ex.~~X \sim exp(\lambda)
\begin{align*}
f_X(x) &= \lambda e^{-\lambda x}
,~~x\ge0,~~\lambda\ge0
\\[10pt]
F_X(x) &= 1 - e^{-\lambda x},~~x\ge0
\end{align*}
- X_{(1)}의 확률함수
\begin{align*}
f_{X_{(1)}}(n) =&
~n~\{1-(1-e^{-\lambda x})\}\cdot
\lambda e^{-\lambda x}
\\[10pt]
=&~n~(e^{-\lambda x})^{n-1}~\cdot~
\lambda e^{-\lambda x}
\\[10pt]
=&~n~\lambda ~e^{-n\lambda x}
\\[20pt]
\therefore~~X_{(1)} \sim & ~exp(n\lambda)
\end{align*}
- X_{(n)}의 확률함수
\begin{align*}
f_{X_2}(x) =&~
n(n-1)~P_{X}~
(1-P_{X})^{n-2}~\cdot~{P_{X}}^{\prime}
\\[10pt]
=&~n(n-1)~F_{X_1}(x)
\{1-F_{X_1}(x)\}^{n-2}
~\cdot~f_{x_1}(x)
\end{align*}
- X_{(3)}의 확률함수
\begin{align*}
&F_{X_3}(x)=~P(Y \le 3)
= 1 - \sum_{i=0}^{2}~P(Y=i)
\\[10pt]
=&~1-\binom{n}{0}p^0(1-p)^n
- \binom{n}{1}p^1(1-p)^{n-1}
- \binom{n}{2}p^2(1-p)^{n-2}
\end{align*}
\begin{align*}
f_{X_2}(x) =&~
n(n-1)~P_{X}~
(1-P_{X})^{n-2}~\cdot~{P_{X}}^{\prime}
\\[10pt]
=&~n(n-1)~F_{X_1}(x)
\{1-F_{X_1}(x)\}^{n-2}
~\cdot~f_{x_1}(x)
\end{align*}
-
X_{(1)} , X_{(n)} 의 결합확률함수
(n-1) \cdot \dfrac {f_{X_1}(x_1)}{F_{X_1}(x_1)} \cdot \bigg\{ \dfrac {F_{X_1}(x_2)-F_{X_1}(x_1)} {F_{X_1}(x_2)} \bigg\}^{n-2}
(n-1) \cdot
\dfrac
{\lambda e^{-\lambda x_1}}
{1 - e^{-\lambda x_2}}
\cdot
\bigg\{
\dfrac
{e^{-\lambda x_1}-e^{-\lambda x_2}}
{1 - e^{-\lambda x_2}}
\bigg\}^{n-2}