F분포
F = \dfrac{U_1/v_1}{U_2/v_2} \sim ~ F_{v_1,~v_2}
X_1~\sim~\chi^{2}(d_1) \kern{10pt}
X_2~\sim~\chi^{2}(d_2)
\\[10pt]
U = \dfrac{X_1/d_1}{X_2/d_2} \sim
F(d_1, d_2)
- 카이제곱분포의 비율을 나타낸 것
- 분산분석에 활용한다.
CDF
\begin{align*}
&F_U(U) = P(U \le w) = P
\bigg(\dfrac{X_1/d_1}{X_2/d_2} \le \mu\bigg)
\\[20pt]
% 2\text{번째 줄} %
&= P\bigg(~X_1~\le~
\dfrac{d_2}{d_1}~\mu~X_2\bigg)
\\[20pt]
&= \int_{0}^{\infty}
~\int_{0}^{\frac{d_2}{d_1}~\mu~t}
~f_{X_1, X_2}(x_1, x_2)~dx_1dx_2
\\[20pt]
&= \int_{0}^{\infty}~f_{X_1}(t)
\bigg\{
~\int_{0}^{\frac{d_2}{d_1}~\mu~t}
f_{X_2}(s)ds\bigg\}~~dt
\\[20pt]
&= \int_{0}^{\infty}~f_{X_1}(t)
\cdot F_{X_1}
\bigg(\frac{d_2}{d_1}\mu t\bigg) ~dt
\\[20pt]
&= \int_{0}^{\infty}~
\dfrac
{t \raisebox{0.7em}
{$ \textstyle \frac{d_1}{2}\scriptstyle -1$}
~e \raisebox{0.7em}
{$ \scriptstyle - \textstyle \frac{d_1}{2}$}}
{\Gamma(\frac{d_1}{2})~~
2 \raisebox{0.7em}
{$ \textstyle \frac{d_1}{2}$} }
~~\cdot~~
\dfrac
{\big( \frac{d_2}{d_1}\mu t \big) \raisebox{0.9em}
{$ \textstyle \frac{d_1}
{2}\scriptstyle -1$}
~e~ \raisebox{0.9em}
{$ \scriptstyle - \textstyle \frac{1}{2}(\frac{d_2}{d_1}
\mu t) $} }
{\Gamma(\frac{d_2}{2})~~
2 \raisebox{0.7em}
{$ \textstyle \frac{d_1}{2}$} }
~~dt
\\[30pt]
% \text{마지막 줄} %
&= \dfrac
{\frac{d_2}{d_1} \raisebox{0.7em}
{$ \textstyle \frac{d_1}{2} $}~~
\mu \raisebox{0.7em}
{$ \textstyle \frac{d_1}{2}
\scriptstyle - 1$} }
{ \Gamma(\frac{d_1}{2})~
\Gamma(\frac{d_2}{2})~
2 \raisebox{0.7em}
{$ \textstyle \frac{d_1+d_2}{2}$}
}
~~\cdot~~
\int_{0}^{\infty}~
t \raisebox{0.9em}
{$ \textstyle \frac{d_1+d_2}{2}
\scriptstyle -1 $}
~e~ \raisebox{0.9em}
{$ \scriptstyle -
\textstyle \frac{1}{2}
\scriptstyle -
\textstyle \frac{d_2}{2d_1}\mu t $}~~dt
\end{align*}
여기서 exp 안의 분수를 정리한다.
-\frac{1}{2} + \frac{d_2}{2d_1}
= -t~\Big\{ \frac{1}{2} +
\frac{d_2 \mu}{2d_1} \Big\}
= - \frac{t}
{2d_1 / (d_1 + d_2 \mu)}
또 감마함수의 테크닉을 이용한다.
