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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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Key Concepts

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

Geometric series
In mathematics, a geometric series is a series with a constant ratio between successive terms. Each term is derived by multiplying the previous term by a fixed, non-zero number called the common ratio. When considering branching processes with offspring distributions, the geometric series arises naturally. Specifically, when calculating the expected population across generations, we encounter an infinite series. The formula to find the sum of an infinite geometric series is:
  • First term \(a\)
  • Common ratio \(r\)
The sum \(S\) is given by:\[S = \frac{a}{1-r}\]This formula is valid only when the common ratio \(r\) satisfies \(-1 < r < 1\). As the formula shows, the series converges to a specific value that reflects the cumulative sum of all the terms. This is critical for understanding the behavior of branching processes in populations that stabilize rather than growing unbounded.
Expected population
When talking about the expected population in a branching process, we're referring to the average number of individuals that can be anticipated, based on a given average offspring production rate per individual. If each individual contributes an average of \(\mu\) offspring, then we can calculate the expected number of individuals in any future generation using this rate.For instance, given an initial population, \(X_0 = 1\), the expected number of individuals in generation \(n\) is expressed as:\[E[X_n] = \mu^n\]This calculation builds on the key concept that future populations depend heavily on this constant offspring rate multiplied across generations. However, it is also concerned with potential convergence to a total expected population when summed over an infinite span. If \(\mu\) is below 1, it indicates a declining or stabilizing population. This figure also adjusts proportionately as the initial population scales, as when \(X_0 = n\).
Offspring distribution
Offspring distribution in branching processes is about understanding the probabilities associated with the number of offspring produced by each individual. It plays an essential part in predicting how a population evolves over time. In our context, the average number of offspring each individual produces is \(\mu\), reinforcing the expected number of individuals in succeeding generations. The branching process model highlights:
  • Predicting whether a population will grow, decline, or stabilize.
  • Calculating expected population dynamics based on offspring averages.
  • Adjusting the calculations when \(X_0\) changes, reflecting on different initial conditions.
With a consistently low \(\mu\), potential for extinction or stabilization is high, implying that understanding offspring distribution is vital to predicting long-term population trends.
Convergence in series
Convergence in series refers to the behavior of a series approaching a finite limit as more terms are considered. For our branching process case, convergence is crucial; it ensures that the expected total population reaches a calculable figure despite potentially infinite generations. A series converges when
  • The sequence of its partial sums approaches a limit.
  • It satisfies the condition that the absolute value of the common ratio \(r\), is less than one: \(|r| < 1\).
Thus, for a convergent geometric series with initial term \(a\) and common ratio \(\mu < 1\), the sum is \(\frac{a}{1-\mu}\). In our population model, this means that despite continuing indefinitely, the expected total number of individuals becomes a finite value, ensuring that calculations for population management can be practical and actionable even in theoretical infinite scenarios.

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Most popular questions from this chapter

For the random walk of Example \(4.13\) use the strong law of large numbers to give another proof that the Markov chain is transient when \(p \neq \frac{1}{2}\). Note that the state at time \(n\) can be written as \(\sum_{i=1}^{n} Y_{i}\) where the \(Y_{i}^{\prime}\) s are independent and \(\left.P \mid Y_{i}=1\right\\}=p=1-P\left\\{Y_{i}=-1\right\\}\). Argue that if \(p>\frac{1}{2}\), then, by the strong law of large numbers, \(\sum_{1}^{n} Y_{j} \rightarrow \infty\) as \(n \rightarrow \infty\) and hence the initial state 0 can be visited only finitely often, and hence must be transient. A similar argument holds when \(p<\frac{1}{2}\).

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.

Suppose that on each play of the game a gambler either wins 1 with probability \(p\) or loses 1 with probability \(1-p\). The gambler continues betting until she or he is either winning \(n\) or losing \(m\). What is the probability that the gambler quits a winner?

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 \(l\), \(i=0,1,2,3\), if the first urn contains \(i\) white balls. At cach 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.

A Markov chain is said to be a tree process if (i) \(P_{U}>0\) whenever \(P_{y}>0\). (ii) for every pair of states \(i\) and \(j, i \neq j\), there is a unique sequence of distinct states \(i=i_{0}, i_{1}, \ldots, i_{n-1}, i_{n}=j\) such that $$ P_{h \cdot h+1}>0, \quad k=0,1, \ldots, n-1 $$ In other words, a Markov chain is a tree process if for every pair of distinet states \(i\) and \(j\) there is a unique way for the process to go from \(i\) to / without reentering a state (and this path is the reverse of the unique path from \(j\) to \(l\) ). Argue that an ergodic tree process is time reversible.

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