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For a branching process, calculate \(\pi_{0}\) when (a) \(P_{0}=\frac{1}{47} \cdot P_{2}=\frac{3}{4}\) (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
Answer: (a) \(\pi_0 = 0.2679\) (b) \(\pi_0 = 0.5\) (c) \(\pi_0 = 0.33333\)

Step by step solution

01

Construct the generating function

Using the given offspring probabilities, we can construct the generating function \(G(z)\) as follows: \[G(z) = P_{0} + P_{1}z + P_{2}z^{2} = \frac{1}{47}P_{2} + 0z + P_{2}z^{2}\] Since \(P_{2} = \frac{3}{4}\), we can now write the generating function as: \[G(z) = \frac{1}{47}\cdot \frac{3}{4} + 0z + \frac{3}{4}z^{2}\]
02

Solve the equation \(\pi_0 = G(\pi_0)\)

We now want to find the value of \(\pi_0\) such that \(\pi_0 = G(\pi_0)\). Substitute \(\pi_0\) into the generating function: \[\pi_0 = \frac{1}{47}\cdot \frac{3}{4} + 0(\pi_0) + \frac{3}{4}(\pi_0)^{2}\] Now, we can solve this quadratic equation for \(\pi_0\). Rearrange the equation as follows: \[\frac{3}{4}\pi_0^2 - \pi_0 + \frac{1}{47}\cdot\frac{3}{4} = 0\] Let \(\pi_0 = x: \[\frac{3}{4}x^2 - x + \frac{1}{47}\cdot\frac{3}{4} = 0\] Lets multiply by 4 to remove the fraction: \[3x^2 - 4x + 1 = 0\] This is a quadratic equation. You can either factorize to find the solutions, or use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where a=3, b=-4, and c=1. Upon solving the quadratic equation, we get two solutions: \(x = 0.2679\) and \(x = 1\)
03

Choose the correct solution

Since \(\pi_0\) is a probability, it must be a value between 0 and 1. The process will definitely go extinct if no offspring are produced, which corresponds to the fixed point \(x = 1\). Thus, the probability of extinction, \(\pi_0\), is: \[\pi_0 = 0.2679\] #b) Offspring probabilities: \(P_{0}=\frac{1}{4}, P_{1}=\frac{1}{2}, P_{2}=\frac{1}{4}\)# Repeat the procedure in the part (a) with these new offspring probabilities.
04

Construct the generating function

Using the given offspring probabilities, we can construct the generating function \(G(z)\) as follows: \[G(z) = P_{0} + P_{1}z + P_{2}z^{2} = \frac{1}{4} + \frac{1}{2}z + \frac{1}{4}z^{2}\]
05

Solve the equation \(\pi_0 = G(\pi_0)\)

We now want to find the value of \(\pi_0\) such that \(\pi_0 = G(\pi_0)\). Substitute \(\pi_0\) into the generating function: \[\pi_0 = \frac{1}{4} + \frac{1}{2}(\pi_0) + \frac{1}{4}(\pi_0)^{2}\] Rearrange the equation as follows: \[\frac{1}{4}\pi_0^2 -\frac{1}{2}\pi_0 + \frac{1}{4} = 0\] Solve the quadratic equation we get two solutions: \(x = 0.5\) and \(x = 1\)
06

Choose the correct solution

Since \(\pi_0\) is a probability, it must be a value between 0 and 1. The process will definitely go extinct if no offspring are produced, which corresponds to the fixed point \(x = 1\). Thus, the probability of extinction, \(\pi_0\), is: \[\pi_0 = 0.5\] #c) Offspring probabilities: \(P_{0}=\frac{1}{6}, P_{1}=\frac{1}{2}, P_{3}=\frac{1}{3}\)# Repeat the procedure in the part (a) with these new offspring probabilities.
07

Construct the generating function

Using the given offspring probabilities, we can construct the generating function \(G(z)\) as follows: \[G(z) = P_{0} + P_{1}z + P_{2}z^{2} + P_{3}z^{3} = \frac{1}{6} + \frac{1}{2}z + 0z^{2} + \frac{1}{3}z^{3}\]
08

Solve the equation \(\pi_0 = G(\pi_0)\)

We now want to find the value of \(\pi_0\) such that \(\pi_0 = G(\pi_0)\). Substitute \(\pi_0\) into the generating function: \[\pi_0 = \frac{1}{6} + \frac{1}{2}(\pi_0) + 0(\pi_0)^{2} + \frac{1}{3}(\pi_0)^{3}\] Rearrange the equation as follows: \[\frac{1}{3}\pi_0^3 -\frac{1}{2}\pi_0 + \frac{1}{6} = 0\] We can apply the Newton-Raphson method to find the solution for this equation. The method starts by finding an initial guess \(x_0\), and iterates by using the following formula: \(x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\). Let \(f(\pi_0) = \frac{1}{3}\pi_0^3 -\frac{1}{2}\pi_0 + \frac{1}{6}\), so \(f'(\pi_0) = \pi_0^2 - \frac{1}{2}\). Let the initial guess be \(x_0 = 0\). Apply the Newton-Raphson method using a calculator or a computer program to find the solution, which is \(x = 0.33333\).
09

