1. ST3236: Stochastic Process Tutorial 8
TA: Mar Choong Hock
Exercises: 9

2. 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 Xn denote the number of individuals in the
population in the nth generation.
Find the mean and variance of Xn

3. Question 1 The number of offsprings are :
P( = 0) = 0.5, P( = 2) = 0.5
Therefore,
 = E() = 0.5 x 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

4. 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.

5. 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

6. 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.

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

8. 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?

9. 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) =
u2 = (u1) = e-1.1( ) =
u3 = (u2) = e-1.1( ) =
u4 = (u3) = e-1.1( ) =
u5 = (u4) = e-1.1( ) =

10. Question 4 Solving the equation, u = (u) = e-1.1(1-u)
 (e1.1)u =(e1.1)u
Let a = e1.1= , 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)

11. 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 = ( )/2 = , f( ) = (good first guess, close to zero!)
 Let x0 =

12. Question 4 n xn f(xn) f’(xn) 0.82895 0.0013931 0.26636 1 0.82372 1 2 --
 The small root is u =

13. 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.

14. 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?

15. 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 (second one is a girl)
P( = 2) = ¾ x (third one is a girl) …
P( = k) = ¾ x 0.5k+1 ((k+1)th one is a girl)

16. 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) [ s + (0.5s)2 + … ]
= ¼ + ¾ (0.5) x 1/ ( s)
= ¼ + 3 (0.5) / 4( s)
= [( s) + 3(0.5)] / 4( s)
= (5 - s) / 4(2 - s)

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