Home page for accesible maths 3 Probability Models 3.4.2 Gamma distribution 3.6 The Weibull Distribution:

\operatorname{\mathsf{Weib}}(\alpha,\beta)

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3.5 The Beta Distribution: $\operatorname{\mathsf{Beta}}(\alpha_{1},\alpha_{2})$

Parameters: $\boldsymbol{\theta}=(\alpha_{1},\alpha_{2})$ with $\alpha_{1}>0$ and $\alpha_{2}>0$ both shape parameters.

\displaystyle f_{X}(x;\boldsymbol{\theta})=\frac{1}{B(\alpha_{1},\alpha_{2})}x% ^{\alpha_{1}-1}(1-x)^{\alpha_{2}-1}

for $0\leq x\leq 1$ , where

$B(\alpha_{1},\alpha_{2})=\frac{\Gamma(\alpha_{1})\Gamma(\alpha_{2})}{\Gamma(% \alpha_{1}+\alpha_{2})}$ ,
$\operatorname{\mathsf{E}}\left[{X}\right]=\frac{\alpha_{1}}{\alpha_{1}+\alpha_% {2}}$ ,
${\operatorname{\mathsf{Var}}}\left[{X}\right]=\frac{\alpha_{1}\alpha_{2}}{(% \alpha_{1}+\alpha_{2})^{2}(\alpha_{1}+\alpha_{2}+1)}$ .

We write $X\sim\operatorname{Beta}(\alpha_{1},\alpha_{2})$ .

A proof that $\int_{0}^{1}x^{\alpha_{1}-1}(1-x)^{\alpha 2-1}=B(\alpha_{1},\alpha_{2})$ is given in the appendix.

Usage: The family of Beta distributions constitutes a flexible class of distributions on $[0,1]$ used for modelling. The Beta $(1,1)$ distribution is the uniform distribution on $[0,1]$ .

Example 3.5.1.

Prove that if $X\sim\operatorname{\mathsf{Beta}}(2,5)$ that $\operatorname{\mathsf{E}}\left[{X}\right]=\frac{2}{2+5}$ using the unit integrability property of the pdf.

Solution.

	$\displaystyle\operatorname{\mathsf{E}}\left[{X}\right]$	$\displaystyle=\int_{0}^{1}x\frac{\Gamma(2+5)}{\Gamma(2)\Gamma(5)}x^{2-1}(1-x)^% {5-1}\,\mathrm{d}x$
		$\displaystyle=\frac{\Gamma(2+5)}{\Gamma(2)\Gamma(5)}\int_{0}^{1}x^{3-1}(1-x)^{% 5-1}\,\mathrm{d}x$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{\Gamma(2+5)}{\Gamma(2)\Gamma(5)}% \frac{\Gamma(3)\Gamma(5)}{\Gamma(3+5)}\times\frac{\Gamma(3+5)}{\Gamma(3)\Gamma% (5)}\int_{0}^{1}x^{3-1}(1-x)^{5-1}\,\mathrm{d}x}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{\Gamma(2+5)}{\Gamma(2)\Gamma(5)}% \frac{\Gamma(3)\Gamma(5)}{\Gamma(3+5)}\times 1}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{2}{2+5},}$

using the recurrence relations $\Gamma(\alpha+1)=\alpha\Gamma(\alpha)$ .

Unnumbered Figure: First link, Second Link

Example 3.5.2.

Express the probability that a rv $X\sim\operatorname{\mathsf{Beta}}(2,3)$ is less than $0.5$ as an integral. Evaluate the integral and verify your answer by simulation. Verify by simulation that the expected value of $X$ is about $0.4$ .

Solution.

\displaystyle f_{X}(x)=\frac{\Gamma(2+3)}{\Gamma(2)\Gamma(3)}x^{2-1}(1-x)^{3-1}

for $0\leq x\leq 1$ and since $\frac{\Gamma(5)}{\Gamma(2)\Gamma(3)}=12$ ,

\displaystyle\operatorname{\mathsf{P}}\left({X\leq 0.5}\right)

\displaystyle={\color[rgb]{0.76,0.01,0}\int_{x=0}^{0.5}12x^{1}(1-x)^{2}\,% \mathrm{d}x}

\displaystyle={\color[rgb]{0.76,0.01,0}\int_{x=0}^{0.5}12x-24x^{2}+12x^{3}\,% \mathrm{d}x,}

\displaystyle\operatorname{\mathsf{P}}\left({X\leq 0.5}\right)={\color[rgb]{% 0.76,0.01,0}\left[6x^{2}-8x^{3}+3x^{4}\right]_{0}^{1/2}=11/16=0.6875.}

xsample = rbeta(10000,2,3); mean(xsample<0.5)

[1] 0.6837

mean(xsample)

[1] 0.401731

Note that $\operatorname{\mathsf{E}}\left[{X}\right]=\frac{\alpha_{1}}{\alpha_{1}+\alpha_% {2}}=2/5$ .

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3.5 The Beta Distribution: 𝖡𝖾𝗍𝖺⁡(α1,α2)

Example 3.5.1.

Example 3.5.2.

3.5 The Beta Distribution: $\operatorname{\mathsf{Beta}}(\alpha_{1},\alpha_{2})$