homework answer-10

Prévia do material em texto

MATH424 HOMEWORK ANSWER
Homework 10. (Chapter 6)
Exercise 2.1
Solution: We have P3(t) = e−µ3t = e−5t,
P2(t) = µ3
(
A22e−µ2t + A32e−µ3t
)
= 5
(
1
µ3 − µ2 e
−µ2t +
1
µ2 − µ3 e
−µ3t
)
=
5
3
(
e−2t − e−5t
)
,
P1(t) = µ2µ3
(
A11e−µ1t + A21e−µ2t + A31e−µ3t
)
= µ2µ3
(
1
(µ2 − µ1)(µ3 − µ1)e
−µ1t +
1
(µ3 − µ2)(µ1 − µ2)e
−µ2t +
1
(µ2 − µ3)(µ1 − µ3)e
−µ3t
)
= 10
(
−1
2
e−3t +
1
3
e−2t +
1
6
e−5t
)
and P0(t) = 1 − P1(t) − P2(t) − P3(t). 2
Exercise 2.3
Solution: We have P3(t) = e−µ3t = e−t,
P2(t) = µ3
(
A22e−µ2t + A32e−µ3t
)
= 5
(
1
µ3 − µ2 e
−µ2t +
1
µ2 − µ3 e
−µ3t
)
= −e−2t + e−t
P1(t) = µ2µ3
(
A11e−µ1t + A21e−µ2t + A31e−µ3t
)
= µ2µ3
(
1
(µ2 − µ1)(µ3 − µ1)e
−µ1t +
1
(µ3 − µ2)(µ1 − µ2)e
−µ2t +
1
(µ2 − µ3)(µ1 − µ3)e
−µ3t
)
= 2
(
1
2
e−3t − e−2t + 1
2
e−t
)
and P0(t) = 1 − P1(t) − P2(t) − P3(t). 2
Problem 2.2
Solution: Let’s define a new process Y(t)
de f
= N − X(t). Since X(t) is a
pure death process with constant death rate θ, Y(t) can be viewed as a
truncated Poisson process below N, namely, Y(t) has the same distribution
1
2 MATH424 HOMEWORK ANSWER
as Z(t) ∧ N where Z(t) is a Poisson process with parameter θ. Therefore,
for 1 ≤ n ≤ N,
Pn(t) = P(X(t) = n)
= P(N − Z(t) ∧ N = n)
= P(Z(t) ∧ N = N − n)
= P(Z(t) = N − n)
=
(θt)N−n
(N − n)!e
−θt
and for n = 0,
Pn(t) = P(X(t) = 0) = P(Z(t) ∧ N = N) = P(Z(t) ≥ N) = 1 −
N−1∑
k=0
(θt)k
k!
e−θt. 2
Problem 2.6
Proof: (a). We know S k is an exponential time with parameter αk, there-
fore
E(T ) = E(S 1 + · · · + S N)
= E(S 1) + · · · + E(S N)
=
1
α · N + · · · +
1
α · 1
=
1
α
(
1
N
+
1
N − 1 + · · · +
1
1
)
(b).
E(T ) =
∫ ∞
0
P(T > t)dt
=
∫ ∞
0
(1 − FT (t))dt
=
∫ ∞
0
(
1 − (1 − e−αt)N) dt(
Let y = 1 − e−αt, then dt = 1
α(1 − y)dy
)
=
∫ ∞
0
1 − yN
α(1 − y)dy
=
1
α
∫ ∞
0
(
1 + y + · · · + yN−1
)
dy
=
1
α
(
1
N
+
1
N − 1 + · · · +
1
1
)
2
MATH424 HOMEWORK ANSWER 3
Exercise 3.2
Solution: Notice that the treatment time only depends on the number of pa-
tients at that time but not how long the patient has been waiting, therefore,
N(t) has Markovian property. Furthermore, since there is only one doctor
who can only serve one patient at a time and the arrival of patients is a
Poisson process. Therefore, N(t) might be modeled as a birth and death
process. The necessary assumption should be: for any t,
P( Particular patient finish the treatment in [t, t + h)|k patients at time t)
=
1
mk
h + o(h). 2
Problem 3.1
Solution: First of all, let’s show that X(t) is a Markov Chain. To show this,
for s, t > 0, n ∈ N and any 0 ≤ s1 < s2 < · · · < sn = s, we have
P(X(s + t) = j|X(s1) = i1, . . . , X(sn) = in)
= P(ξN(s+t) = j|ξN(s1) = i1, . . . , ξN(sn) = in)
= P(ξN(s+t) = j|ξN(sn) = in) (0.1)
= P(X(s + t) = j|X(sn) = in)
where ( 0.1) holds due to the fact that 0 ≤ N(s1) < N(s2) < · · · < N(sn) and
ξn is a Markov Chain. Hence, we have X(t) is a two state Markov Chain,
too.
