ST3236: Stochastic ProcessTutorial 8

TA: Mar Choong Hock

Email: g0301492@nus.edu.sg

Exercises: 9

Question 1A 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 Xn denote 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 = 12 = Var() = E(2) - E()2 = 2 – 1 = 1

Thus E(Xn) = n = 1,

And Var(Xn) = 2[n-1 + … + 2n-2] = n2 = 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 3Suppose 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 3The 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 = 0.5u2 = (u1) = 0.5 + 0.5(0.5)2 = 0.6250u3 = (u2) = 0.5 + 0.5(0.6250)2 = 0.6953u4 = (u3) = 0.5 + 0.5(0.6953)2 = 0.7417u5 = (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 u1 of 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) = 0.3329u2 = (u1) = e-1.1(1-0.3329) = 0.4801u3 = (u2) = e-1.1(1-0.4801) = 0.5644u4 = (u3) = e-1.1(1-0.5644) = 0.6193u5 = (u4) = e-1.1(1-0.6193) = 0.6579

Question 4Solving 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 4To 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

n xn f(xn) f’(xn)

0 0.82895 0.0013931 0.26636

1 0.82372 -0.000041255 0.28207

2 0.82387 0.0000011670 --

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 5bThe 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) = 0.6250u2 = (u1) = 0.7955u3 = (u2) = 0.8726u4 = (u3) = 0.9153u5 = (u4) = 0.9414

The probability that the husband's male line descent will cease to exist by the 5th generation is 0.9414