감마분포
- 먼저 감마함수에 대해 아는게 필요하다. 클릭!
\begin{align*}
X &\sim gamma(\alpha, \beta) \\\\
f_{X}(x) &= \frac{1}
{\Gamma(\alpha)\beta^\alpha}~
x^{\alpha-1}e^{-\frac{x}{\beta}}_~,~ I(x>0)
\end{align*}
확률함수 조건 확인
\int_{-\infty}^{\infty}f_X(x)dx
=
\int_{0}^{\infty}
\dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}
~x^{\alpha-1}
~e^{-\frac{x}{\beta}}
\\[15pt]
(y~=~\dfrac{x}{\beta},
~x~=~\beta y~,
~J~=~\beta )
\\[15pt]
\begin{align*}
&=\int_{0}^{\infty}
\dfrac{1}
{\Gamma(\alpha)\beta^{\alpha}}
~(\beta y)^{\alpha-1}
~e^{-y}\beta
\\[15pt]
&=\int_{0}^{\infty}
\dfrac{1}
{\Gamma(\alpha)}
~y^{\alpha-1}
~e^{-y}
\\[15pt]
&=1
\end{align*}
여기서 중요한 특징을 볼 수 있다.
\int_{0}^{\infty}
\dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}
~x^{\alpha-1}
~e^{-\frac{x}{\beta}}~dx = 1
\\[20pt]
\Gamma(\alpha)\beta^{\alpha}
=
\int_{0}^{\infty}
~x^{\alpha-1}
~e^{-\frac{x}{\beta}}
~dx
이를 이용해 적률 계산에 활용한다.
1차 적률 = 평균
\begin{align*}
E(X) &= \int_{0}^{\infty}
x\cdot f_X(x)dx
\\[15pt]
&= \int_{0}^{\infty}
x\cdot \dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}
~x^{\alpha-1}~e^{-\frac{x}{\beta}}
\\[15pt]
&= \dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}
\int_{0}^{\infty}
~x^{(\alpha+1)-1}
~e^{-\frac{x}{\beta}}
\\[15pt]
&= \dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}
\cdot \Gamma(\alpha+1)
\cdot \beta^{\alpha+1}
\\[15pt]
&= \alpha\beta
\end{align*}
2차 적률과 분산
\begin{align*}
E(X^2) &= \int_{-\infty}^{\infty}
~x^2 \cdot f_X(x)dx
\\[15pt]
&= \int_{-\infty}^{\infty}
~x^2 \cdot \dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}
~x^{\alpha-1}~e^{-\frac{x}{\beta}}
\\[15pt]
&= \dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}
\cdot \Gamma(\alpha+2)
\cdot \beta^{\alpha+2}
\\[15pt]
&= \alpha(\alpha+1)\beta^2
\end{align*}
\begin{align*}
Var(X) &= E(X^2) - E(X^2)
\\[15pt]
&= \alpha(\alpha+1)\beta^2~-~
\alpha^2\beta^2
\\[15pt]
& = \alpha\beta^2
\end{align*}
MGF
\begin{align*}
M_Y(y) &= E(e^{ty})
\\[15pt]
&= \int_{0}^{\infty}
e^{ty} \cdot
\dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}
~y^{\alpha-1}~e^{-\frac{y}{\beta}}
\\[20pt]
&= \dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}
\cdot\int_{0}^{\infty}
y^{\alpha-1}
\exp \bigg\{
-y~\Big(\frac{1}{\beta} - t\Big) \bigg\}
\\[20pt]
&= \dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}
\cdot\int_{0}^{\infty}
y^{\alpha-1}
\exp \bigg\{
-y~\Big(\frac{1-\beta t}{\beta} \Big) \bigg\}
\\[20pt]
&= \dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}
\cdot\int_{0}^{\infty}
y^{\alpha-1}
\exp \bigg\{
-y \div \Big(\frac{\beta}
{1-\beta t}\Big) \bigg\}
\\[20pt]
\end{align*}
\begin{align*}
M_Y(y) &= \dfrac{1}{\Gamma(\alpha)~\beta^{\alpha}}
\cdot
\int_{0}^{\infty}
y^{\alpha-1}
\exp \bigg\{
- \dfrac{y}{\beta_*}
\bigg\} \kern{50pt}
\\[20pt]
&=\dfrac
{1}
{\Gamma(\alpha)~\beta^{\alpha}}
\cdot
{\Gamma(\alpha)~{\beta_*}^{\alpha}}
\\[20pt]
&= \bigg(
\dfrac{\beta_*}{\beta}
\bigg)
^{\alpha}
= \bigg( \dfrac
{\beta}{1-\beta t}~\cdot
~\dfrac{1}{\beta}
\bigg)^{\alpha}
\\[40pt]
\therefore~
M_Y(y) &= (1-\beta t)^{-\alpha}
~~I
\bigg(t < \dfrac{1}{\beta}\bigg)
\end{align*}
계산
\begin{align*}
E(X) &= \dfrac
{\Gamma(\alpha + k)~\beta^{a+k}}
{\Gamma(\alpha)~\beta^{a}}
\\[20pt]
E(X^2) &= \dfrac
{\Gamma(\alpha + 2)~\beta^{a+2}}
{\Gamma(\alpha)~\beta^{a}}
= (\alpha + 1)\alpha \cdot
\beta^{\alpha}
\\[20pt]
E(X^{-1}) &= \dfrac
{\Gamma(\alpha-1)~\beta^{a-1}}
{\Gamma(\alpha)~\beta^{a}} =
\dfrac {1}
{(\alpha-1)\beta}
\\[20pt]
E(X^{-2}) &= \dfrac
{\Gamma(\alpha-2)~\beta^{a-2}}
{\Gamma(\alpha)~\beta^{a}} =
\dfrac {1}
{(\alpha-1)(\alpha-1)\beta^2}
\\[20pt]
\end{align*}