Secondly, we want to show X(t) is a time homogeneous Markov Chain:
P(X(s + t) = i|X(s) = j)
= P(ξN(s+t) = i|ξN(s) = j)
= P(ξN(s+t)−N(s) = i|ξ0 = j) (0.2)
= P(ξN(t) = i|ξ0 = j) (0.3)
= P(X(t) = i|X(0) = j)
where ( 0.2) holds because ξn is a time homogeneous Markov Chain and
ξn and N(t) are independent; ( 0.3) holds since N(t) is time homogeneous.
4 MATH424 HOMEWORK ANSWER
Finally, we will show X(t) is a birth and death process. In fact,
P(X(t + h) = 1|X(t) = 0)
= P(X(h) = 1)|X(0) = 0
= P(ξN(h) = 1|ξ0 = 0)
=
∞∑
k=1
P(ξk = 1|ξ0 = 0)P(N(h) = k)
=
∞∑
k=1
P(ξk = 1|ξ0 = 0)(λh)
k
k!
e−λh
=
∞∑
k=1
P(ξk = 1|ξ0 = 0)(λh)
k
k!
(1 − λh + o(h))
= P(ξ1 = 1|ξ0 = 0)λh + o(h)
= λh + o(h);
P(X(t + h) = 0|X(t) = 0)
= P(X(h) = 0)|X(0) = 0
= P(ξN(h) = 0|ξ0 = 0)
=
∞∑
k=0
P(ξk = 0|ξ0 = 0)P(N(h) = k)
=
∞∑
k=0
P(ξk = 0|ξ0 = 0)(λh)
k
k!
e−λh
=
∞∑
k=0
P(ξk = 0|ξ0 = 0)(λh)
k
k!
(1 − λh + o(h))
= P(ξ0 = 0|ξ0 = 0)(1 − λh) + o(h)
= 1 − λh + o(h);
P(X(t + h) = 0|X(t) = 1)
= P(X(h) = 0)|X(0) = 1
= P(ξN(h) = 0|ξ0 = 1)
=
∞∑
k=1
P(ξk = 0|ξ0 = 1)P(N(h) = k)
=
∞∑
k=1
P(ξk = 0|ξ0 = 1)(λh)
k
k!
e−λh
MATH424 HOMEWORK ANSWER 5
=
∞∑
k=1
P(ξk = 0|ξ0 = 1)(λh)
k
k!
(1 − λh + o(h))
= P(ξ1 = 0|ξ0 = 1)λh + o(h)
= (1 − α)λh + o(h);
and
P(X(t + h) = 1|X(t) = 1)
= P(X(h) = 1)|X(0) = 1
= P(ξN(h) = 1|ξ0 = 1)
=
∞∑
k=0
P(ξk = 1|ξ0 = 1)P(N(h) = k)
=
∞∑
k=0
P(ξk = 1|ξ0 = 1)(λh)
k
k!
e−λh
=
∞∑
k=0
P(ξk = 1|ξ0 = 1)(λh)
k
k!
(1 − λh + o(h))
= P(ξ0 = 1|ξ0 = 1)(1 − λh) + P(ξ1 = 1|ξ0 = 1)λh + o(h)
= 1 − λh + αλh + o(h)
= 1 − (1 − α)λh + o(h).
hence, we obtain that X(t) is a two state birth and death process with pa-
rameter λ0 = λ and µ1 = (1 − α)λ. 2
Problem 3.2
Solution: Since we want to model the position of the beetle at time t as a
birth and death process and the mean staying time at position 0 and N are
m0 and mN , respectively, the assumption at the ends 0 and N should be:
P(The beetle move right to 1 in [t, t + h)| the beetle at position 0 at time t)
=
1
m0
h + o(h)
P(The beetle stays at 0 in [t, t + h)| the beetle at position 0 at time t)
= 1 − 1
m0
h + o(h)
6 MATH424 HOMEWORK ANSWER
P(The beetle move left to N − 1 in [t, t + h)| the beetle at position N at time t)
=
1
mN
h + o(h)
P(The beetle stays at N in [t, t + h)| the beetle at position N at time t)
= 1 − 1
mN
h + o(h) 2