Tutorial 5
1
1.1
Review of Chapter 3
Introduction
[Def ] A discrete-time Markov chain is a stochastic process cfw_Xn , n = 0, 1, . . . with a discrete state
space such that S such that
P (Xn+1 = j |Xn = i, Xn1 = in1 , Xn2 = in2 , . . . , X1 = i1 , X0 = i0
Tutorial 7
1
Review of Chapter 3 (III)
1.1
Limiting probabilities
(n)
[Def ] Let d(i) denote the greatest common divisor of all positive integers n for which pii > 0. The
integer d(i) is called the period of state i. A state with period of 1 is called ape
T4 Solution
1. (a) Suppose the location of the ant just after nth stage is (Xn , Yn ), where Xn and Yn represent
the displacements in West-East direction and North-South direction, respectively. Then,
2
2
2
E (Zn ) = E (Xn + Yn ).
Let
1
1
Ui =
0
1
1
Vi =
Tutorial 6
1
Review of Chapter 3 (II)
1.1
Calculation of P n
When |z | < 1,
(I zP )1 = I + zP + z 2 P 2 + . . .
Hence, P n = coecient of z n in expansion of (I zP )1 .
1.2
Classication of states
(n)
State j is said to be accessible from state i if pij >
T2 solution
1. M follows binomial distribution with parameters (8, 1 ).
2
1
Hence, E (M ) = (8)( 2 ) = 4
1
When M = m, N follows binomial distribution with parameters (m, 2 ).
1
When N = n, Z follows binomial distribution with parameters (n, 2 ).
Hence,
E
Tutorial 4
1
Review of Chapter 2
1.1
Random walks
The stochastic process X = cfw_X0 , X1 , X2 , . . . is a random walk in one dimension if
X0 = c where c is a constant.
For n 1, Xn = Xn1 + Zn where cfw_Zn , n = 1, 2, . . . are i.i.d. with P (Zn = 1) = p
T1 solution
1. (a)
P (X < x, y1 < Y < y2 , Z > z )
=P (X < x)P (y1 < Y < y2 )P (Z > z ) (Because X, Y and Z are independent)
( x
) ( y2
) (
)
=
et dt
et dt
et dt
0
y1
=(1 e
x
)(e
y1
z
y2
e
)e
z
(b) Let T = maxcfw_X, Y, Z , then for t 0 we have
FT (t) = P
Tutorial 2
1
Review of Chapter 1 (II)
1.1
Conditional expectations (II)
[Def ] Suppose X is a random variable and Y is a discrete random variable. Dene a random variable
E (X |Y ) (conditional expectation of X given Y ) as
E (X |Y ) = E (X |Y = y ) when Y
T7 solution
1. (a) This is a nite irreducible Markov chain with period 1.
Note that p11 = 0.6 > 0, that means starting from state 1, the chain can go back to state
1 in 1, 2, . . . steps. The period of state 1 is 1. Also, this is an irreducible Markov cha
T3 solution
1. Note that X is a nonnegative random variable.
(1 F (x)dx
x
=
dx +
e dx
0
[
]
x
= + e
E (X ) =
0
=+
2. Method 1: Note that X is a nonnegative random variable.
E (X ) =
0
1
(1 F (x)dx
(1 F (x)dx +
(1 F (x)dx
0
1
1
1
x
(1 (1 2 )dx
(1 )dx +
=
Tutorial 3
1
Review of Chapter 1 (III)
1.1
Random variables which are neither discrete nor continuous
[Thm1.11] For any nonnegative random variable X , if FX (x) denotes the distribution function of
X , then
+
E (X ) = 0 [1 FX (x)]dx
provided that E [X ]
Tutorial 10
1
Review of Chapter 5
1.1
Brownian motion
1
[Lemma] The moment generating function (with t being the argument) of N (, 2 ) is et+ 2
2 t2
.
[Def ] A stochastic process cfw_Xt , t 0 is a Brownian motion (process) or Wiener process
if
X0 = 0,
STAT2303 & STAT3603
Assignment 2
(Due at 17:00, 10 Oct. 2013)
Q1
Let q 1 p and S be the random sum X1 X2 XN. Then
G X1 (z) p z q.
