ST3236: Stochastic Process Tutorial 8

ST3236: Stochastic ProcessTutorial 8 TA: Mar Choong Hock Email: g0301492@nus.edu.sg Exercises: 9

Question 1 A population begins with a single individual. In each generation, each individual in the population dies with probability 0.5 or double with probability 0.5. Let Xndenote the number of individuals in the population in the nth generation. Find the mean and variance of Xn

Question 1 The number of offsprings are : P(= 0) = 0.5, P(= 2) = 0.5 Therefore,  = E() = 0.5 x 0 + 0.5 x 2 = 1 2 = Var() = E(2) - E()2 = 2 – 1 = 1 Thus E(Xn) = n= 1, And Var(Xn) = 2[n-1 + … + 2n-2] = n2 = n

Question 2 The number of offspring's of an individual in a population is 0, 1, 2 with respectively probabilities a > 0, b > 0 and c > 0, where a + b + c = 1. Express the mean and variance of the offspring distribution in terms of b and c.

Question 2 • The number of offsprings are : • P(= 0) = a, P(= 1) = b, P(= 2) = c • It is easy to see that • = E() = a x 0 + b x 1 + c x 2 = b + 2c 2 = E(2) - E()2 = (a x 02 + b x 12+ c x 22) - (b + 2c)2 = b + 4c - (b + 2c)2

Question 3 Suppose a parent has no offspring with probability 1/2 and has two with probability 1/2. If a population of such individuals begins with a single parent and evolves as a branching process, determine un, the probability that the population is extinct by the nth generation, for n = 1, 2, 3, 4, 5.

Question 3 The number of offsprings are : P(= 0) = 0.5, P(= 2) = 0.5 We have the p.g.f.: (s) = 0.5 + 0s + 0.5s2 = 0.5 + 0.5s2 Note that u0 = P(X0 = 0 |X0 = 1) = 0, u1 = (u0) = 0.5 + 0.5(0)2 = 0.5 u2 = (u1) = 0.5 + 0.5(0.5)2 = 0.6250 u3 = (u2) = 0.5 + 0.5(0.6250)2 = 0.6953 u4 = (u3) = 0.5 + 0.5(0.6953)2 = 0.7417 u5 = (u4) = 0.5 + 0.5(0.7417)2 = 0.7751

Question 4 Suppose that the offspring distribution is Poisson with mean = 1.1. Compute the extinction probabilities un= P(Xn= 0 | X0 = 1) for n = 0, 1, 2, 3, 3, 4, 5. What is the probability u1of ultimate extinction?

Question 4 The p.g.f of the offspring distribution of an individual is (s) = e-1.1(1-s). Note that u0 = P(X0 = 0 | X0 = 1) = 0, u1 = (u0) = e-1.1(1-0) = 0.3329 u2 = (u1) = e-1.1(1-0.3329) = 0.4801 u3 = (u2) = e-1.1(1-0.4801) = 0.5644 u4 = (u3) = e-1.1(1-0.5644) = 0.6193 u5 = (u4) = e-1.1(1-0.6193) = 0.6579

Question 4 Solving the equation, u= (u) = e-1.1(1-u)  (e1.1)u =(e1.1)u Let a = e1.1= 3.004166, x = u ax - ax = 0 No known closed form solution, solve by Newton’s Method. Let: f(x)=ax - ax ,f’(x)=a – 1.1ax, xn represents the nth iterations. Aim : Solve for f(x) =0. Then: xn+1 = xn – f(xn)/f’(xn)

Question 4 To get a good first guess for the root, we can plot the graph or do a simple bisection search. Since we know solution must be, 0.6579(see u5) < x < 1. Try x = (0.6579 +1)/2 = 0.82895, f(0.82895) = 0.0013931 (good first guess, close to zero!)  Let x0 = 0.82895

Question 4  The small root is u= 0.8239

Question 5 One-fourth of the married couples in a distant society has no children at all. The other three- fourths of families continue to have children until the first girl and then cease childbearing. Assume that each child is equally likely be boy and girl.

Question 5 (a) For k = 0, 1, 2, … what is the probability that a particular husband will have k male offspring? (b) What is the probability that the husband's male line descent will cease to exist by the 5th generation?

Question 5a The number of male offsprings are distributed as: P( = 0) = ¼ + ¾ x 0.5 (no child or first one is a girl) P( = 1) = ¾ x 0.52 (second one is a girl) P( = 2) = ¾ x 0.53 (third one is a girl) … P( = k) = ¾ x 0.5k+1 ((k+1)th one is a girl)

Question 5b The p.g.f. is, (s) = P( = 0) s0 + P( = 1) s1 + P( = 2) s2 + … = [¼ + ¾ (0.5)]s0 + ¾ (0.5)2s1 + ¾ (0.5)3s2 + … = ¼ + ¾ (0.5) [1 + 0.5s + (0.5s)2 + … ] = ¼ + ¾ (0.5) x 1/ (1 - 0.5s) = ¼ + 3 (0.5) / 4(1 - 0.5s) = [(1 - 0.5s) + 3(0.5)] / 4(1 - 0.5s) = (5 - s) / 4(2 - s)

Question 5b Note that u0 = P(X0 = 0 | X0 = 1) = 0, u1 = (u0) = 0.6250 u2 = (u1) = 0.7955 u3 = (u2) = 0.8726 u4 = (u3) = 0.9153 u5 = (u4) = 0.9414 The probability that the husband's male line descent will cease to exist by the 5th generation is 0.9414

ST3236: Stochastic Process Tutorial 8