewhere}} \end{matrix} \right)$	$\alpha < \beta$		$\mu = \frac{\alpha + \beta}{2}$ $\sigma^{2} = \frac{\left( \beta - \alpha \right)^{2}}{12}$	$X=\alpha +(\beta -\alpha )U(0,1)$	The beta distribution with $\alpha = \beta = 1$		[dpqr]unif(x, alpha, beta)
beta distribution	$f\left( x \right) = \left\{ \begin{matrix} \frac{\Gamma\left( \alpha + \beta \right)}{\Gamma\left( \alpha \right)\Gamma\left( \beta \right)}x^{\alpha - 1}\left( 1 - x \right)^{\beta - 1} & \mathrm{\text{for }}0 < x < 1 \\ 0 & \mathrm{\text{elsewhere}} \end{matrix} \right)$	$\alpha > 0$ $\beta > 0$		$\mu = \frac{\alpha}{\alpha + \beta}$ $\sigma^{2} = \frac{\text{αβ}}{\left( \alpha + \beta \right)^{2}\left( \alpha + \beta + 1 \right)}$		¹		[dpqr]beta(x, alpha, beta)
exponential distribution	$g(x;\ \lambda) = \left\{ \begin{matrix} \frac{1}{\lambda}e^{- \frac{x}{\lambda}} & \mathrm{\text{for }}x > 0 \\ 0 & \mathrm{\text{elsewhere}} \\ \end{matrix} \right)$	$\lambda > 0$	$M_{X}\left( t \right) = \left( 1 - \text{λt} \right)^{- 1}$	$\mu = \frac{1}{\lambda}$ $\sigma^{2} = \frac{1}{\lambda^{2}}$	$X=-\frac{1}{\lambda}\ln(1-U(0,1))$ or $X=-\frac{1}{\lambda}\ln(U(0,1))$	²		[dpqr]exp(x, lambda)
chi-square distribution	$f\left( x \right) = \left\{ \begin{matrix} \frac{1}{\lambda^{\nu/2}\Gamma\left( \nu/2 \right)}x^{\frac{\nu}{2} - 1}e^{- \frac{x}{2}} & \mathrm{\text{for }}x > 0 \\ 0 & \mathrm{\text{elsewhere}} \\ \end{matrix} \right)$	$\nu > 0$ (no. of degrees of freedom = degrees of freedom)	$M_{X}\left( t \right) = \left( 1 - 2t \right)^{- \nu/2}$	$\mu = \nu$ $\sigma^{2} = 2\nu$		³
gamma distribution	$g(x;\ \alpha,\ \lambda) = \left\{ \begin{matrix} \frac{1}{\lambda^{\alpha}\Gamma\left( \alpha \right)}x^{\alpha - 1}e^{- \frac{x}{\lambda}} & \mathrm{\text{for }}x > 0 \\ 0 & \mathrm{\text{elsewhere}} \\ \end{matrix} \right)$	$\alpha > 0$ $\lambda > 0$	$M_{X}\left( t \right) = \left( 1 - \text{λt} \right)^{- \alpha}$	$\mu = \frac{\alpha}{\lambda}$ $\sigma^{2} = \frac{\alpha}{\lambda^{2}}$		⁴		[dpqr]gamma(x, alpha, lambda)
normal distribution (= Gaussian distribution)	$n(x;\ \mu,\ \sigma) = \frac{1}{\sigma\sqrt{2\pi}}e^{- \frac{1}{2}\left( \frac{x - \mu}{\sigma} \right)^{2}}$	$- \infty < \mu < \infty$ $\sigma > 0$	$M_{X}\left( t \right) = e^{\text{μt} + \frac{1}{2}\sigma^{2}t^{2}}$	$\mu = \mu$ $\sigma^{2} = \sigma^{2}$	$X=\sqrt{2}\text{erf}^{-1}(2U(0,1)-1)$	⁵	Binomial distribution $\mu = \text{np}$ $\sigma^{2} = \text{np}(1 - p)$	[dpqr]norm(x, mu, sigma)
standard normal distribution	$n(x;0,\ 1) = \frac{1}{\sqrt{2\pi}}e^{- \frac{1}{2}x^{2}}$		$M_{X}\left( t \right) = e^{\frac{t^{2}}{2}}$	$\mu = 0$ $\sigma^{2} = 1$
Cauchy distribution	$f\left( x \right) = \frac{\beta/\pi}{\beta^{2} + \left( x - \alpha \right)^{2}}$ for $- \infty < x < \infty$	$- \infty < \alpha < \infty$ $\beta > 0$		$\mu$ and $\sigma^{2}$ does not exist.		Normal $\div$ Normal
Rayleigh distribution	$f\left( x \right) = \left\{ \begin{matrix} 2\text{αx}e^{- \alpha x^{2}} & \mathrm{\text{for }}x > 0 \\ 0 & \mathrm{\text{elsewhere}} \\ \end{matrix} \right)$	$\alpha > 0$		$\mu = \frac{1}{2}\sqrt{\frac{\pi}{\alpha}}$ $\sigma^{2} = \frac{1}{\alpha}\left( 1 - \frac{\pi}{4} \right)$
Pareto distribution	$f\left( x \right) = \left\{ \begin{matrix} \frac{\alpha}{x^{\alpha + 1}} & \mathrm{\text{for }}x > 1 \\ 0 & \mathrm{\text{elsewhere}} \\ \end{matrix} \right)$	$\alpha > 0$		$\mu = \frac{\alpha}{\alpha - 1}$ provided $\alpha > 1$ $\sigma^{2} = \frac{\alpha}{\left( \alpha - 1 \right)^{2}\left( \alpha - 2 \right)}$ provided $\alpha > 2$ * The $r$-th moment about the origin $\mu_{r}^{‘}$ exists only if $r < \alpha$	$X=(1-U(0,1))^{-\frac{1}{\alpha}}$ or $X=U(0,1)^{-\frac{1}{\alpha}}$
Weibull distribution	$f\left( x \right) = \left\{ \begin{matrix} kx^{\beta - 1}e^{- \alpha x^{\beta}} & \mathrm{\text{for }}x > 0 \\ 0 & \mathrm{\text{elsewhere}} \\ \end{matrix} \right)$	$\alpha > 0$ $\beta > 0$		$\mu = \alpha^{- \frac{1}{\beta}}\Gamma\left( 1 + \frac{1}{\beta} \right)$		⁶

