조건 파악
X_0 - 1 \sim bernoulli(p)
\\[20pt]
\begin{align*}
P(X_0-1 = 0) &= P(X_0=1) = 1-p
\\
P(X_0-1 = 1) &= P(X_0=2) = p
\end{align*}
조건부로 풀기
\begin{align*}
E(X_1X_2X_4) &= E(X_1~E(X_2X_4|X_1))
\\[10pt]
E(X_2X_4|X_1)
&= E(E(X_2X_4|X_1,~X_2))
\\
&= E(X_2E(X_4|X_1,~X_2))
\\
&= E(X_2E(X_4|X_2))
\end{align*}
경우의 수 따져주기
X_2 → X_3 → X_4 로 갈 때의 순서를 보자.
-
0 → 0 → 0
-
0 → 0 → 1
-
0 → 1 → 0
-
0 → 1 → 2
-
1 → 0 → 1
-
1 → 2 → 1
-
1 → 2 → 3
-
2 → 1 → 2
-
2 → 3 → 2
-
3 → 2 → 1
-
3 → 2 → 3
\begin{align*}
E(X_4|X_2=0) &=
\sum_{y}~yP(X_4=y|X_2=0)
\\[5pt]
&= 1 \cdot P(X_4=1|X_2=0) +
2 \cdot P(X_4=2|X_2=0)
\\[5pt]
&= (1-p)\cdot p + 2p^2
\\[5pt]
&= (p^2 + p)~I(X_2=0)
\\[20pt]
E(X_4|X_2=1) &=
\sum_{y}~yP(X_4=y|X_2=1)
\\[5pt]
&= 1 \cdot P(X_4=1|X_2=1) +
3 \cdot P(X_4=3|X_2=1)
\\[5pt]
&= 2p\cdot(1-p) + 3p^2
\\[5pt]
&= (p^2 + 2p)~I(X_2=1)
\\[20pt]
E(X_4|X_2=2) &=
\sum_{y}~yP(X_4=y|X_2=2)
\\[5pt]
&= 2 \cdot P(X_4=2|X_2=2)
\\[5pt]
&= 2\cdot(p-p^2+p)
\\[5pt]
&= (-2p^2 + 4p)~I(X_2=2)
\\[20pt]
E(X_4|X_2=3) &=
\sum_{y}~yP(X_4=y|X_2=3)
\\[5pt]
&= 1 \cdot P(X_4=1|X_2=3) +
3 \cdot P(X_4=3|X_2=3)
\\[5pt]
&= (1-p) + 3p
\\[5pt]
&= (2p + 1)~I(X_2=3)
\end{align*}
\begin{align*}
&E(X_2E(X_4|X_2))
= E\bigg[X_2\big\{
(p^2 + p)~I(X_2=0) +
(p^2 + 2p)~I(X_2=1) +
(-2p^2 + 4p)~I(X_2=2) +
(2p + 1)~I(X_2=3)\big\}
\bigg]
\\[15pt]
&=~ E[X_2~(p^2 + p)~I(X_2=0)] +
E[X_2~(p^2 + 2p)~I(X_2=1)] +
E[X_2~(-2p^2 + 4p)~I(X_2=2)] +
E[X_2~(2p + 1)~I(X_2=3)]
\\[30pt]
&=~(p^2 + p)~E\big[0~I(X_2=0)\big]
+ (p^2 + 2p)~E\big[1~I(X_2=1)\big]
+ (-2p^2 + 4p)
~E\big[2~I(X_2=2)\big]
+ (2p + 1)
~E\big[3~I(X_2=3)\big]
\\[15pt]
&=~
(p^2 + 2p)~P(X_2=1)
+~(-4p^2 + 8p)~P(X_2=2) +
(6p + 3)~P(X_2=3)
\\[15pt]
&=E(X_2X_4|X_1)
\end{align*}
P(X_2 = x_2) 계산
추이행렬을 이용해서 계산.
\begin{bmatrix}
0 & 1-p & p & 0
\end{bmatrix}
\begin{bmatrix}
1-p & p & 0 & 0 \\
1-p & 0 & p & 0 \\
0 & 1-p & 0 & p \\
0 & 0 & 1 & 0 \\
\end{bmatrix}
\\[10pt]
=\begin{bmatrix}
(1-p)^2 & p(1-p) & p(1-p) & p^2
\end{bmatrix}
\\[10pt]
\phi_0(x)~\underline{P} = \phi_1(x)
\begin{bmatrix}
(1-p)^2 & p(1-p) & p(1-p) & p^2
\end{bmatrix}
\begin{bmatrix}
1-p & p & 0 & 0 \\
1-p & 0 & p & 0 \\
0 & 1-p & 0 & p \\
0 & 0 & 1 & 0 \\
\end{bmatrix}
\\[10pt]
=\begin{bmatrix}
(1-p)^2 & p(1-p) & p(1-p) & p^2
\end{bmatrix}
\\[10pt]
\phi_1(x)~\underline{P} = \phi_2(x)
이를 토대로 기댓값을 마저 계산해보자.
\begin{align*}
E(X_2X_4|X_1)=&
~(p^2 + 2p)~P(X_2=1) +
~(-4p^2 + 8p)~P(X_2=2) +
(6p + 3)~P(X_2=3)
\\[10pt]
=&~(p^2 + 2p)\cdot 2p(1-p)^2 +
(-4p^2 + 8p)\cdot p^2(2-p) +
(6p + 3)\cdot p^2(1-p)
\\[10pt]
=&~2p^2\cdot(p+2)\cdot(1-p)^2 + 4p^3 \cdot(2-p)^2 +
3p^2 \cdot(2p+1)(1-p)
\\[10pt]
=&~2p^5 - 6p^3 + 4p^2 +
4p^5 - 16p^4 + 16p^3 -
6p^4 + 3p^3 + 3p^2
\\[10pt]
=&~6p^5 - 22p^4 + 13p^3 + 7p^2
\end{align*}
E(X_1) 계산
앞에서 구한 행렬을 이용한다.
E(X_1) = \sum_x x \cdot P(X_1=x)
\\[5pt]
= 1 \cdot p(1-p) + 2p(1-p) + 3p^2
\\[5pt]
= 3p(1-p) + 3p^2 = 3p
E(X_1X_2X_4) 계산
\begin{align*}
&E(X_1X_2X_4) = E(X_1~E(X_2X_4|X_1))
\\[5pt]
&=E\big[X_1 \cdot
(6p^5 - 22p^4 + 13p^3 + 7p^2)\big]
\\[5pt]
&=E(X_1) \cdot
(6p^5 - 22p^4 + 13p^3 + 7p^2)
\\[5pt]
&=(3p) \cdot
(6p^5 - 22p^4 + 13p^3 + 7p^2)
\\[5pt]
&= 18p^6 - 66p^5 + 39p^4 + 21p^3
\end{align*}