Question

Given the transition matrix P and the? initial-state matrix Upper S 0S0 ?below, find Upper P Superscript 4P4 and use Upper P Superscript 4P4 to find Upper S 4S4. A B Upper PPequals= A B left bracket Start 2 By 2 Matrix 1st Row 1st Column 0.8 2nd Column 0.2 2nd Row 1st Column 0.3 2nd Column 0.7 EndMatrix right bracket 0.8 0.2 0.3 0.7 ?; Upper S 0S0equals= left bracket Start 1 By 2 Matrix 1st Row 1st Column 0.1 2nd Column 0.9 EndMatrix right bracket

Answer 1

Answer:

$S^4=\left[\begin{array}{ccc}0.56875&0.43125\end{array}\right]$

Step-by-step explanation:

$\text{Initial-State Matrix}, S_0=\left[\begin{array}{ccc}0.1&0.9\end{array}\right]$

$\text{Transition Matrix}, P=\left[\begin{array}{ccc}0.8&0.2\\0.3&0.7\end{array}\right]$

First, we are to determine $P^4$.

$P^2=\left[\begin{array}{ccc}0.8&0.2\\0.3&0.7\end{array}\right]\left[\begin{array}{ccc}0.8&0.2\\0.3&0.7\end{array}\right]=\left[\begin{array}{ccc}0.8*0.8+0.2*0.3&0.8*0.2+0.2*0.7\\0.3*0.8+0.7*0.3&0.3*0.2+0.7*0.7\end{array}\right]\\\\=\left[\begin{array}{ccc}0.7&0.3\\0.45&0.55\end{array}\right]$

$P^4=(P^2)^2=\left[\begin{array}{ccc}0.7&0.3\\0.45&0.55\end{array}\right]\left[\begin{array}{ccc}0.7&0.3\\0.45&0.55\end{array}\right]\\=\left[\begin{array}{ccc}0.7*0.7+0.3*0.45&0.7*0.3+0.3*0.55\\0.45*0.7+0.55*0.45&0.45*0.3+0.55*0.55\end{array}\right]\\\\=\left[\begin{array}{ccc}0.625&0.375\\0.5625&0.4375\end{array}\right]$

Therefore:

$S^4=S_0P^4$

$=\left[\begin{array}{ccc}0.1&0.9\end{array}\right]\left[\begin{array}{ccc}0.625&0.375\\0.5625&0.4375\end{array}\right]\\=\left[\begin{array}{ccc}0.1*0.625+0.9*0.5625&0.1*0.375+0.9*0.4375\end{array}\right]\\\\S^4=\left[\begin{array}{ccc}0.56875&0.43125\end{array}\right]$