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Chapter 4: Problem 5

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 \(E\left[X_{3}\right]\).

Short Answer

Expert verified
Based on the provided steps, we can calculate the expected value of the Markov chain at step 3 by following these steps: 1. Write down the initial state probabilities as \(V_0=[\frac{1}{4}, \frac{1}{4}, \frac{1}{2}]\). 2. Set up the transition matrix \(P = \left[\begin{array}{ccc} \frac{1}{2} & \frac{1}{3} & \frac{1}{6} \\\ 0 & \frac{1}{3} & \frac{2}{3} \\\ \frac{1}{2} & 0 & \frac{1}{2} \end{array}\right] \). 3. Compute the distribution after 3 steps by calculating \(V_{3} = V_{0}P^{3}\). 4. Calculate the expected value using the formula \(E[X_{3}] = \sum xP[X_{3} = x]\), where \(P[X_{3} = x]\) is the probability that variable \(X_{3}\) takes on the value x.

Step by step solution

01

Write Down the Initial State Probabilities

The initial state probabilities are given as: \(P\left\\{X_{0}=0\right\\}=\frac{1}{4}\) and \(P\left\\{X_{0}=1\right\\} = \frac{1}{4}\). Since the total probability must sum up to 1, then the probability that \(X_{0} = 2\) must be \(1 - P\left\\{X_{0}=0\right\\} - P\left\\{X_{0}=1\right\\} = \frac{1}{2}\). So we can write the initial state probabilities as a vector \(V_0=[\frac{1}{4}, \frac{1}{4}, \frac{1}{2}]\).
02

Set Up the Transition Matrix

The Transition matrix given in the problem is denoted as \(P\). Let's call the matrix \(P\) and represent it as: \[ P = \left[\begin{array}{ccc} \frac{1}{2} & \frac{1}{3} & \frac{1}{6} \\\ 0 & \frac{1}{3} & \frac{2}{3} \\\ \frac{1}{2} & 0 & \frac{1}{2} \end{array}\right] \]
03

Compute the Distribution after 3 Steps

To compute the probability distribution of the states after 3 steps, we use the formula \(V_{n}=V_{0}P^{n}\). Here, \(n = 3\), so we calculate \(V_{3} = V_{0}P^{3}\). To get \(P^{3}\), we multiply the matrix \(P\) by itself twice (i.e. squaring the matrix), then multiply this by the initial state probability vector \(V_{0}\).
04

Calculate the Expected Value

The expected value of a random variable is calculated as \(E[X] = \sum xP[X = x]\) where \(P[X = x]\) is the probability that variable X takes on the value x. Here, we want to find \(E[X_{3}]\), so we simply multiply each state by their respective probabilities after 3 steps (i.e. elements of the vector \(V_{3}\)), and then sum up these values to get the expected value \(E[X_{3}]\).

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Most popular questions from this chapter

For a Markov chain \(\left\\{X_{n}, n \geqslant 0\right\\}\) with transition probabilities \(P_{i, j}\), consider the conditional probability that \(X_{n}=m\) given that the chain started at time 0 in state \(i\) and has not yet entered state \(r\) by time \(n\), where \(r\) is a specified state not equal to either \(i\) or \(m\). We are interested in whether this conditional probability is equal to the \(n\) stage transition probability of a Markov chain whose state space does not include state \(r\) and whose transition probabilities are $$ Q_{i, j}=\frac{P_{i, j}}{1-P_{i, r}}, \quad i, j \neq r $$ Either prove the equality $$ P\left\\{X_{n}=m \mid X_{0}=i, X_{k} \neq r, k=1, \ldots, n\right\\}=Q_{i, m}^{n} $$ or construct a counterexample.

It follows from Theorem \(4.2\) that for a time reversible Markov chain $$ P_{i j} P_{j k} P_{k i}=P_{i k} P_{k j} P_{j i}, \quad \text { for all } i, j, k $$ It turns out that if the state space is finite and \(P_{i j}>0\) for all \(i, j\), then the preceding is also a sufficient condition for time reversibility. [That is, in this case, we need only check Equation (4.26) for paths from \(i\) to \(i\) that have only two intermediate states.] Prove this.

Consider three urns, one colored red, one white, and one blue. The red urn contains 1 red and 4 blue balls; the white urn contains 3 white balls, 2 red balls, and 2 blue balls; the blue urn contains 4 white balls, 3 red balls, and 2 blue balls. At the initial stage, a ball is randomly selected from the red urn and then returned to that urn. At every subsequent stage, a ball is randomly selected from the urn whose color is the same as that of the ball previously selected and is then returned to that urn. In the long run, what proportion of the selected balls are red? What proportion are white? What proportion are blue?

A DNA nucleotide has any of 4 values. A standard model for a mutational change of the nucleotide at a specific location is a Markov chain model that supposes that in going from period to period the nucleotide does not change with probability \(1-3 \alpha\), and if it does change then it is equally likely to change to any of the other 3 values, for some \(0<\alpha<\frac{1}{3}\). (a) Show that \(P_{1,1}^{n}=\frac{1}{4}+\frac{3}{4}(1-4 \alpha)^{n}\). (b) What is the long run proportion of time the chain is in each state?

Suppose that coin 1 has probability \(0.7\) of coming up heads, and \(\operatorname{coin} 2\) has probability \(0.6\) of coming up heads. If the coin flipped today comes up heads, then we select coin 1 to flip tomorrow, and if it comes up tails, then we select coin 2 to flip tomorrow. If the coin initially flipped is equally likely to be coin 1 or coin 2 , then what is the probability that the coin flipped on the third day after the initial flip is coin \(1 ?\)
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