베타분포

choyunjeong·2024년 12월 25일

2.6.6 베타분포

B(a,b)B(a,b)로 표기되는 베타함수는 a>0, b>0a>0,\ b>0에서

B(a,b)=01xa1(1x)b1dxB(a,b)=\int_{0}^{1}x^{a-1}(1-x)^{b-1}dx

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1) 확률밀도함수

f(x;a, b)=1B(a,b)xa1(1x)b1(0<x<1)f(x; a,\ b)=\dfrac{1}{B(a,b)}x^{a-1}(1-x)^{b-1}\quad (0<x<1)

XBETA(a, b)X\sim \text{BETA}(a,\ b)로 표기.

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2) 기댓값과 분산

E(xk)=1B(a,b)01xk+a1(1x)b1dx=B(a+k,b)B(a,b)=Γ(a+k)Γ(b)Γ(a+b+k)Γ(a+b)Γ(a)Γ(b)(B(a,b)=Γ(a)Γ(b)Γ(a+b))=Γ(a+k)Γ(a+b)Γ(a)Γ(a+b+k)\begin{aligned} E(x^k) &=\dfrac{1}{B(a,b)}\int_{0}^{1}x^{k+a-1}(1-x)^{b-1}dx \\[10pt] &=\dfrac{B(a+k,b)}{B(a,b)} =\dfrac{\Gamma(a+k)\Gamma(b)}{\Gamma(a+b+k)}\cdot\dfrac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\quad (\because B(a,b)=\dfrac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)})\\[10pt] &=\dfrac{\Gamma(a+k)\Gamma(a+b)}{\Gamma(a)\Gamma(a+b+k)} \end{aligned}

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E(X)=Γ(a+1)Γ(a+b)Γ(a)Γ(a+b+1)=aa+b\therefore E(X)=\dfrac{\Gamma(a+1)\Gamma(a+b)}{\Gamma(a)\Gamma(a+b+1)}=\dfrac{a}{a+b}\\[10pt]
Var(X)=Γ(a+2)Γ(a+b)Γ(a)Γ(a+b+2)(aa+b)2=(a+1)a(a+b+1)(a+b)(aa+b)2=ab(a+b+1)(a+b)2\begin{aligned}\therefore \text{Var}(X)&=\dfrac{\Gamma(a+2)\Gamma(a+b)}{\Gamma(a)\Gamma(a+b+2)}-\left(\dfrac{a}{a+b}\right)^2 \\[10pt] &=\dfrac{(a+1)a}{(a+b+1)(a+b)}-\left(\dfrac{a}{a+b}\right)^2\\[10pt] &=\dfrac{ab}{(a+b+1)(a+b)^2}\end{aligned}

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