EXM.1.AHL.TZ0.18

pestleMathematicsAIHLPaper 1EXM· ahl-4-19-transition-matrices-markov-chainssource ↗

A $2 \times 2$ transition matrix for a Markov chain will have the form ${\mathbf{M}} = \left( {\begin{array}{*{20}{c}} a&{1 - b} \\ {1 - a}&b \end{array}} \right){\text{,}}\,\,0 < a < 1{\text{,}}\,\,0 < b < 1$ .

Show that $\lambda = 1$ is always an eigenvalue for M and find the other eigenvalue in terms of $a$ and $b$ .

[4]

Find the steady state probability vector for M in terms of $a$ and $b$ .

[5]

Markscheme / solution

$\left| {\begin{array}{*{20}{c}} {a - \lambda }&{1 - b} \\ {1 - a}&{b - \lambda } \end{array}} \right| = 0 \Rightarrow \left( {a - \lambda } \right)\left( {b - \lambda } \right) - \left( {1 - b} \right)\left( {1 - a} \right) = 0$ M1A1

$\Rightarrow {\lambda ^2} - \left( {a + b} \right)\lambda + a + b - 1 = 0 \Rightarrow \left( {\lambda - 1} \right)\left( {\lambda + \left( {1 - a - b} \right)} \right) = 0$ A1

$\Rightarrow \lambda = 1\,\,{\text{or}}\,\,\lambda = a + b - 1$ AGA1

[4 marks]

$\left( {\begin{array}{*{20}{c}} a&{1 - b} \\ {1 - a}&b \end{array}} \right)\,\left( {\begin{array}{*{20}{c}} p \\ {1 - p} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} p \\ {1 - p} \end{array}} \right) \Rightarrow ap + 1 - b - p + bp = p$ M1A1

$\Rightarrow 1 - b = \left( {2 - a - b} \right)p \Rightarrow p = \frac{{1 - b}}{{2 - a - b}}$ M1

So vector is $\left( {\begin{array}{*{20}{c}} {\frac{{1 - b}}{{2 - a - b}}} \\ {\frac{{1 - a}}{{2 - a - b}}} \end{array}} \right)$ A1A1

[5 marks]

Examiners’ report

[N/A]