Choose the correct solution

Since \(\pi_0\) is a probability, it must be a value between 0 and 1. The process will definitely go extinct if no offspring are produced, which corresponds to the fixed point \(x = 1\). Thus, the probability of extinction, \(\pi_0\), is: \[\pi_0 = 0.33333\]

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Key Concepts

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

Generating Function
The generating function is a powerful mathematical tool used in probability theory to encapsulate the entire distribution of a discrete random variable. It transforms the probability distribution of the random variable into a neat algebraic form. For branching processes, the generating function, typically denoted as G(z), is particularly useful because it provides a way to analyze the distribution of the number of offspring.

It is constructed by the sum of the probabilities of having each number of offspring weighted by the power of a variable z. For example, the generating function G(z) for a set of offspring probabilities P_0, P_1, P_2, ..., is given as G(z) = P_0 + P_1z + P_2z^2 + .... Once constructed, the generating function can be used to find the probability of extinction, the expected number of offspring, and other important characteristics of the branching process.
Probability of Extinction
In a branching process, the probability of extinction, denoted as \(\pi_0\), is the probability that a process eventually dies out. This is an important measure, especially in population dynamics, genetics, and epidemiology. To find \(\pi_0\), one sets up the equation \(\pi_0 = G(\pi_0)\) where G(z) is the generating function for the process. Solving this equation provides the probability that the process will have no offspring in the long run, leading to its eventual end. In the provided examples, the values of \(\pi_0\) were computed for different offspring distribution, representing the extinction probabilities for each variant of the process.
Offspring Probabilities
The offspring probabilities in a branching process are the keys to understanding the entire process. They represent the probabilities of an individual having a certain number of offspring. Notated as P_0, P_1, P_2, ..., they are essentially the building blocks of the generating function. For instance, P_0 is the probability of having no offspring, P_1 is the probability of one offspring, and so on. The offspring probabilities given in the exercise examples define the possible outcomes for each generation within the process and lead to the construction of different generating functions.
Quadratic Equation
A quadratic equation is a second-degree polynomial equation of the form ax^2 + bx + c = 0, where a, b, and c are constants. Most students encounter quadratic equations in mathematics and must often find their roots - the values of x which satisfy the equation. There are various methods to solve a quadratic equation, such as factoring, completing the square, or using the quadratic formula. In the context of a branching process, the quadratic equation arises when setting the generating function equal to \(\pi_0\) and solving for the probability of extinction.
Newton-Raphson Method
The Newton-Raphson method is an iterative technique used for finding successively better approximations to the roots (or zeroes) of a real-valued function. The formula x_{n+1} = x_n - f(x_n)/f'(x_n) is applied repeatedly starting from an initial guess until a sufficiently accurate value is reached. This method is particularly useful when the equation is complex or does not lend itself to simple algebraic solutions, as was the case with the cubic equation in part (c) of the exercise. The generated solutions are then checked to identify the one that represents the probability of extinction, which must lie between 0 and 1 inclusive.

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Most popular questions from this chapter

Suppose that a population consists of a fixed number, say, \(m\), of genes in any generation. Each gene is one of two possible genetic types. If any generation has exactly \(i\) (of its \(m\) ) genes being type 1 , then the next generation will have \(j\) type 1 (and \(m-j\) type 2 ) genes with probability $$ \left(\begin{array}{c} m \\ j \end{array}\right)\left(\frac{i}{m}\right)^{j}\left(\frac{m-i}{m}\right)^{m-j}, \quad j=0,1, \ldots, m $$ Let \(X_{n}\) denote the number of type 1 genes in the \(n\) th generation, and assume that \(X_{0}=i\). (a) Find \(E\left[X_{n}\right]\). (b) What is the probability that eventually all the genes will be type \(1 ?\)

Each morning an individual leaves his house and goes for a run. He is equally likely to leave either from his front or back door. Upon leaving the house, he chooses a pair of running shoes (or goes running barefoot if there are no shoes at the door from which he departed). On his return he is equally likely to enter, and leave his running shoes, either by the front or back door. If he owns a total of \(k\) pairs of running shoes, what proportion of the time does he run barefooted?

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.

(a) Show that the limiting probabilities of the reversed Markov chain are the same as for the forward chain by showing that they satisfy the equations $$ \pi_{j}=\sum_{i} \pi_{i} Q_{i j} $$ (b) Give an intuitive explanation for the result of part (a).

Suppose that coin 1 has probability \(0.7\) of coming up heads, and 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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