* gamma function

$\Gamma\left( \alpha \right) = \int_{0}^{\infty}{x^{\alpha - 1}e^{-x}\text{dx}}$ for $\alpha > 0$

$\Gamma\left( \alpha \right) = \left( \alpha - 1 \right)!$

In recent year, the beta distribution has found important applications in Bayesian inference, where parameters are looked upon as random variables, and there is a need for a fairly "flexible" probability density for the parameter $\theta$ of the binomial distribution, which takes on nonzero values only on the interval from 0 to 1.
If $\alpha > 1$ and $\beta > 1$, the beta density has a relative maximum at $x = \frac{\alpha - 1}{\alpha + \beta - 2}$ ↩
Waiting time between random events ($\lambda$ = the rate at which the events occur) = The gamma distribution with $\alpha = 1$ ↩
The gamma distribution with $\alpha = \nu/2$ and $\lambda = 2$. The chi-square distribution plays a very important role in sampling theory. ↩
The total waiting time for all $\alpha$ events (independent and identically distributed $\text{Exp}(\lambda)$) to occur distribution = $\text{Exp}\left( \lambda \right) + \text{Exp}\left( \lambda \right) + \cdots$ $\rightarrow$ sum of $\alpha$ exponential distributions = used to model positive-valued, continuous quantities whose distribution is right-skewed = As $\alpha$ increases, the gamma distribution more closely resembles the normal ↩
Limiting distribution of sums (and averages) of random variables (Central Limit Theorem)
* 99% of the probability mass is concentrated within $3\sigma$ ↩
Lieblein and Zelen proposed. This is useful for modeling the number of revolutions to failure. Weibull distributions with $\beta = 1$ are exponential distributions. ↩

