Exercise 5 (10pts) Let \(X\) be uniformly distributed on \(\{0,1,2\}\). Find the PGF \(\Pi_X(s)\) of \(X\).

Solution We have \[ P(X=0)=P(X=1)=P(X=2)=\frac{1}{3}. \] Therefore, for the PGF of \(X\) we obtain: \[ \Pi_X(s)=\frac{1}{3}s^0+\frac{1}{3}s^1+\frac{1}{3}s^2= \frac{1}{3}(1+s+s^2). \]

Exercise 6 (10pts) Suppose that the random variable \(X\) has a probability mass function given by \[ p_j=\frac{1}{2^{j+1}}, \qquad j=0,1,2,\ldots. \] Find the PGF \(\Pi_X(s)\) of \(X\).

Solution By the definition of a PDF: \[ \Pi_X(s)=\sum_{x=0}^\infty \frac{1}{2^{x+1}}\cdot s^x = \frac{1}{2}\sum_{x=0}^\infty \left(\frac{s}{2}\right)^x = \frac{\frac{1}{2}}{1-\frac{s}{2}}=\frac{1}{2-s}. \]
Exercise 7 (20pts) Let \(X\) be a Poisson random variable, i.e., for \(\lambda>0\) \[ P(X=k)=\frac{\lambda^k}{k!}e^{-\lambda}, \qquad k=0,1,2,\ldots \]

  1. Find the PGF of \(X\).

  2. Assume that \(Y\) is another Poisson random variable with parameter \(\mu\). If \(X\) and \(Y\) are independent, find the PGF of \(X+Y\).

  3. What is the distribution of \(X+Y\)?

Solution

  1. For the PDF of \(X\) we have: \[ \Pi_X(s)=\sum_{x=0}^\infty \frac{\lambda^x}{x!}e^{-\lambda}s^x =e^{-\lambda}\sum_{x=0}^\infty \frac{(\lambda s)^x}{x!}=e^{-\lambda}e^{\lambda s}=e^{\lambda (s-1)}. \]

  2. Using the result in (a) and the independence of \(X\) and \(Y\), we obtain \[ \Pi_{X+Y}(s)=\Pi_X(s)\Pi_Y(s)=e^{\lambda(s-1)}\cdot e^{\mu(s-1)}=e^{(\lambda+\mu)(s-1)}. \]

  3. The sum \(X+Y\) follows a Poisson distribution with parameter \((\lambda+\mu)\).

Exercise 8 (20pts) Let \(Y_1, Y_2, \ldots\) be a sequence of i.i.d. Bernoulli random variables with parameter \(p\). Let \(N\) be a Poisson random variable with parameter \(\lambda\), which is independent of the \(Y_i\).

  1. Find the PGF of \(X=\sum_{i=1}^N Y_i\).

  2. Use (a) to identify the distribution of \(X\).

Solution
Similarly to the beginning of Lecture~6, for the PGF of \(X\) we have \[ \Pi_X(s)=\Pi_N\left(\Pi_Y(s)\right). \] On the other hand \[ \Pi_N(s)=e^{\lambda(s-1)}\qquad \mbox{and}\qquad \Pi_Y(s)=91-p)+ps. \] Therefore, \[ \Pi_X(s)=e^{\lambda(1-p+ps-1)}=e^{\lambda p(s-1)}. \] (b) The PGF in (a) is the PGF of a Poisson random variable with parameter \(\lambda\).

Exercise 9 (10pts) Given a branching process with the following offspring distributions, determine the extinction probability \(p_e\). Suppose \(X_0=1\).

  1. \(p_0=0.25\), \(p_1=0.4\), \(p_2=0.35\).

  2. \(p_0=0.5\), \(p_1=0.1\), \(p_3=0.4\).

Solution

  1. The offspring PGF is: \[ \Pi_{X_0}(s)=0.25+0.40s+0.35s^2. \] Thus, the \(p_e\) satisfies the equation \[ 0.25+0.40p_e+0.35p_e^2=p_e. \] One can check that this equation has as solutions \(5/7\) an \(1\). So, \(p_e=5/7\).

  2. The equation for \(p_e\) is \[ 0.5+0.1p_e+0.4p_e^3=p_e, \] or, after factorization, \[ (p_e-1)(4p_e^2+4p_e-5)=0. \] The solutions are -1.725, 1, and 0.725. Therefore, \(p_e=0.725\), which is the smallest solution between 0 and 1.

Exercise 10 (10pts) Consider the branching process with offspring distribution as in Exercise 9 and suppose \(X_0=1\). What is the probability that the population is extinct in the second generation \((X_2=0)\), given that it did not die out in the first generation \((X_1>0)\)?

Solution We will consider the branching process in Exercise 9(b) only. Using the definition of conditional probability we have: \[ P(X_2=0\ |\ X_1>0)=\frac{P(X_2=0, X_1>0)}{P(X_1>0)}=\frac{P(X_2=0, X_1=1)+P(X_2=0, X_1=3)}{1-P(X_1=0)}\\ = \frac{p_{01}\cdot p_{10}+p_{03}\cdot p_{30}}{1-p_{10}}=\frac{(0.1)((0.5)+(0.4)(0.5)^3}{0.5}=0.2. \]

Exercise 11 (20pts) A branching process has offspring distribution \((p_0, p_1, p_3)=(0.25, 0.25, 0.5)\). Find the following:

  1. The offspring PGF \(\Pi_{X_0}(s)\).

  2. The extinction probability \(p_e\).

  3. \(\Pi_{X_2}(s)\).

  4. \(P(X_2=0)\).

Solution

  1. By definition \[ \Pi_{X_0}(s)=0.25+0.25s+0.50s^3. \]

  2. Since \[ 0.25+0.25p_e+0.50 p_e^3=p_e, \] or, equivalently, \[ (p_e-1)(2p_e^2+2p_e-1)=0, \] we have \(p_e=0.366\).

  3. Now, using the properties of branching processes, we write \[ \Pi_{X_2}(s)=\Pi_{X_0}\left(\Pi_{X_0}\right)=0.25+0.25(0.25+0.25s+0.5s^3)+0.5(0.25+0.25s+0.25s^3)^3. \]

  4. Using teh result in (c), we calculate \[ P(X_2=0)=\Pi_{X_2}(0)=0.3203. \]