주어진 추이행렬은 다음과 같다.
\underline{\mathbf{p}} =
\begin{bmatrix}
\dfrac{3}{4} & \dfrac{1}{4}
\\[10pt]
\dfrac{1}{3} & \dfrac{2}{3}
\end{bmatrix}
=
\begin{bmatrix}
p & 1-p
\\[10pt]
1-q & q
\end{bmatrix}
계산의 편리함을 위해 다음 부분을 미리 증명함.
\begin{align*}
&\sum_{k=2}^{\infty}~
kq^{k-1}
= \sum_{k=2}^{\infty}~
\dfrac{d}{dq}~q^k
\\[20pt]
&= \dfrac{d}{dq}
~\sum_{k=2}^{\infty}~q^k =
\dfrac{d}{dq}
\bigg(q^2 \cdot \dfrac{1}{1-q} \bigg)
\\[20pt]
&= \dfrac{d}{dq}
\bigg(-(q+1) + (1+q)^{-1} \bigg)
\\[20pt]
&= -1 + (1-q)^{2}
\end{align*}
1. h_{11}
\begin{align*}
f_{11}(1) &= \dfrac{3}{4} \\[10pt]
f_{11}(2) &=
\dfrac{1}{4} \cdot
\bigg(\dfrac{2}{3} \bigg)^{0}
\cdot \dfrac{1}{3} \\[10pt]
f_{11}(3) &=
\dfrac{1}{4} \cdot
\bigg(\dfrac{2}{3} \bigg)^{1}
\cdot\dfrac{1}{3} \\[10pt]
\vdots \\[10pt]
f_{11}(k) &=
\dfrac{1}{4} \cdot
\bigg(\dfrac{2}{3} \bigg)^{k-2}
\cdot\dfrac{1}{3}
\end{align*}
\begin{align*}
h_{11} &= \sum_{m=1}^{\infty}
~m \cdot f_{11}(m) \\[15pt]
&= \dfrac{3}{4} +
\sum_{m=2}^{\infty}~m \cdot
\dfrac{1}{12} \cdot
\bigg(\dfrac{2}{3}\bigg)^{m-2}
\\[15pt]
&= \dfrac{3}{4} +
\dfrac{1}{12} \cdot
\dfrac{3}{2} \cdot
\sum_{m=2}^{\infty}~m \cdot
\bigg(\dfrac{2}{3}\bigg)^{m-1}
\\[15pt]
&= \dfrac{3}{4} +
\dfrac{1}{8} \cdot
\Bigg\{-1 + \bigg(1-
\dfrac{2}{3}\bigg)^{-2} \Bigg\}
\\[15pt]
&= \dfrac{3}{4} +
\dfrac{1}{8} \cdot (-1 + 9)
\\[15pt]
&= \dfrac{7}{4}
\end{align*}
2. h_{12}
\begin{align*}
f_{12}(y) &= P(X_m=2, X_{m-1}=1, X_{m-2}=1, \cdots X_1=1 | X_0 = 1)
\\[10pt]
&= {1 \over 4} \cdot
\bigg({3 \over 4}\bigg)^{y-1}
\kern{.5cm}
Y \sim Geo\bigg(\dfrac{1}{4}\bigg)
\end{align*}
E(Y) = \sum_{y=1}^{\infty}~y \cdot
{1 \over 4} \cdot
\bigg({3 \over 4}\bigg)^{y-1}
= 4
3. h_{21}
\begin{align*}
f_{21}(y) &= P(X_m=1, X_{m-1}=2, X_{m-2}=2, \cdots X_1=2 | X_0 = 2)
\\[10pt]
&= {1 \over 3} \cdot
\bigg({2 \over 3}\bigg)^{y-1}
\kern{.5cm}
Y \sim Geo\bigg(\dfrac{1}{3}\bigg)
\end{align*}
E(Y) = \sum_{y=1}^{\infty}~y \cdot
{1 \over 3} \cdot
\bigg({2 \over 3}\bigg)^{y-1}
= 3
4. h_{22}
\begin{align*}
f_{22}(1) &= \dfrac{2}{3} \\[10pt]
f_{22}(2) &=
\dfrac{1}{3} \cdot
\bigg(\dfrac{3}{4} \bigg)^{0}
\cdot \dfrac{2}{3} \\[10pt]
f_{22}(3) &=
\dfrac{1}{3} \cdot
\bigg(\dfrac{3}{4} \bigg)^{1}
\cdot\dfrac{2}{3} \\[10pt]
\vdots \\[10pt]
f_{22}(k) &=
\dfrac{1}{3} \cdot
\bigg(\dfrac{3}{4} \bigg)^{k-2}
\cdot\dfrac{2}{3}
\end{align*}
\begin{align*}
h_{22} &= \sum_{m=1}^{\infty}
~m \cdot f_{22}(m) \\[15pt]
&= \dfrac{2}{3} +
\sum_{m=2}^{\infty}~m \cdot
\dfrac{1}{12} \cdot
\bigg(\dfrac{3}{4}\bigg)^{m-2}
\\[15pt]
&= \dfrac{2}{3} +
\dfrac{1}{12} \cdot
\dfrac{4}{3} \cdot
\sum_{m=2}^{\infty}~m \cdot
\bigg(\dfrac{3}{4}\bigg)^{m-1}
\\[15pt]
&= \dfrac{2}{3} +
\dfrac{1}{9} \cdot
\Bigg\{-1 + \bigg(1-
\dfrac{3}{4}\bigg)^{-2} \Bigg\}
\\[15pt]
&= \dfrac{3}{4} +
\dfrac{1}{9} \cdot (-1 + 16)
\\[15pt]
&= \dfrac{7}{3}
\end{align*}