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