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Chapter 4: Problem 66

For a branching process, calculate \(\pi_{0}\) when (a) \(P_{0}=\frac{1}{4}, P_{2}=\frac{3}{4}\). (b) \(P_{0}=\frac{1}{4}, P_{1}=\frac{1}{2}, P_{2}=\frac{1}{4}\). (c) \(P_{0}=\frac{1}{6}, P_{1}=\frac{1}{2}, P_{3}=\frac{1}{3}\).

Short Answer

Expert verified
In the given branching processes, the probabilities of extinction \(\pi_0\) are: (a) \(\pi_0 = \frac{3}{4}\) (b) \(\pi_0 = \frac{3}{4}\) (c) \(\pi_0 = \frac{1}{2}\)

Step by step solution

01

a) Calculate \(\pi_0\) for \(P_0=\frac{1}{4}\) and \(P_2=\frac{3}{4}\)

Given that \(P_0=\frac{1}{4}\) and \(P_2=\frac{3}{4}\), we can write the extinction probability as: \[\pi_0 = P_0 + P_2 \cdot \pi_0^2\] Plugging in the given probabilities: \[\pi_0 = \frac{1}{4} + \frac{3}{4} \cdot \pi_0^2\] Now, we need to solve this quadratic equation to find the smallest non-negative solution for \(\pi_0\). Rewriting the equation in the standard quadratic form: \[\pi_0^2 - (4\pi_0 - 3) = 0\] Solving for \(\pi_0\), we get two possible solutions: \[\pi_0 = 1\] \[\pi_0 = \frac{3}{4}\] Since \(\pi_0\) represents a probability, we choose the smallest non-negative solution, thus: \[\pi_0 = \frac{3}{4}\]
02

b) Calculate \(\pi_0\) for \(P_0=\frac{1}{4}\), \(P_1=\frac{1}{2}\), and \(P_2=\frac{1}{4}\)

Given that \(P_0=\frac{1}{4}\), \(P_1=\frac{1}{2}\), and \(P_2=\frac{1}{4}\), we can write the extinction probability as: \[\pi_0 = P_0 + P_1\pi_0 + P_2 \cdot (\pi_0)^2\] Plugging in the given probabilities: \[\pi_0 = \frac{1}{4} + \frac{1}{2}\pi_0 + \frac{1}{4}(\pi_0)^2\] Now, we need to solve this equation to find the smallest non-negative solution for \(\pi_0\). Rearranging the equation: \[(\pi_0)^2 - (4\pi_0 - 3) = 0\] Solving for \(\pi_0\), we get two possible solutions: \[\pi_0 = 1\] \[\pi_0 = \frac{3}{4}\] Since \(\pi_0\) represents a probability, we choose the smallest non-negative solution, thus: \[\pi_0 = \frac{3}{4}\]
03

c) Calculate \(\pi_0\) for \(P_0=\frac{1}{6}\), \(P_1=\frac{1}{2}\), and \(P_3=\frac{1}{3}\)

Given that \(P_0=\frac{1}{6}\), \(P_1=\frac{1}{2}\), and \(P_3=\frac{1}{3}\), we can write the extinction probability as: \[\pi_0 = P_0 + P_1\pi_0 + P_3 \cdot (\pi_0)^3\] Plugging in the given probabilities: \[\pi_0 = \frac{1}{6} + \frac{1}{2}\pi_0 + \frac{1}{3}(\pi_0)^3\] Now, we need to solve this equation to find the smallest non-negative solution for \(\pi_0\). Rearranging the equation: \[(\pi_0)^3 - (6\pi_0 - 4)(\pi_0-1) = 0\] Solving for \(\pi_0\), we get one possible solution: \[\pi_0 = \frac{1}{2}\] Thus, in this case: \[\pi_0 = \frac{1}{2}\]

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Most popular questions from this chapter

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 \(T\).

Trials are performed in sequence. If the last two trials were successes, then the next trial is a success with probability \(0.8\); otherwise the next trial is a success with probability \(0.5\). In the long run, what proportion of trials are successes?

Suppose in the gambler's ruin problem that the probability of winning a bet depends on the gambler's present fortune. Specifically, suppose that \(\alpha_{i}\) is the probability that the gambler wins a bet when his or her fortune is \(i .\) Given that the gambler's initial fortune is \(i\), let \(P(i)\) denote the probability that the gambler's fortune reaches \(N\) before \(0 .\) (a) Derive a formula that relates \(P(i)\) to \(P(i-1)\) and \(P(i+1)\). (b) Using the same approach as in the gambler's ruin problem, solve the equation of part (a) for \(P(i)\). (c) Suppose that \(i\) balls are initially in urn 1 and \(N-i\) are in urn 2, and suppose that at each stage one of the \(N\) balls is randomly chosen, taken from whichever urn it is in, and placed in the other urn. Find the probability that the first urn becomes empty before the second.

A group of \(n\) processors is arranged in an ordered list. When a job arrives, the first processor in line attempts it; if it is unsuccessful, then the next in line tries it; if it too is unsuccessful, then the next in line tries it, and so on. When the job is successfully processed or after all processors have been unsuccessful, the job leaves the system. At this point we are allowed to reorder the processors, and a new job appears. Suppose that we use the one- closer reordering rule, which moves the processor that was successful one closer to the front of the line by interchanging its position with the one in front of it. If all processors were unsuccessful (or if the processor in the first position was successful), then the ordering remains the same. Suppose that each time processor \(i\) attempts a job then, independently of anything else, it is successful with probability \(p_{i}\) (a) Define an appropriate Markov chain to analyze this model. (b) Show that this Markov chain is time reversible. (c) Find the long-run probabilities.

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.
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