\Gamma(\alpha)\beta^{\alpha}
=
\int_{0}^{\infty}
~x\raisebox{0.7em}
{$ \scriptstyle \alpha-1 $}
~e\raisebox{0.9em}
{$ \scriptstyle -
\textstyle \frac{x}{\beta} $}
~dx
\\[20pt]
\Gamma \Big(
\frac{d_1+d_2}{2}
\Big)
\cdot \Big\{
\frac{2d_1}{d_1+d_2\mu}
\Big\}
\raisebox{1.3em}
{$\frac{d_1+d_2}{2}$}
=
\int_{0}^{\infty}~
t \raisebox{0.9em}
{$ \textstyle \frac{d_1+d_2}{2}
\scriptstyle -1 $}
~e~ \raisebox{0.9em}
{$ \textstyle \frac{t}
{2d_1 / (d_1 + d_2 \mu)}$}~~dt
\begin{align*}
&= \dfrac
{ \Gamma(\frac{d_1+d_2}{2}) }
{
\Gamma(\frac{d_1}{2})~
\Gamma(\frac{d_2}{2})~
2 \raisebox{0.7em}
{$ \textstyle \frac{d_1+d_2}{2}$}
}
~~\cdot~~
\Gamma \Big(
\frac{d_1+d_2}{2}
\Big)
\cdot \Big\{
2 \cdot \frac{d_1}{d_1+d_2\mu}
\Big\}
\raisebox{1.3em}
{$\frac{d_1+d_2}{2}$}
\\[30pt]
&= \dfrac
{ \Gamma(\frac{d_1+d_2}{2}) }
{
\Gamma(\frac{d_1}{2})~
\Gamma(\frac{d_2}{2})}
~~\cdot~~
\Gamma \Big(
\frac{d_1+d_2}{2}
\Big)
\cdot \Big\{
1 + \frac{d_2}{d_1}\mu
\Big\}
\raisebox{1.3em}
{$\scriptstyle -
\frac{d_1+d_2}{2}$}
\end{align*}
f_{\mu}(a) = \dfrac
{ \Gamma(\frac{d_1+d_2}{2}) }
{
\Gamma(\frac{d_1}{2})~
\Gamma(\frac{d_2}{2})~
}
\cdot
\mu \raisebox{0.9em}
{$\textstyle \frac{d_1}{2}$}
\cdot
\Big( 1+
\textstyle\frac{d_2}{d_1}\mu \Big)
\raisebox{0.9em}
{$ \scriptstyle -
\textstyle\frac{d_1+d_2}{2}$}
1차 적률 = 평균
\begin{align*}
E(u) &= E\Big(
\dfrac{X_1 / d_1}{X_2 / d_2}
\Big)
= \dfrac{d_2}{d_1}\cdot
E(X_1)\cdot
E\Big(\frac{1}{d_2}\Big)
\\[18pt]
&= \dfrac{d_2}{d_1}~\cdot~
d_1~\cdot~\frac{1}
{(\frac{d_2}{2} - 1) \cdot 2}
\\[18pt]
&= \frac{d_2}{d_2-2}
= 1 + \frac{2}{d_2-2}
\end{align*}
2차 적률과 분산
\begin{align*}
E(u^2) &=
\dfrac{{d_2}^2}{{d_1}^2}~\cdot~
E(X_1)~\cdot~
E\Big(\frac{1}{{X_2}^2}\Big)
\\[20pt]
&=
\dfrac{{d_2}^2}{{d_1}^2}~\cdot~
\Big( \dfrac{d_2}{2} + 1 \Big)~\cdot
\dfrac{d_1}{2}\cdot 4\cdot
\dfrac{1}
{(\frac{d_2}{2}-1)
(\frac{d_2}{2}-2)\cdot 4}
\\[20pt]
&=
\dfrac{{d_2}^2}{{d_1}^2}~\cdot~
(d_2 + 1)\cdot~d_1~\cdot~
\dfrac{1}
{(d_2-2)(d_2-4)}
\\[20pt]
&=
\dfrac
{{d_2}^2 (d_1+2)}
{d_1 (d_2-2)(d_2-4)}
\end{align*}
이제 분산을 구해보자.
\begin{align*}
Var(u) &= \dfrac{{d_2}^2 (d_1+2)}
{d_1 (d_2-2)(d_2-4)}
- \frac{{d_2}^2}{(d_2-2)^2}
\\[20pt]
&= \dfrac{{d_2}^2}
{d_1 (d_2-2)^2(d_2-4)}
\cdot [(d_1+2)(d_2-2)-d_1(d_2-4)]
\\[20pt]
&= \dfrac{{d_2}^2}
{d_1 (d_2-2)^2(d_2-4)}~
(2d_1+2d_2-1)
\\[20pt]
&= \dfrac{2\cdot{d_2}^2(d_1+d_2-2)}
{d_1\cdot(d_2-2)^2\cdot(d_2-4)}>0
\end{align*}