2019-01-23[Stay Sharp]The beta d

2019-01-23 本文已影响4人三千雨点

The beta distribution is distribution of probabilities.

$\operatorname { Beta } ( \mu | a , b ) = \frac { \Gamma ( a + b ) } { \Gamma ( a ) \Gamma ( b ) } \mu ^ { a - 1 } ( 1 - \mu ) ^ { b - 1 }$
where the $\Gamma$ is the gamma function： $\Gamma ( n ) = ( n - 1 ) !$

It's normalized

verification:
to verify it's normalized, i.e.

$\int _ { 0 } ^ { 1 } Beta(\mu|a, b) d\mu = 1$
so,
$\int _ { 0 } ^ { 1 } \mu ^ { a - 1 } ( 1 - \mu ) ^ { b - 1 } \mathrm { d } \mu = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$
that's what we want to derive.

$\begin{aligned} \Gamma ( a ) \Gamma ( b ) & = \int _ { 0 } ^ { \infty } e ^ {- x } x ^ { a - 1 } d x \int _ { 0 } ^ { \infty } e ^ {- y } y ^ { b - 1 } d y \\ & = \int _ { 0 } ^ { \infty } \int _ { 0 } ^ { \infty } e ^{ - x } x ^ { a - 1 } e^ { - y } y ^ { b - 1 } d x d y \\ & = \int _ { 0 } ^ { \infty } \int _ { 0 } ^ { \infty } e ^ { - t } x ^ { a - 1 } ( t - x ) ^ { b - 1 } d t d x \\ &\\ & \stackrel{t=x + y}{=} \int _ { 0 } ^ { \infty } e^ {- t } \left( \int _ { 0 } ^ { t } x ^ { a - 1 } ( t - x ) ^ { b - 1 } \mathrm { d } x \right) \mathrm { d } t \\ & \stackrel{x=\mu t}{=} \int_{0}^{\infty} e^{-t}\left( \int_{0} ^{t} {(\mu t)}^{a-1}{(t - \mu t)}^{b-1} \mathrm{d}x\right) \mathrm{d}t \\&= \int_{0}^{\infty} e^{-t}\left( {( t)}^{a + b -1} \int_{0} ^{t} {(\mu)}^{a-1}{(1 - \mu)}^{b-1} \mathrm{d}\mu\right) \mathrm{d}t \\&=\int _ { 0 } ^ { \infty } t ^ { a + b - 1 } \mathrm { e } ^ { - t } \mathrm { d } t \cdot \int _ { 0 } ^ { 1 } \mu ^ { a - 1 } ( 1 - \mu ) ^ { b - 1 } \mathrm { d } \mu \\&= \Gamma(a+b) \int _ { 0 } ^ { 1 } \mu ^ { a - 1 } ( 1 - \mu ) ^ { b - 1 } \mathrm { d } \mu \end{aligned}$
so, the beta distribution is normalized.

Ref

https://stats.stackexchange.com/questions/47771/what-is-the-intuition-behind-beta-distribution

http://rads.stackoverflow.com/amzn/click/0387310738

2019-01-23[Stay Sharp]The beta d

It's normalized

Ref

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