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MATH424 HOMEWORK ANSWER
Exercise 3.3
Proof:
P(V(t) = 1) = P(V(t) = 1|V(0) = 0)P(V(0) = 0) + P(V(t) = 1|V(0) = 1)P(V(0) = 1)
= p01(t)(1 − pi) + p11(t)pi
= (pi − pie−τt)(1 − pi) + (pi + (1 − pi)e−τt)pi
= pi 2
Problem 3.3
Proof: For 0 < s < t,
E[V(s)V(t)] = E[V(s)V(t)|V(s) = 0]P(V(s) = 0) + E[V(s)V(t)|V(s) = 1]P(V(s) = 1)
= E[0 · V(t)|V(s) = 0]P(V(s) = 0) + E[1 · V(t)|V(s) = 1]P(V(s) = 1)
= pip11(t − s)
= pi − pip10(t − s)
hence,
Cov[V(s)V(t)] = E[V(s)V(t)] − E[V(s)]E[V(t)]
= pi − pip10(t − s) − pi2
= pi(1 − pi) − pi(1 − pi)(1 − e−(α+β)(t−s))
= pi(1 − pi)e−(α+β)(t−s) 2
Problem 3.4
Solution: Let pi =
α
α + β
and τ = α + β. We have
E[S (t)] =
∫ t
0
E[V(u)]du
=
∫ t
0
p01(u)du
=
∫ t
0
(pi − pie−τu)du
= pit − pi
τ
(1 − e−τt) 2
1
2 MATH424 HOMEWORK ANSWER
Exercise 4.2
Solution: We have θ0 = 1 and for 1 ≤ n ≤ N,
θn =
λ0λ1 · · · λn−1
µ1µ2 · · · µn =
αnN(N − 1) · · · (N − n + 1)
βnn!
=
(
N
n
) (
α
β
)n
.
therefore, we have
pin =
θn∑N
m=0 θm
=
θn∑N
m=0
(
N
m
) (
α
β
)m =
(
N
n
) (
α
β
)n(
β+α
β
)N = (Nn
) (
α
α + β
)n (
β
α + β
)N−n
2
Problem 4.3
Solution: Since the operating time for each machine is i.i.d. exponentially
distributed and the repair time is also exponentially distributed, this prob-
lem can be modeled as a birth-death process with state space {0, 1, 2, 3, 4, 5}
where state k (0 ≤ k ≤ 5) means that k machine is broken at that moment.
Since the operating time for each machine is exponentially distributed with
parameter 0.2, we know under the condition that no machine is broken
right now, the first time for a machine breaks is an exponential time with
parameter 0.2 × 5 = 1, i.e. λ0 = 1. Similar argument leads to the fact that
λk = 0.2(5 − k). As to µk, they are all equal to 0.5. Hence, we have the
infinitesmal generator as

0 1 2 3 4 5
0 −1 1
1 0.5 −1.3 0.8
2 0.5 −1.1 0.6
3 0.5 −0.9 0.4
4 0.5 −0.7 0.2
5 0.5 −0.5

Therefore, θ0 = 1, θ1 = 2, θ2 = θ1 · λ1
µ2
=
16
5
, θ3 = θ2 · λ2
µ3
=
96
25
,
θ4 = θ3 · λ3
µ4
=
384
125
, θ5 = θ4 · λ4
µ5
=
768
625
.
Hence we have ‘The fraction of time which the repairman is idle’= pi0 =
θ0
θ0 + θ1 + · · · + θ5 � 0.0697 2