Chapter 4: Problem 15
Prove that if the number of states in a Markov chain is \(M\), and if state \(j\) can be reached from state \(i\), then it can be reached in \(M\) steps or less.
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Chapter 4: Problem 15
Prove that if the number of states in a Markov chain is \(M\), and if state \(j\) can be reached from state \(i\), then it can be reached in \(M\) steps or less.
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A particle moves among \(n+1\) vertices that are situated on a circle in the following manner. At each step it moves one step either in the clockwise direction with probability \(p\) or the counterclockwise direction with probability \(q=1-p\). Starting at a specified state, call it state 0 , let \(T\) be the time of the first return to state \(0 .\) Find the probability that all states have been visited by time \(T\).
For a series of dependent trials the probability of success on any trial is \((k+1) /(k+2)\) where \(k\) is equal to the number of successes on the previous two trials. Compute \(\lim _{n \rightarrow \infty} P\) \\{success on the \(n\) th trial\\}
Show that if state \(i\) is recurrent and state \(i\) does not communicate with state \(j\), then \(P_{i j}=0 .\) This implies that once a process enters a recurrent class of states it can never leave that class. For this reason, a recurrent class is often referred to as a closed class.
Let \(\pi_{i}\) denote the long-run proportion of time a given irreducible Markov chain is in state \(i\). (a) Explain why \(\pi_{i}\) is also the proportion of transitions that are into state \(i\) as well as being the proportion of transitions that are from state \(i\). (b) \(\pi_{i} P_{i j}\) represents the proportion of transitions that satisfy what property? (c) \(\sum_{i} \pi_{i} P_{i j}\) represent the proportion of transitions that satisfy what property? (d) Using the preceding explain why $$ \pi_{j}=\sum_{i} \pi_{i} P_{i j} $$
A Markov chain \(\left\\{X_{n}, n \geqslant 0\right\\}\) with states \(0,1,2\), has the transition probability matrix $$ \left[\begin{array}{ccc} \frac{1}{2} & \frac{1}{3} & \frac{1}{6} \\ 0 & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{2} & 0 & \frac{1}{2} \end{array}\right] $$ If \(P\left\\{X_{0}=0\right\\}=P\left\\{X_{0}=1\right\\}=\frac{1}{4}\), find \(E\left[X_{3}\right]\).
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