Exercises

Jun 09'24

Let [math]Z_1[/math], [math]Z_2[/math], ..., [math]Z_N[/math] describe a branching process in which each parent has [math]j[/math] offspring with probability [math]p_j[/math]. Find the probability [math]d[/math] that the process eventually dies out if

[math]p_0 = 1/2[/math], [math]p_1 = 1/4[/math], and [math]p_2 = 1/4[/math].
[math]p_0 = 1/3[/math], [math]p_1 = 1/3[/math], and [math]p_2 = 1/3[/math].
[math]p_0 = 1/3[/math], [math]p_1 = 0[/math], and [math]p_2 = 2/3[/math].
[math]p_j = 1/2^{j + 1}[/math], for [math]j = 0[/math], 1, 2, ....
[math]p_j = (1/3)(2/3)^j[/math], for [math]j = 0[/math], 1, 2, ....
[math]p_j = e^{-2} 2^j/j![/math], for [math]j = 0[/math], 1, 2, ... (estimate [math]d[/math] numerically).

BBy Bot

Jun 09'24

Let [math]Z_1[/math], [math]Z_2[/math], ..., [math]Z_N[/math] describe a branching process in which each parent has [math]j[/math] offspring with probability [math]p_j[/math]. Find the probability [math]d[/math] that the process dies out if

[math]p_0 = 1/2[/math], [math]p_1 = p_2 = 0[/math], and [math]p_3 = 1/2[/math].
[math]p_0 = p_1 = p_2 = p_3 = 1/4[/math].
[math]p_0 = t[/math], [math]p_1 = 1 - 2t[/math], [math]p_2 = 0[/math], and [math]p_3 = t[/math], where [math]t \leq 1/2[/math].

BBy Bot

Jun 09'24

In the chain letter problem (see Example) find your expected profit if

[math]p_0 = 1/2[/math], [math]p_1 = 0[/math], and [math]p_2 = 1/2[/math].
[math]p_0 = 1/6[/math], [math]p_1 = 1/2[/math], and [math]p_2 = 1/3[/math].

Show that if [math]p_0 \gt 1/2[/math], you cannot expect to make a profit.

BBy Bot

Jun 09'24

Let [math]S_N = X_1 + X_2 +\cdots+ X_N[/math], where the [math]X_i[/math]'s are independent random variables with common distribution having generating function [math]f(z)[/math]. Assume that [math]N[/math] is an integer valued random variable independent of all of the [math]X_j[/math] and having generating function [math]g(z)[/math]. Show that the generating function for [math]S_N[/math] is [math]h(z) = g(f(z))[/math]. Hint: Use the fact that

[[math]] h(z) = E(z^{S_N}) = \sum_k E(z^{S_N} | N = k) P(N = k)\ . [[/math]]

BBy Bot

Jun 09'24

We have seen that if the generating function for the offspring of a single parent is [math]f(z)[/math], then the generating function for the number of offspring after two generations is given by [math]h(z) = f(f(z))[/math]. Explain how this follows from the result of Exercise.

BBy Bot

Jun 09'24

Consider a queueing process such that in each minute either 1 or 0 customers arrive with probabilities [math]p[/math] or [math]q = 1 - p[/math], respectively. (The number [math]p[/math] is called the arrival rate.) When a customer starts service she finishes in the next minute with probability [math]r[/math]. The number [math]r[/math] is called the service rate.) Thus when a customer begins being served she will finish being served in [math]j[/math] minutes with probability [math](1 - r)^{j -1}r[/math], for [math]j = 1[/math], 2, 3, ....

Find the generating function [math]f(z)[/math] for the number of customers who arrive in one minute and the generating function [math]g(z)[/math] for the length of time that a person spends in service once she begins service.
Consider a customer branching process by considering the offspring of a customer to be the customers who arrive while she is being served. Using Exercise, show that the generating function for our customer branching process is [math]h(z) = g(f(z))[/math].
If we start the branching process with the arrival of the first customer, then the length of time until the branching process dies out will be the busy period for the server. Find a condition in terms of the arrival rate and service rate that will assure that the server will ultimately have a time when he is not busy.

BBy Bot

Jun 09'24

Let [math]N[/math] be the expected total number of offspring in a branching process. Let [math]m[/math] be the mean number of offspring of a single parent. Show that

[[math]] N = 1 + \left(\sum p_k \cdot k\right) N = 1 + mN [[/math]]

and hence that [math]N[/math] is finite if and only if [math]m \lt 1[/math] and in that case [math]N = 1/(1 - m)[/math].

BBy Bot

Jun 09'24

Consider a branching process such that the number of offspring of a parent is [math]j[/math] with probability [math]1/2^{j + 1}[/math] for [math]j = 0[/math], 1, 2, ....

Using the results of Example show that the probability that there are [math]j[/math] offspring in the [math]n[/math]th generation is
[[math]] p_j^{(n)} = \left \{ \begin{array}{ll} \frac{1}{n(n + 1)} (\frac {n}{n + 1})^j, & \mbox{if $ j \geq 1$}, \\ \frac {n}{n + 1}, & \mbox{if $ j = 0$}.\end{array}\right. [[/math]]
Show that the probability that the process dies out exactly at the [math]n[/math]th generation is [math]1/n(n + 1)[/math].
Show that the expected lifetime is infinite even though [math]d = 1[/math].