Stat 333 Summer 2007
Renewal theory: Practice Problems
1. Below are some hypothetical frst waitingtime distributions
f
n
=
P
(
T
λ
=
n
)whe
re
λ
is
a renewal event oF a stochastic process. In each case determine whether
λ
is transient or
recurrent; iF recurrent, Further determine whether
λ
is positive recurrent or null recurrent by
computing
E
(
T
λ
).
(a)
P
(
T
λ
=
n
)=(2
/
3)
n
,
n
=3
,
4
,
5
...
(b)
P
(
T
λ
=
n
)=(1
/
2)
n
,
n
=1
,
2
,
3
, ...
(c) Let
c
=
∑
∞
n
=4
1
/n
3
/
2
. Using any legitimate arguments you might have From calculus (and
you do have them), determine whether
c
is fnite or infnite. IF
c
is fnite and you defne
P
(
T
λ
=
n
)=1
/
(
cn
3
/
2
)For
n
=4
,
5
,
6
, ...
, determine whether
λ
is transient, positive recurrent,
or null recurrent.
Solution:
(a)
P
(
T
λ
=
n
)=(
2
3
)
n
for k
,
4
,...
P
(
T
λ
<
∞
)=
∞
±
n
=3
(
2
3
)
n
=
(2
/
3)
3
1
−
2
/
3
=
8
9
<
1
,
so
λ
is transient.
(b)
P
(
T
λ
=
n
)=2
−
n
for n
,
2
,
3
P
(
T
λ
<
∞
∞
±
n
=1
P
(
T
λ
=
n
∞
±
n
=1
2
−
n
=
1
/
2
1
/
2
.
So
f
λ
=
P
(
T
λ
<
∞
)=1,then
λ
is recurrent.
E
(
T
λ
∞
±
n
=1
n
2
−
n
=
E
(geometric distribution with
p
/
2) =
1
1
/
2
=2
.
So
λ
is positive recurrent.
(c) ±irstly, we introduce a Famous result For series:
∞
±
n
=4
1
n
p
=
²
∞
iF
p
≤
1
±inite number
<
∞
iF
p>
1
.
±
romtheaboveresu
lt
,weknow
c
is a fnite number. Since
P
(
T
λ
<
∞
³
∞
±
n
=4
1
n
3
/
2
´µ
c
,
then
λ
is recurrent. Note that
E
(
T
λ
c
−
1
∞
±
n
=4
n
·
1
n
(3
/
2)
=
c
−
1
∞
±
n
=4
1
n
(1
/
2)
=
∞
,
then
λ
is null recurrent.
1
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View Full Document2. Suppose
λ
is a renewal event and
V
λ
= total number of occurrences of
λ
in the sequence.
(a) If
E
(
V
λ
) = 7, Fnd the probability that
λ
occurs at least once in the sequence.
Is
λ
recurrent or transient?
(b) If
E
(
V
λ
) = 7, what is the probability of exactly 5 occurrences of
λ
in the sequence? Of
at least 5 occurrences? (Hint: use the probability distribution of
V
λ
.)
(c) ±or a general transient event
λ
, determine the smallest value of
f
λ
which will yield
E
(
V
λ
)
≥
50.
Solution:
(a)
E
(
V
λ
)=
f
λ
1
−
f
λ
=7 (
finite, so transient
)
f
λ
=7
−
7
f
λ
,
so 8
f
λ
.
So
f
λ
/
8
<
1, then
λ
is transient.
(b)
P
(
V
λ
=5)=
f
5
λ
(1
−
f
λ
)=(7
/
8)
5
(1
/
8)
∼
= 0
.
0641
P
(
V
λ
≥
5) =
f
5
λ
=(7
/
8)
5
∼
= 0
.
5129
.
(c)
E
(
V
λ
f
λ
1
−
f
λ
≥
50
⇒
51
f
λ
≥
50
⇒
f
λ
≥
50
51
.
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 Fall '10
 MenZhongxian
 Probability, Probability theory, Recurrence relation, renewal sequence

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