Transcript ST3236: Stochastic Process Tutorial

ST3236: Stochastic Process
Tutorial 8
TA: Mar Choong Hock
Email: [email protected]
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 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 = 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 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)
= 0.3329
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
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 threefourths 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
P( = 1) = ¾ x 0.52
P( = 2) = ¾ x 0.53
…
P( = k) = ¾ x 0.5k+1
(no child or first one is a girl)
(second one is a girl)
(third one is a girl)
((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