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Chapter 4: Problem 64

Consider a branching process having \(\mu<1 .\) Show that if \(X_{0}=1\), then the expected number of individuals that ever exist in this population is given by \(1 /(1-\mu) .\) What if \(X_{0}=n ?\)

Short Answer

Expert verified
When \(X_0 = 1\), the expected number of individuals that ever exist in this population is \(\frac{1}{1-\mu}\). For the generalized case, when the initial population is \(X_0 = n\), the expected number of individuals is given by \(\frac{n}{1-\mu}\).

Step by step solution

01

Computing the Expected Population at Each Generation

Using the given expectation of \(\mu\), we can compute the expected population at each generation. The expected number of individuals in generation \(n\) can be expressed as: \(E[X_n] = \mu^n\) The factor of \(\mu^n\) comes because each individual produces \(\mu\) offspring on average in each generation.
02

Computing the Expected Total Population

Now let's find the expected total number of individuals that ever exist in the population. We will do this by summing the expected population at each generation from \(0\) to infinity: \(E[\text{Total Population}] = E[X_0] + E[X_1] + E[X_2] + \cdots\) Substitute the values from Step 1: \(E[\text{Total Population}] = 1 + \mu + \mu^2 + \cdots\)
03

Evaluating the Infinite Geometric Series

The summation we have here is an infinite geometric series with a first term equal to 1 and a common ratio of \(\mu\). Since \(\mu < 1\), this geometric series converges. The sum of an infinite geometric series can be computed using the formula: \(\text{Sum} = \frac{a}{1 - r}\), where \(a\) is the first term and \(r\) is the common ratio. Using this formula with \(a = 1\) and \(r = \mu\), we get: \(E[\text{Total Population}] = \frac{1}{1 - \mu}\) Hence, when \(X_0 = 1\), the expected number of individuals that ever exist in this population is \(1/(1-\mu)\).
04

Generalization of the Result for \(X_0 = n\)

Now we generalize the result for the case when the initial population is \(X_0 = n\). The only difference in this case is that there are \(n\) times more individuals in each generation. Thus, the expected number of individuals in the total population will be \(n\) times larger: \(E[\text{Total Population}] = n \left(1 + \mu + \mu^2 + \cdots \right)\) Now we use the result we have just found for the case \(X_0 = 1\): \(E[\text{Total Population}] = n \times \frac{1}{1 - \mu} = \frac{n}{1 - \mu}\) So when the initial population is \(X_0 = n\), the expected number of individuals that ever exist in the population is \(\frac{n}{1-\mu}\).

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Most popular questions from this chapter

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.

Let \(\pi_{i}\) denote the long-run proportion of time a given irreducible Markov chain is in state \(i\). (a) Explain why \(\pi_{i}\) is also the proportion of transitions that are into state \(i\) as well as being the proportion of transitions that are from state \(i\). (b) \(\pi_{i} P_{i j}\) represents the proportion of transitions that satisfy what property? (c) \(\sum_{i} \pi_{i} P_{i j}\) represent the proportion of transitions that satisfy what property? (d) Using the preceding explain why $$ \pi_{j}=\sum_{i} \pi_{i} P_{i j} $$

For the Markov chain with states \(1,2,3,4\) whose transition probability matrix \(\mathbf{P}\) is as specified below find \(f_{i 3}\) and \(s_{i 3}\) for \(i=1,2,3\). $$ \mathbf{P}=\left[\begin{array}{llll} 0.4 & 0.2 & 0.1 & 0.3 \\ 0.1 & 0.5 & 0.2 & 0.2 \\ 0.3 & 0.4 & 0.2 & 0.1 \\ 0 & 0 & 0 & 1 \end{array}\right] $$

A taxi driver provides service in two zones of a city. Fares picked up in zone \(A\) will have destinations in zone \(A\) with probability \(0.6\) or in zone \(B\) with probability \(0.4 .\) Fares picked up in zone \(B\) will have destinations in zone \(A\) with probability \(0.3\) or in zone \(B\) with probability \(0.7 .\) The driver's expected profit for a trip entirely in zone \(A\) is 6 ; for a trip entirely in zone \(B\) is \(8 ;\) and for a trip that involves both zones is \(12 .\) Find the taxi driver's average profit per trip.

Let \(\left\\{X_{n}, n \geqslant 0\right\\}\) denote an ergodic Markov chain with limiting probabilities \(\pi_{i} .\) Define the process \(\left\\{Y_{n}, n \geqslant 1\right\\}\) by \(Y_{n}=\left(X_{n-1}, X_{n}\right) .\) That is, \(Y_{n}\) keeps track of the last two states of the original chain. Is \(\left\\{Y_{n}, n \geqslant 1\right\\}\) a Markov chain? If so, determine its transition probabilities and find $$ \lim _{n \rightarrow \infty} P\left\\{Y_{n}=(i, j)\right\\} $$
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