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Chapter 3: Problem 30

Let \(X(n)\) be the number of individuals in the \(n\)th generation of a branching process \((X(0)=1)\) with reproduction mean \(m(=\) \(E X(1)\) ). Set $$ Z_{n}=\frac{X(n)}{m^{n}}, \quad n \geq 1 . $$ Show that \(Z_{1}, Z_{2}, \ldots\) is a martingale.

Short Answer

Expert verified
\(Z_1, Z_2, \ldots\) is a martingale since \(E[Z_{n+1} \mid Z_n, Z_{n-1}, \ldots] = Z_n\). The sequence satisfies the martingale property.

Step by step solution

01

Define a Martingale

A sequence \(Z_1, Z_2, \ldots\) is a martingale if for every \(n \geq 1\), \(E[|Z_n|] < \infty\) and \(E[Z_{n+1} \mid Z_1, Z_2, \ldots, Z_n] = Z_n\). This is the condition we need to verify for \(Z_n\).
02

Calculate Expected Value of \(X(n+1)\)

Given that the process starts with one individual, the expected number of individuals in the generation \(n+1\) is given by the reproduction mean, hence, \(E[X(n+1) \mid X(n)] = X(n) m\).
03

Express \(Z_{n+1}\) Using \(X(n+1)\)

By definition, \(Z_{n+1} = \frac{X(n+1)}{m^{n+1}}\). Using the expectation from the previous step, \(E\left[Z_{n+1} \mid X(n)\right] = \frac{E[X(n+1) \mid X(n)]}{m^{n+1}} = \frac{X(n) m}{m^{n+1}} = \frac{X(n)}{m^n}\).
04

Establish Martingale Property from Expectation

From step 3, we have \(E\left[Z_{n+1} \mid Z_n, Z_{n-1}, \ldots, Z_1\right] = E\left[Z_{n+1} \mid X(n)\right] = \frac{X(n)}{m^n} = Z_n\). This satisfies the martingale condition since the expected value of \(Z_{n+1}\) given the past is equal to \(Z_n\).
05

Conclude with the Martingale Verification

Since all the conditions of a martingale are satisfied, i.e., the expected value condition \(E[Z_{n+1} \mid Z_1, Z_2, \ldots, Z_n] = Z_n\) holds, and we can assume \(E[|Z_n|] < \infty\) due to the assumption of a finite mean, \(Z_1, Z_2, \ldots\) forms a martingale.

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Key Concepts

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

Branching Process
A branching process is a mathematical model that describes how populations evolve over generations. It often begins with a single individual, as seen in our exercise, with the future generations growing based on certain probabilistic rules.
  • The branching process assumes that each individual in one generation produces offspring according to a fixed probability distribution.
  • This distribution does not change over time, meaning that every generation has the same chance of producing a given number of offspring.
The branching process is notably used to study population dynamics, spread of diseases, and even certain financial models. By considering the number of individuals at each generation, we can describe how the population changes over time.
Expected Value
Expected value is a fundamental concept in probability and statistics that represents the average outcome of a random process over the long run. In the context of the branching process, it helps determine the expected number of individuals in each generation.
  • The expected value is calculated by taking the sum of all possible outcomes, each weighted by their probability of occurring.
  • For the reproduction mean, which is the expected value of the offspring number of one individual, it sets the stage for predicting the number of individuals in future generations.
Hence, the expected value in our exercise deals with finding how the population size is likely to develop, using the reproduction mean as the cornerstone of prediction.
Reproduction Mean
The reproduction mean is a key parameter in a branching process. It is represented as \(m = E[X(1)]\) in our problem. This value speaks volumes about the expected output of offspring per individual in a generation.
  • If the reproduction mean \(m > 1\), the population tends to grow exponentially because each individual, on average, produces more than one offspring.
  • Conversely, if \(m < 1\), the population is likely to shrink as each individual produces less than one offspring on average.
  • When \(m = 1\), the process is said to be critical, meaning that the population size remains roughly constant over generations.
Understanding the reproduction mean is essential for predicting the long-term behavior of the population in a branching process.
Sequence
A sequence is simply an ordered list of numbers or terms. In the exercise at hand, we deal with a sequence \(Z_1, Z_2, \ldots\) characterized by our previous discussions.
  • This sequence is mapped out over different generations, measuring the population size normalized by powers of \(m\), the reproduction mean.
  • By defining \(Z_n = \frac{X(n)}{m^n}\), we standardize the successive generations' sizes in relation to their expected growth under the reproduction mean assumption.
  • In proving that \(Z_n\) is a martingale, we verify that changes in this sequence occur in a particular, predictable manner.
In essence, the sequence plays a pivotal role in determining how the branching process translates into a stable martingale form, reflecting a balance of expectation over time.

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Most popular questions from this chapter

Old MacDonald had a farm with (among other animals) 300 hens. The number of eggs laid by a hen during one day can be considered to be a Po(3)-distributed random variable. Every egg gives (independently of the others) rise to 0,1 , or 2 chickens with probabilities \(0.3,0.6\), and \(0.1\), respectively. (a) Find the expected value and variance of the number of chickens stemming from one mother during one day. (b) What is the probability that at least 700 chickens are produced from the total egg production during one day?

The probability of success in a certain game is \(p\). The results in successive trials are independent. Let \(Z\) be the number of turns required in order to obtain \(r\) successes. Determine the generating function of \(Z\), and use the result to compute \(E Z\) and \(\operatorname{Var} Z\).

Let \(X_{1}, X_{2} \ldots\) be independent \(C(0,1)\)-distributed random variables, and set \(S_{n}=\sum_{k=1}^{n} X_{k}, n \geq 1\). Show that (a) \(\frac{S_{\mathrm{n}}}{n} \in C(0,1)\), and (b) \(\frac{1}{n} \sum_{k=1}^{n} \frac{S_{k}}{k} \in C(0,1)\). Remark. If \(\left\\{\frac{S_{k}}{k}, k \geq 1\right\\}\) were independent, then (b) would follow immediately from (a).

Philip throws a fair die until he obtains a four. Diane then tosses a coin as many times as Philip threw his die. Determine the expected value and variance of the number of (a) heads, (b) tails, and (c) heads and tails obtained by Diane.

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