각종 예시들을 모아놓은 자료.
변수 2개 case
예시 1
- P(X_2|X_1 = x_1) 구하기
- E(X_2~|~X_1 = x_1) 구하기
- E(X_1X_2) 구하기
해답 1
P(X_2|X_1 = x_1) =
{P(X_2=x_2,~X_1 = x_1)
\over P(X_1 = x_1)}
\\[10pt]
P(X_2=x_2,~X_1 = x_1)
=P()
해답 2
\begin{align*}
&E(X_2|X_1=x_1) =
E(X_2-X_1+X_1|X_1=x_1)
\\[10pt]
&=E(X_2-X_1|X_1=x_1)E(X_1|X_1=x_1)
\\[10pt]
&=\lambda + x_1 = g(x_1)
\end{align*}
해답 3
\begin{align*}
&E(X_1X_2) = E\Big(X_1(\lambda+X_1)\Big)
\\[10pt]
&=\lambda E(X_1) + E({X_1}^2) =
\lambda + 2\lambda^2
\end{align*}
예시 2
- P(X_{1} = x_{1}~|~X_{10} = x_{10})
- P(X_{10} = x_{10}~|~X_{1} = x_{1})
해답 1
P(X_1=x_1,~X_{10}=x_{10})
\\[10pt]
= {e^{-\lambda}\cdot\lambda^{x_1} \over (x_1)!}
\cdot
{e^{-9\lambda}\cdot
\lambda^{x_{10}-x_{9}}
\over (x_{10}-x_{9})!}
해답 2
P(X_1=x_1~|~X_{10}=x_{10})
해답 3
P(X_{10}=x_{10}~|~X_1=x_1)
예시 3 : 시점 비교
E(X_2X_3) = E\Big(E(X_2X_3|X_2)\Big)
=E\Big(E(X_2X_3|X_3)\Big)
P(X_3=x_3~|~X_{2}=x_{2})
P(X_2=x_2~|~X_{3}=x_{3})
E(X_2X_3)
변수 3개 case
예시 1
E(X_1X_3X_5) 구하기 : Ver.1
\begin{align*}
&E(X_1X_3X_5) = E\big(X_1~E(X_3X_5|X_1)\big)
=E(X_3X_5|X_1=x_1)
\\[15pt]
& = \sum_{x_3} \sum_{x_5}~x_3\cdot x_5\cdot
~P(X_3=x_3,X_5=x_5|X_1=x_1)
\\[25pt]
& = \sum_{x_3} \sum_{x_5}~x_3
\cdot~x_5~\cdot
{e^{-2\lambda}~(2\lambda)^{x_3-x_1}
\over (x_3-x_1)!}~\cdot~
{e^{-2\lambda}~(2\lambda)^{x_5-x_3}
\over (x_5-x_3)!}
\end{align*}
E(X_1X_3X_5) 구하기 : Ver.2
\begin{align*}
&E(X_1X_3X_5) = E\Big[E(X_1X_5X_7|X_7)\Big]
\\[10pt]
&=E\Big[X_1(X_5-X-1)|X_7\Big] +
E({X_1}^2|X_7)
\\[10pt]
&= x_7~(x_7-1) \cdot {1 \over 7} \cdot {4 \over 7} +
{x_7} \cdot {1 \over 7} \cdot
{6 \over 7} +
{x_7}^2 \cdot
\bigg({1 \over 7}\bigg)^2
\end{align*}
X_1|X_7=x_7 \sim binominal\bigg(x_{7,}~~
{1 \over 7}\bigg)
예시 2
E(X_1X_5X_9) 구하기 : 정상독립증분을 이용한다.
\begin{align*}
E(&X_1X_5X_9) =
E\Big[X_1X_5(X_7-X_5)
+ X_1X_5(X_5-0)\Big]
\\[15pt]
=& ~E(X_1X_5)~E(X_7-X_5) + E(X_1{X_5}^2)
\\[15pt]
=& ~E[X_1(X_5-X_1)+E({X_1}^2)]
~E(X_2) + E[X_1(X_5-X_1+X_1)^2]
\\[15pt]
=& ~[E(X_1)E(X_4) + E({X_1}^2)]~E(X_2)
\\[10pt]
&\kern{15pt}
+ E\Big\{X_1~(X_5-X_1)^2
+ 2X_1(X_5-X_1) + {X_1}^2\Big\}
\\[15pt]
=&~E(X_1)E(X_2)E(X_4) + E({X_1}^2) E(X_2)
\\[10pt]
&\kern{15pt}
+ 2E({X_1}^2)E(X_5-X_1)
+ E({X_1}^3)
\\[15pt]
=&~\lambda(2\lambda)(4\lambda)
+ (\lambda + \lambda^2)~2\lambda
+ \lambda (4\lambda + 16\lambda^2)
\\[10pt]
& \kern{10pt}
+ 2(\lambda + \lambda^2)4\lambda
+ (\lambda^3+3\lambda^2+\lambda)
\\[15pt]
=&~35\lambda^3 + 17\lambda^2 +
\lambda
\end{align*}
예시 4
최근 시점이 동률인 경우 : 정리가 된다.
\begin{align*}
&P(X_3 = x_3~|~X_2 =
x_{2,}~X_1\le x_1) =
P(X_3 = x_3~|~X_2 = x_2)
\\[20pt]
&={P(X_3=x_{3,}~X_2=x_{2,}~X_1=x_1)
\over P(X_2=x_{2,}~X_1 \le x_1)}
\\[20pt]
&=~ {P(X_3-X_2=x_{3}-x_{2}),
~P(X_2=x_{2,~}X_1\le x_1)
\over P(X_2=x_{2,}~X_1 \le x_1)}
\\[20pt]
&= P(X_3-X_2=x_3-x_2)
= P(X_3=x_3~|~X_2=x_2)
\end{align*}
최근 시점이 동률이 아닌 경우 : 정리가 안 된다.
\begin{align*}
P(&X_3 = x_3~|~X_2 \le x_2) =
{P(X_3 = x_{3,}~X_2 \le x_2)
\over P(X_2 \le x_2)}
\\[15pt]
&= {\sum_{k=0}^{x_2}
P(X_3 = x_{3,}~X_2 \le k)
\over P(X_2 \le x_2)}
\\[15pt]
&= {\sum_{k=0}^{x_2}
P(X_3 = x_{3,}~X_2 \le k)
\over P(X_2 \le x_2)}
\\[15pt]
\end{align*}