homework answer-9

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MATH424 HOMEWORK ANSWER
Homework 9. (Chapter 6)
Exercise 1.1
Solution:
Method 1. We have
P0(t) = e−λ0t = e−t
P1(t) = λ0e−λ1t
∫ t
0
eλ1 xP0(x)dx
= e−3t
∫ t
0
e2xdx
=
1
2
e−t − 1
2
e−3t
P2(t) = λ1e−λ2t
∫ t
0
eλ2 xP1(x)dx
= 3e−2t
∫ t
0
(
1
2
ex − 1
2
e−x
)
dx
= 3
(
1
2
e−t +
1
2
e−3t − e−2t
)
P3(t) = λ2e−λ3t
∫ t
0
eλ3 xP2(x)dx
= 2e−5t
∫ t
0
3
(
1
2
e−x +
1
2
e−3x − e−2x
)
dx
= 6
(
1
8
e−t +
1
4
e−3t − 1
3
e−2t − 1
24
e−5t
)
Method 2.
Based on the formula on page 338, we can calculate B0,1 = 12 , B1,1 =
− 12 , B0,2 = 12 , B1,2 = 12 , B2,2 = −1, B0,3 = 18 , B1,3 = 14 , B2,3 = − 13 , B3,3 = − 124 .
Therefore, we have
P0(t) = e−t;
P1(t) =
1
2
e−t − 1
2
e−3t;
1
2 MATH424 HOMEWORK ANSWER
P2(t) = 3
(
1
2
e−t +
1
2
e−3t − e−2t
)
;
P3(t) = 6
(
1
8
e−t +
1
4
e−3t − 1
3
e−2t − 1
24
e−5t
)
. 2
Exercise 1.2
Solution: Let S 1, S 2, S 3 be the first, second and third sojourn time, respec-
tively. Therefore, we have S 1, S 2 and S 3 are independent and exponentially
distributed with parameter 1, 1/3, 1/2, respectively. W1 = S 1, W2 = S 1 +S 2
and W3 = S 1 + S 2 + S 3. Therefore, we have
(a). E(W3) = E(S 1) + E(S 2) + E(S 3) = 1λ0 +
1
λ1
+ 1
λ2
= 1 + 12 +
1
3 =
11
6 .
(b). E(W1 + W2 + W3) = 3E(S 1) + 2E(S 2) + E(S 3) = 3 + 2 · 13 + 12 = 256 .
(c). Var(W3) = Var(S 1) + Var(S 2) + Var(S 3) = 1 + 132 +
1
22 =
49
36 . 2
Exercise 1.3
Solution:
P(X(t + h) − X(t) = 1|X(t) = n)
= P(only one of the n organisms splits in the time interval (t, t + h])
=
(
n
1
)
(θh + o(h)) (1 − θh + o(h))n−1
= nθh + on(h)
Therefore, X(t) is a Yule Process. 2
Problem 1.4
Solution: (a). Let’s consider the first sojourn time, it will be exponentially
distributed with parameter 1. About the second sojourn time, it will be
exponentially distributed with parameter 3(=1+2, where 1 from the media
advertising and 2 from the word-of-mouth advertising from the previous
one customer.)∗ In general, the nth sojourn time will be exponentially dis-
tributed with parameter 1+2(n−1). Therefore, treating X(t) as a pure jump
process, we have the parameters λk = 1 + 2k.
Remark: * Here we use the fact that if S and T are two independent ex-
ponentially distributed random variables with parameter α and β, respec-
tively, then min{S ,T } is still an exponentially distributed random variable
with parameter α + β.
MATH424 HOMEWORK ANSWER 3
(b). λ0 = 1, λ1 = 3, λ2 = 5, . . . In order to find P2(1), we need to compute
P2(t). Given P0(t) = e−t, we have
P1(t) = λ0e−λ1t
∫ t
0
eλ1 xP0(x)dx
= e−3t
∫ t
0
e2xdx
=
1
2
e−t − 1
2
e−3t
P2(t) = λ1e−λ2t
∫ t
0
eλ2 xP1(x)dx
=
3
8
e−t − 3
4
e−3t +
3
8
e−5t
hence, P2(1) =
3
8
e−1 − 3
4
e−3 +
3
8
e−5 � 0.103 2
Problem 1.11
Solution: Clearly we have P0(t) = e−λt and
P1(t) = λe−λt
∫ t
0
eλxP0(x)dx = λe−λt
∫ t
0
eλxe−λxdx = λte−λt.
Using induction, suppose Pn−1(t) =
(λt)n−1
(n − 1)!e
−λt, then
Pn(t) = λe−λt
∫ t
0
eλxPn−1(x)dx
= λe−λt
∫ t
0
eλx
(λx)n−1
(n − 1)!e
−λxdx
= λe−λt
∫ t
0
(λx)n−1
(n − 1)!dx
=
(λt)n
n!
e−λt.
Therefore,
Pn(t) =
(λt)n
n!
e−λt, n = 0, 1, . . . 2