Soonkyu Jeong

Table of Probability Density Functions

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	probability density function	argument and parameter(s)	Moment-generating function $M_{X}\left( t \right) = E(e^{\text{tX}})$	Moments $\mu = E\left( X \right) = \int_{- \infty}^{\infty}{x \cdot f\left( x \right)\text{dx}}$<$\sigma^{2} = \text{Var}\left( X \right) = \int_{- \infty}^{\infty}{\left( x - \mu \right)^{2} \cdot f\left( x \right)\text{dx}}$	Random variable generator	Remark	Corresponding of discrete distributions	R
uniform distribution	\(u\left( x;\ \alpha,\ \beta \right) = \left\{ \begin{matrix} \frac{1}{\beta - \alpha} & \mathrm{\text{for }}\alpha < x < \beta \\ 0 & \mathrm{\text{elsewhere}} \end{matrix} \right)\)	$\alpha < \beta$		$\mu = \frac{\alpha + \beta}{2}$ $\sigma^{2} = \frac{\left( \beta - \alpha \right)^{2}}{12}$	$X=\alpha +(\beta -\alpha )U(0,1)$	The beta distribution with $\alpha = \beta = 1$		[dpqr]unif(x, alpha, beta)
beta distribution	\(f\left( x \right) = \left\{ \begin{matrix} \frac{\Gamma\left( \alpha + \beta \right)}{\Gamma\left( \alpha \right)\Gamma\left( \beta \right)}x^{\alpha - 1}\left( 1 - x \right)^{\beta - 1} & \mathrm{\text{for }}0 < x < 1 \\ 0 & \mathrm{\text{elsewhere}} \end{matrix} \right)\)	$\alpha > 0$ $\beta > 0$		$\mu = \frac{\alpha}{\alpha + \beta}$ $\sigma^{2} = \frac{\text{αβ}}{\left( \alpha + \beta \right)^{2}\left( \alpha + \beta + 1 \right)}$		¹		[dpqr]beta(x, alpha, beta)
exponential distribution	\(g(x;\ \lambda) = \left\{ \begin{matrix} \frac{1}{\lambda}e^{- \frac{x}{\lambda}} & \mathrm{\text{for }}x > 0 \\ 0 & \mathrm{\text{elsewhere}} \\ \end{matrix} \right)\)	$\lambda > 0$	$M_{X}\left( t \right) = \left( 1 - \text{λt} \right)^{- 1}$	$\mu = \frac{1}{\lambda}$ $\sigma^{2} = \frac{1}{\lambda^{2}}$	$X=-\frac{1}{\lambda}\ln(1-U(0,1))$ or $X=-\frac{1}{\lambda}\ln(U(0,1))$	²		[dpqr]exp(x, lambda)
chi-square distribution	\(f\left( x \right) = \left\{ \begin{matrix} \frac{1}{\lambda^{\nu/2}\Gamma\left( \nu/2 \right)}x^{\frac{\nu}{2} - 1}e^{- \frac{x}{2}} & \mathrm{\text{for }}x > 0 \\ 0 & \mathrm{\text{elsewhere}} \\ \end{matrix} \right)\)	$\nu > 0$ (no. of degrees of freedom = degrees of freedom)	$M_{X}\left( t \right) = \left( 1 - 2t \right)^{- \nu/2}$	$\mu = \nu$ $\sigma^{2} = 2\nu$		³