1A
GS(z) GN(G X1 (z) exp(G X1 (z) 1) exp(p z q 1) ep(z 1),
1M
which is the p.g.f. of the Poisson distribution with mean p. Th
STAT2303 & STAT3603
Assignment 4
(Due at 17:00, 28 Nov. 2013)
1. Assume that X1 and X2 are random variables following Exp(1) and Exp(2) respectively.
Suppose Y is a nonnegative continuous random variable, and X1, X2 and Y are independent.
Prove, for u 0 a
STAT2303 & STAT3603
Assignment 4
(Due at 17:00, 28 Nov. 2013)
Suppose Y has a density f(y). For u 0 and v 0,
Q1
P(X1 Y u, X2 Y v)
0
0
f(y)dy
y u
1exp(1x1)dx1
y v
2exp(2x2) dx2
f(y)exp(1y 1u 2y 2v)dy
exp(1u2v)
0
3A
1M
f(y)exp(1y 2y)dy.
1A
Letting u v 0
STAT2303 Probability Modelling
Class Test (1st semester, 2012~13)
Q1 Q1 of Assignment 2 this year.
Q2 Two gamblers, A and B, initially have capital $k and $(a k) respectively, where k and a
k are positive integers. At each round of the game, player A
wi
THE UNIVERSITY OF HONG KONG
DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCES
STAT 2303 Probability Modeling (2012-2013)
Example Class 1
Q1. Supp ose that the numb er of road accidents p er week is Poisson distributed. If N, the
numb er of injured in each a
STAT2303 & 3603 PROBABILITY MODELLING
Final Exam
What to be examined: all chapters except options in Section 5.2
How many questions: 7
Any results in lecture notes or assignments can be quoted without proofs unless
otherwise specified.
Final numerical ans
STAT2303 & STAT3603
Assignment 3
(Due at 17:00, 11 Nov. 2013)
Q1
(a) Since p01, p10, p12, p21, p23, p32 are all greater than 0, we know that 0 1 2 3. Therefore
there is only one class, which is closed. Since the chain is irreducible, all states are recurr
Tutorial 9
1
Review of Chapter 4
1.1
Denition of Poisson processes
[Def ] A stochastic process cfw_Nt , t 0 is a counting process if Nt represents the total number of
events that have occurred up to time t. Hence a counting process must satisfy
Nt 0,
Nt
STAT2303 & STAT3603
Assignment 1
(Due at 17:00, 26 Sept. 2013)
Q1
[Solution 1]
Let X 1A, Y 1B. Let F(x, y) be the joint distribution function of X and Y, and FX(x)
and FY(y) be the marginal distribution functions of X and Y respectively. Then
for x 0 or y
Tutorial 8
1
Review of Chapter 3 (IV)
1.1
Absorbing states
Assume that S = cfw_1, 2, . . . , u, u + 1, . . . , u + v and states 1, 2, . . . , u are absorbing. Then
)
(
Iuu 0uv
P=
Rvu Qvv
Let SA = cfw_1, 2, . . . , u be the set consisting of all absorbing
T8 Solution
1. (a) Dene 5 states where state i represents that there are i red balls in the urn. The state
space S = cfw_0, 1, 2, 3, 4.
Initial distribution: 0 = (0, 0, 1, 0, 0)
The transition probability matrix is as follows:
0
1
2
3
4
0
1
0
0
0
0
1 0.25
STAT2303 & STAT3603 Probability Modelling
Class Test (28 Oct 2013)
Q1
h11 P(last shot is by player 1 | the first shot is by player 1 and does not hit the target)
P(the first shot is by player 1 and does not hit the target)
P(last shot is by player 1 | t
STAT2303 & STAT3603
Assignment 1
(Due at 17:00, 26 Sept. 2013)
1. Suppose there are two events A and B. Prove that
A and B are independent 1A and 1B are independent.
2. Manuscripts are sent to a typing firm consisting of 3 typists. If a manuscript is type
T9 Solution
1. (a) Use conditional probability:
P[N (t) = n] = P[N (t) = n|Good]P(Good) + P[N (t) = n|Other]P(Other)
0.3(3t)n e3t 0.7(5t)n e5t
+
n!
n!
n e3t + 0.7(5t)n e5t
0.3(3t)
=
.
n!
=
(b) cfw_N (t) : t 0 is not a Poisson process as N (t) does not fol