Consider a general chain with state space $S = \left\{1, 2\right\}$ and write the transition matrix as $$ P = \begin{pmatrix} 1-a & a \\ b & 1-b \end{pmatrix} $$ Use the Markov property to show that $$ \operatorname{Pr}\left\{X_{n+1} = 1\right\} − \frac{b}{a+b} = (1 − a − b)\left[\operatorname{Pr}\left\{X_n = 1\right\} − \frac{b}{a+b}\right] $$ I'm using the fact that $p(1,1)=1-a$ and trying to equate this to $\operatorname{Pr}\left\{X_{n+1}=1\mid X_n=1\right\}$ to try and get $\operatorname{Pr}\left\{X_{n+1}=1\right\}$ and $\operatorname{Pr}\left\{X_n=1\right\}$ individually but I cant work it out.
2026-02-23 13:25:08.1771853108
Can someone help on this Markov Chain question?
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I guess you would want to use the fact that $$ \operatorname{Pr}\left\{X_{n+1}=1\right\} = \operatorname{Pr}\left\{X_{n+1}=1 \mid X_n=1\right\}\operatorname{Pr}\left\{X_n=1\right\}+\operatorname{Pr}\left\{X_{n+1} = 1 \mid X_n = 2\right\}(1-\operatorname{Pr}\left\{X_n=1)\right\} $$ and fill in the conditional probabilities from the transition matrix.