In Example 11.4, let a = 0 and b = 1=2. Find$ P; P^2; and P^3: $What would Pn be? What happens to Pn as n tends to infinity? Interpret this result.
Lets find $ P; P^2; and P^3:$ a = 0 ; b = 1/2
\(P = \begin{bmatrix} 1 & 0 \\0 & \frac {1}{2}\end{bmatrix}\)
From state 1 (the first row), the system transitions to itself with probability 1 (since
a=0), and from state 2 (the second row), the system transitions back to state 1 with probability 0 and stays in state 2 with probability \(\frac {1}{2}\)
\(P^2 = \begin{bmatrix} 1 & 0 \\0 & \frac {1}{2}\end{bmatrix} * \begin{bmatrix} 1 & 0 \\0 & \frac {1}{2}\end{bmatrix}\)
\(P^2 = \begin{bmatrix} 1 & 0 \\0 & \frac {1}{4}\end{bmatrix}\)
\(P^3 = P^2 * P\)
$P^3 = \[\begin{bmatrix} 1 & 0 \\0 & \frac {1}{4}\end{bmatrix}\]\(P^3 = \begin{bmatrix} 1 & 0 \\0 & \frac {1}{8}\end{bmatrix}\)
so \(P^n\) is calculated by: \(P^n = \begin{bmatrix} 1 & 0 \\0 & \frac {1}{2^n}\end{bmatrix}\)
as n tend to infinity \(\frac {1}{2^n}\) tends to end to zero Therefore, the elements of \(P^n\) tend to: \(P^n = \begin{bmatrix} 1 & 0 \\0 & 0\end{bmatrix}\)