Mock Final Exam of MA3238/ST3236
Name:
Matric No.:
INSTRUCTIONS:
• This test paper contains a total of 4 questions and 4 printed pages.
• Answer all questions in the space below the questions. You can use the back side of the
paper as scrap paper. If you write your answers on the back side, please indicate clearly.
• You should lay out systematically the various steps in your answers. Any result or theorem
used to justify the answers should be clearly stated. (For example, you may say a result
is from our tutorial.)
1
1. At the New York State Fair in Syracuse, Larry encounters a carnival game where for one
dollar he may buy a single coupon allowing him to play a guessing game. On each play,
Larry has an even chance of winning or losing a coupon. When he runs out of coupons
he loses the game. However, if he can collect three coupons, he wins a surprise.
(a) Express this problem as a Markov chain.
Solution
This is a Markov chain with 4 states. The transition matrix is
0 1 2
0 1 0 0
1 21 0 12
2 0 21 0
3 0 0 0
3
0
0
1
2
1
(b) What is the probability Larry will win the surprise?
Solution Let h(x) be the probability for Larry to win the surprise if he has x
coupons initially.
h(0) = 0,
h(3) = 1,
1
h(1) = h(0) +
2
1
h(2) = h(1) +
2
We solve that
1
h(1) = ,
3
1
h(2),
2
1
h(3).
2
2
h(2) = .
3
The answer is 1/3.
(c) What is the expected number of plays he needs to win or lose the game.
Solution Let g(x) be the expected time for Larry to exit the game if he has x
coupons initially.
g(0) = 0,
g(3) = 0,
1
g(1) = 1 + g(0) +
2
1
g(2) = 1 + g(1) +
2
1
g(2),
2
1
g(3).
2
We solve that
g(1) = g(2) = 2.
So the answer is 2 plays.
2. Wayne Gretsky scored a Poisson mean 6 number of points per game. 60% of these were
goals and 40% were assists (each is worth one point). Suppose he is paid a bonus of 3K
for a goal and 1K for an assist.
(a) Find the mean and standard deviation for the total revenue he earns per game.
2
Solution Suppose Gretsky plays the game in one hour, and the points occur in
Poisson process with rate 6 per hour. Then the rewards of goals and assists are in a
compound Poisson process, with P(Yi = 3) = 0.6 and P(Yi = 1) = 0.4 (units: $K).
Then
E(Y1 + · · · + YN (1) ) = 6E(Yi ) = 6 × 2.2 = 13.2,
var(Y1 + · · · + YN (1) ) = 6E(Yi2 ) = 6 × 5.8 = 34.8.
Thus the mean and standard deviation are 13.2 and
√
34.8 respectively.
(b) What is the probability that he has 4 goals and 2 assists in one game?
Solution The goals occur in a Poisson process with rate 3.6, the assists occur in
a Poisson process with rate 2.4, and they are independent. The probability that
Gretsky has 4 goals and 2 assists is
P(Poisson(3.6) = 4)P(Poisson(2.4) = 2) = e−3.6
(3.6)4
(2.4)2
× e−2.4
.
24
2
3. A small computer store has room to display up to 3 computers for sale. Customers come
at times of a Poisson process with rate 2 per week to buy a computer and will buy one if
at least 1 is available. When the store has only 1 computer left it places an order for 2
more computers. The order takes an exponentially distributed amount of time with mean
1 week to arrive. Of course, while the stoor is waiting for delivery, sales may reduce the
inventory to 1 and then to 0.
(a) Write down the matrix of transition rates Q.
Solution
The matrix of transition rates is
0
1
2
3
0 −1 0
1
0
1 2 −3 0
1
2 0
2 −2 0
3 0
0
2 −2
(b) If currently the computer is out of stock, what is the expected time for the computer
to be out of stock again? (The next time when the computer is out of stock happens
after the sale of at least one computer.)
Solution Let g(x) be the expected time for the stock to be 0 at the first time.
Then since it takes on average 1 week for the stock to be nonzero, and the stock will
be 2, the answer to the question is 1 + g(2).
g(0) = 0,
1 2
1
g(1) = + g(0) + g(3),
3 3
3
1
g(2) = + g(1),
2
1
g(3) = + g(2).
2
3
We solve that
g(1) = 1,
g(2) = 1.5,
g(3) = 2.
So the answer to this quesiton is 2.5 weeks.
4. A small company maintains a fleet of four cars to be driven by its workers on business
trips. Requests to use cars are a Poisson process with rate 1.5 per day. A car is used
for an exponentially distributed time with mean 2 days. Forgetting about weekends, we
arrive at the following Markov chain for the number of cars in service.
0
1
2
3
4
0 −1.5 1.5
0
0
0
1 0.5 −2 1.5
0
0
2
0
1.0 −2.5 1.5 0
3
0
0
1.5 −3 1.5
4
0
0
0
2 −2
(a) Find the stationary distribution.
Solution
tion is
The stationary distribution is obtained by solving πQ = 0, and the solu
8 24 36 36 27
π=
,
,
,
,
.
131 131 131 131 131
(b) In the long run, what fraction of requests are unfulfilled?
Solution In the long run, the fraction of time that all the 4 cars are on business
trips is 27/131. So in the long run, the fraction that the requests are unfulfilled is
27/131.
(c) Suppose currently 3 cars are on business trips and one car is vacant. What is the
expected time for all the 4 cars to be on business trips the first time?
Solution Let g(x) be the expected time for all the 4 cars to be on business trips
the first time, if initially x cars are on business trips. Then
g(4) = 0,
2
g(0) = + g(1),
3
1 1
g(1) = + g(0) +
2 4
2 2
g(2) = + g(1) +
5 5
1 1
g(3) = + g(2) +
3 2
3
g(2),
4
3
g(3),
5
1
g(4).
2
We solve that
g(0) =
128
,
27
g(1) =
The answer is 52/27.
4
110
,
27
g(2) =
86
,
27
g(3) =
52
.
27