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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Probability Generating Function
In mathematical analysis, especially within probability theory, the probability generating function (PGF) is a tool that encapsulates the probabilities of a random variable into a single function. To put it simply, the PGF translates a distribution of probabilities into an equation, which we can manipulate to extract useful information about the probability distribution.

The PGF for a discrete random variable X, which represents the number of offspring in a branching process, is given by: \[G(s) = \text{E}[s^X] = \text{P}_0 + \text{P}_1s + \text{P}_2s^2 + \text{P}_3s^3 + \text{...},\] where \text{P}_i represents the probability of exactly i offspring and \(s\) is an indeterminate. The coefficients (\(\text{P}_i\) values) in the function are the probabilities of having a certain number of offspring, and the power of \(s\) indicates the number of offspring in each term.

For students, mastering the PGF is crucial as it serves as the basis for more complex calculations, including the extinction probability of the branching process. By analyzing the PGF, one can derive equations relevant for finding \text{probabilities of interest}, often involving taking derivatives or solving equations involving the PGF, as shown in the steps of our example exercise.
Calculation of Extinction Probability
The extinction probability calculation refers to finding the probability that a branching process eventually dies out, meaning that there are no individuals left in some generation. The extinction probability is denoted by \(\pi_0\) and can be found by setting the PGF equal to \(s\), as this represents the scenario where the population does not grow.

To find the extinction probability in a branching process, we use the PGF and look for the smallest non-negative root of the equation \(G(s) = s\). This is because we are interested in the likelihood that the process ceases to exist over time. For cases where there is a finite number of offspring possibilities, we may arrive at a polynomial equation in \(s\), which we can solve, as shown in the exercise. The tricky part sometimes can be discerning which root of the polynomial is the actual extinction probability, as we always select the smallest non-negative solution since probabilities cannot exceed 1.

Improvement from the exercise:

  • To refine understanding, practice setting up and solving different PGF equations for various distributions of offspring.
  • Engage in comparative study by analyzing the extinction probabilities under various scenarios and observing the impact of different distribution parameters on the extinction probability.
Analyzing Branching Processes
A branching process analysis involves evaluating the behavior of a population where each individual in one generation produces a certain number of offspring independently to form the next generation. These models are particularly useful in fields like biology, epidemiology, and family lineage studies.

To analyze a branching process, we typically calculate the expected number of offspring (also referred to as the mean or average reproduction rate) and the extinction probability. These metrics tell us whether the population is expected to grow, remain stable, or decline over time. This is crucial for making projections about future populations or understanding the spread of characteristics within a population.

Exercise Improvement Advice:

  • Illustrate the branching process with diagrams, showing the possible outcomes at each generation. This visual aid helps in internalizing the concept.
  • Develop a deeper understanding of PGF's role in branching processes by exploring scenarios where reproduction rates vary and analyzing how this affects the extinction probability and overall population trends.
  • Encourage the use of computational tools for solving complex PGFs or for simulating branching processes to visualize long-term trends and extinction scenarios.
  • Compare the analytical solutions with simulation results to reinforce concepts and validate understanding.

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Most popular questions from this chapter

Three white and three black balls are distributed in two urns in such a way that each contains three balls. We say that the system is in state \(i, i=0,1,2,3\), if the first urn contains \(i\) white balls. At each step, we draw one ball from each urn and place the ball drawn from the first urn into the second, and conversely with the ball from the second urn. Let \(X_{n}\) denote the state of the system after the \(n\) th step. Explain why \(\left\\{X_{n}, n=0,1,2, \ldots\right\\}\) is a Markov chain and calculate its transition probability matrix.

\(M\) balls are initially distributed among \(m\) urns. At each stage one of the balls is selected at random, taken from whichever urn it is in, and then placed, at random, in one of the other \(M-1\) urns. Consider the Markov chain whose state at any time is the vector \(\left(n_{1}, \ldots, n_{m}\right)\) where \(n_{i}\) denotes the number of balls in urn \(i\). Guess at the limiting probabilities for this Markov chain and then verify your guess and show at the same time that the Markov chain is time reversible.

A transition probability matrix \(\mathbf{P}\) is said to be doubly stochastic if the sum over each column equals one; that is, $$ \sum_{i} P_{i j}=1, \quad \text { for all } j $$ If such a chain is irreducible and aperiodic and consists of \(M+1\) states \(0,1, \ldots, M\), show that the limiting probabilities are given by $$ \pi_{j}=\frac{1}{M+1}, \quad j=0,1, \ldots, M $$

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.

Consider a Markov chain with states \(0,1,2,3,4\). Suppose \(P_{0,4}=1\); and suppose that when the chain is in state \(i, i>0\), the next state is equally likely to be any of the states \(0,1, \ldots, i-1\). Find the limiting probabilities of this Markov chain.

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