gamma distribution	\(g(x;\ \alpha,\ \lambda) = \left\{ \begin{matrix} \frac{1}{\lambda^{\alpha}\Gamma\left( \alpha \right)}x^{\alpha - 1}e^{- \frac{x}{\lambda}} & \mathrm{\text{for }}x > 0 \\ 0 & \mathrm{\text{elsewhere}} \\ \end{matrix} \right)\)	$\alpha > 0$ $\lambda > 0$	$M_{X}\left( t \right) = \left( 1 - \text{λt} \right)^{- \alpha}$	$\mu = \frac{\alpha}{\lambda}$ $\sigma^{2} = \frac{\alpha}{\lambda^{2}}$		⁴		[dpqr]gamma(x, alpha, lambda)
normal distribution (= Gaussian distribution)	\(n(x;\ \mu,\ \sigma) = \frac{1}{\sigma\sqrt{2\pi}}e^{- \frac{1}{2}\left( \frac{x - \mu}{\sigma} \right)^{2}}\)	$- \infty < \mu < \infty$ $\sigma > 0$	$M_{X}\left( t \right) = e^{\text{μt} + \frac{1}{2}\sigma^{2}t^{2}}$	$\mu = \mu$ $\sigma^{2} = \sigma^{2}$	$X=\sqrt{2}\text{erf}^{-1}(2U(0,1)-1)$	⁵	Binomial distribution $\mu = \text{np}$ $\sigma^{2} = \text{np}(1 - p)$	[dpqr]norm(x, mu, sigma)
standard normal distribution	\(n(x;0,\ 1) = \frac{1}{\sqrt{2\pi}}e^{- \frac{1}{2}x^{2}}\)		$M_{X}\left( t \right) = e^{\frac{t^{2}}{2}}$	$\mu = 0$ $\sigma^{2} = 1$
Cauchy distribution	$f\left( x \right) = \frac{\beta/\pi}{\beta^{2} + \left( x - \alpha \right)^{2}}$ for $- \infty < x < \infty$	$- \infty < \alpha < \infty$ $\beta > 0$		$\mu$ and $\sigma^{2}$ does not exist.		Normal $\div$ Normal
Rayleigh distribution	\(f\left( x \right) = \left\{ \begin{matrix} 2\text{αx}e^{- \alpha x^{2}} & \mathrm{\text{for }}x > 0 \\ 0 & \mathrm{\text{elsewhere}} \\ \end{matrix} \right)\)	$\alpha > 0$		$\mu = \frac{1}{2}\sqrt{\frac{\pi}{\alpha}}$ $\sigma^{2} = \frac{1}{\alpha}\left( 1 - \frac{\pi}{4} \right)$
Pareto distribution	\(f\left( x \right) = \left\{ \begin{matrix} \frac{\alpha}{x^{\alpha + 1}} & \mathrm{\text{for }}x > 1 \\ 0 & \mathrm{\text{elsewhere}} \\ \end{matrix} \right)\)	$\alpha > 0$		$\mu = \frac{\alpha}{\alpha - 1}$ provided $\alpha > 1$ $\sigma^{2} = \frac{\alpha}{\left( \alpha - 1 \right)^{2}\left( \alpha - 2 \right)}$ provided $\alpha > 2$ * The $r$-th moment about the origin $\mu_{r}^{‘}$ exists only if $r < \alpha$	$X=(1-U(0,1))^{-\frac{1}{\alpha}}$ or $X=U(0,1)^{-\frac{1}{\alpha}}$
Weibull distribution	\(f\left( x \right) = \left\{ \begin{matrix} kx^{\beta - 1}e^{- \alpha x^{\beta}} & \mathrm{\text{for }}x > 0 \\ 0 & \mathrm{\text{elsewhere}} \\ \end{matrix} \right)\)	$\alpha > 0$ $\beta > 0$		$\mu = \alpha^{- \frac{1}{\beta}}\Gamma\left( 1 + \frac{1}{\beta} \right)$		⁶