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Let \(\varphi(s)\) be the generating function of the number of progeny of a single individual in a branching process that starts with one individual at time zero, and let \(\varphi_{n}(s)\) denote its \(n\)th iterate. Suppose in addition to the ordinary branching process there also exists some immigration into the population during a single generation described by the probability generating function \(h(s)\). Consider the branching process with immigration whose transition probability matrix is defined by $$ \sum_{j=0}^{\infty} P_{i j} s^{\prime}=[\varphi(s)]^{i} \cdot h(s) $$

Short Answer

Expert verified
The short answer for this problem is to find the transition probability matrix \(P_{ij}\) by using the given progeny generating function \(\varphi(s)\) and the immigration generating function \(h(s)\) in the formula: \[P_{i j} =\frac{1}{s^{\prime}}\left([\varphi(s)]^{i} \cdot h(s)\right)\] For specific values of i and j, substitute the given functions \(\varphi(s)\) and \(h(s)\), compute the coefficient of \(s^{\prime j}\) in the resulting series, and assemble the transition probability matrix by calculating \(P_{ij}\) for the desired range of i and j values.

Step by step solution

01

Identify the given functions

We are given the progeny generating function \(\varphi(s)\) and the immigration generating function \(h(s)\) for the branching process with immigration. #Step 2: Find the nth iterate of the progeny generating function#
02

Find the nth iterate of the progeny generating function

We have to find \(\varphi_{n}(s)\), which is the nth iterate of the progeny generating function \(\varphi(s)\). To find this, repeat the function n times: \[\varphi_{n}(s) = \varphi(\varphi(\cdots\varphi(s)\cdots))\] where the function is applied n times. #Step 3: Write the transition probability matrix formula#
03

Write the transition probability matrix formula

We have to express the transition probability matrix with the given PGFs. The formula to achieve this is given by the equation: \[\sum_{j=0}^{\infty} P_{i j} s^{\prime}=[\varphi(s)]^{i} \cdot h(s)\] #Step 4: Calculate the transition probability matrix using the given functions#
04

Calculate the transition probability matrix using the given functions

To find the transition probability matrix \(P_{ij}\) using the formula mentioned above, rewrite the equation: \[P_{i j} =\frac{1}{s^{\prime}}\left([\varphi(s)]^{i} \cdot h(s)\right)\] To find \(P_{ij}\) for specific values of i and j, substitute the given functions \(\varphi(s)\) and \(h(s)\), and the appropriate value of i, then compute the coefficient of \(s^{\prime j}\) in the resulting series. The transition probability matrix can then be assembled by computing \(P_{ij}\) for any desired range of i and j values.

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Most popular questions from this chapter

Let \(X_{n}, n \geq 0\), describe a branching process with associated probability generating function \(\varphi(s)\) Define \(Y_{n}\) as the total number of individuals in the first \(n\) generations, i.e., $$ Y_{n}=X_{0}+X_{1}+\cdots+X_{n}, \quad n=0,1,2, \ldots, \quad X_{0}=1 $$ Let \(F_{n}(s)\) be the probability generating function of \(Y_{n}\). Establish the functional relation $$ F_{n+1}(s)=5 \varphi\left(F_{n}(s)\right), \quad \text { for } \quad n=0,1,2, \ldots $$

Consider a discrete time branching process \(\left\\{X_{n}\right\\}\) with probability generating function $$ \varphi(s)=\frac{1-(b+c)}{1-c}+\frac{b s}{1-c s}, \quad 01\). Assume \(X_{0}=1\). Determine the conditional limit distribution: $$ \lim \operatorname{Pr}\left\\{X_{n}=k \mid X_{n}>0\right\\} $$

Find the generating function \(\varphi(t ; s)\) of the continuous time branching process with infinitesimal generating function $$ u(s)=s^{k}-s \quad(k \geq 2, \text { integer }) $$

Let \(X_{n}\) be a discrete branching process with associated probability generating function \(\varphi(s)\) and let \(\varphi_{n}(s)=\sum_{k=0}^{\infty} \operatorname{Pr}\left\\{X_{n}=k\right\\} s^{k}\). Assume that \(\varphi^{\prime}(1)>1\) Let \(\tilde{X}_{n}\) denote the number of all the particles in the nth generation which have an infinite line of descent. Show that the probability generating function for \(\tilde{X}_{n}\) is $$ \sum_{k=0}^{\infty} \operatorname{Pr}\left\\{\bar{X}_{n}=k \mid \bar{X}_{0}=X_{0}=1\right\\} s^{k}=\frac{\varphi_{n}(s(1-q)+q)-q}{1-q} $$ where \(q\) is the probability of extinction. Hint: Note that for \(k \geq 1\) $$ \operatorname{Pr}\left\\{\tilde{X}_{n}=k \mid \dot{X}_{0}=1, X_{0}=1\right\\}=\frac{\sum_{i=k}^{\infty} \operatorname{Pr}\left\\{\tilde{X}_{n}=k, X_{n}=l \mid X_{0}=1\right\\}}{\operatorname{Pr}\left\\{\bar{X}_{0}=1 \mid X_{0}=1\right\\}} $$

The following model has been introduced to study a urological process. Suppose bacteria grow according to a Yule process of parameter \(\lambda\) (see Section 1, Chapter 4). At each unit of time each bacterium present is eliminated with probability \(p .\) What is the probability generating function of the number of bacteria existing at time \(n ?\)

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