随机过程罗斯第三章答案

E[reward in cycle] E[reward in (t, t + a)] → a
E[time of cycle]
2
Stochastic Process
1nÙSK
4&& 2120502082
Assume that any relevant function is directly Riemann integrable. Proof:
g = h + g ∗ F.
Either iterate the above or use Laplace transforms to show that a renewal-type equation has the
solution
t
g(t) = h(t) + h(t − x)dm(x),
0
where m(x) =
0
t
∞
g(t − s)dF (s) + tdF (s)
0
0
t
g(t − s)dF (s) + tF¯(s).
0
Therefore,
lim g(t) =
t→∞
=
∞ 0
tF¯
(t)dt
=
∞ 0
t
∞ 0
dF
(s)dt
µ
µ
∞ 0
t 0
tdtdF (s)
=
∞ 0
s2dF
(s)
=
E [X 2 ] .
µ
2µ
2E [X ]
P (t) = P (on at t)
t
= P (on at t|X1 > t) · F¯(t) + P (on at t|SN(t) = s)dFSN(t) (s)
0
t
= P (on at t|X1 > t) · F¯(t) + P (on at t|X > t − s)F¯(t − s)dm(s)
lim
t→∞
E
[RN
(t)+1]
=
∞ 0
E[R1|X1
=
x]dF (x)
EX
= E[R1 · X1] EX
When the cycle reward is defined to equal the cycle length, the above yields
lim
t→∞
E [XN (t)+1 ]
=
∞ 0
(a) Prove that
t
mD(t) = G(t) + m(t − x)dG(x),
0
where m(t) =
∞ n=1
Fn(t).
(b) Let AD(t) denote the age at time t. Show that if F is nonlattice with
tG¯(t) → 0 as t → 0, then
1
Stochastic Process
1nÙSK
4&& 2120502082
2 Ex3.18
Consider a delayed renewal process {ND(t), t ≥ 0} whose first interarrival has distribution G and the others have distribution F . Let mD(t) = E[ND(t)].
0
Let h(t) = P (on at t|X1 > t) · F¯(t), then
t
P (t) = h(t) + h(t − s)dm(s).
0
Thus, (b)
1∞
lim P (t) =
h(t)dt =
E[Zn] .
t→∞
µF 0
E[Zn] + E[Yn]
g(t) = = =
∞
E[A(t)|X1 = s]dF (s)
E [X 2 ]
lim
t→∞
E [XN (t)+1 ]
=
, E [X ]
which is always greater than E[X] except when X is constant with probability 1. (Why?) Proof:
t
E[RN(t)+1] = E[RN(t)+1|SN(t) = 0]F¯(t) + E[RN(t)+1|SN(t) = s]F¯(t − s)dm(s)
20
t∞
P (In(t) = 1|X1 = y)dG(y)
20 t∞
P (In(t) = 1|X1 = y)dG(y)
02 t∞
Fn(t − y)dG(y)
01 t
m(t − x)dG(x)
0
3 Ex3.19
Prove Blackwell’s theorem for renewal reward processes. That is, assuming that the cycle distribution is not lattice, show that, as t → ∞,
(a) P (t),the probability an alternating renewal process is on at time t;
(b) g(t) = E[A(t)],the expected age of a renewal process at t.
Apply the key renewal theorem to obtain the limiting values in (a) and (b). Solution: (a)
Stochastic Process
1nÙSK
4&& 2120502082
1 Ex3.14
An equation of the form
t
g(t) = h(t) + g(t − x)dF (x)
0
is called a renewal-type equation. In convolution notation the above states that
∞ n=1
Fn(x).
If
h
is
directly
Riemann
integrable
and
F
nonlattice
with
finite
mean,
one
can then apply the key renewal theorem to obtain
lim g(t) =
t→∞
∞ 0
h(t)dt
∞ 0
F¯
(t)dt
0
t
= E[R1|X1 > t]F¯(t) + E[Rn|Xn > t − s]F¯(t − s)dm(s)
0
Let h(t) = E[R1|X1 > t]F¯(t). ThenE[RN(t)+1] = h + h ∗ m. It is easily to know that h(t) is directly Riemann integrable. According to the key renewal theorem,
E [AD (t)]
→
2
∞ 0
x2dF
(x)
∞ 0
xdF
(x)
.
(c) Show that if G has a finite mean, then tG¯(t) → 0 as t → ∞.
x2dF (x) < ∞ and
Proof: (a) Let
Thus,ND(t) =
1, the n-th renewal occurs in[0,t]
E[R1|X1
=
x]dF (x)
=
E [X 2 ] .
EX
Ee V ar(X) = E[X2] − (E[X])2 ≥ 0, i.e.E[X2] ≥ (E[X])2 and therefore
E [X 2 ]
lim
t→∞
E [XN (t)+1 ]
=
E [X ]
≥ E[X].
Renewal-type equations for g(t) are obtained by conditioning on the time at which the process probabilistically starts over. Obtain a renewal-type equation for:
Except when X is constant with probability 1, E[X2] = (E[X])2 and therefore
E [X 2 ]
lim
t→∞
E [XN (t)+1 ]
=
E [X ]
= E[X]
3
4 Ex3.20
For a renewal reward process show that
lim
t→∞
E
[RN
(t)+1
]
=
E[R1 · X1] , E[X1]
Assume the distribution of Xi is nonlattice and that any relevant function is directly Riemann integrable. When the cycle reward is defined to equal the cycle length, the above yields
In(t) = 0,
otherwise
∞ n=1
In(t).
∞
mD(t) = E[ND(t)] = E[In(t)]
n=1
∞
=
P {In(t) = 1}
n=1
∞
= G(t) + P {In(t) = 1}
n=2
∞
P {In(t) = 1} =
n=2
=
=
=
=
∞∞
P (In(t) = 1|X1 = y)dG(y)
E[reward in (t, t + a)] = E[renewals in (t, t + a)] · E[reward in cycle]
E[renewals in (t, t + a)]
= a·
· E[reward in cycle]
a
E[reward in cycle] = a·
E[time of cycle]