HMWK 4 Solutions
Problems from the Text
Ch. 5, Page 346: 4, 37, 44 (Hint: Wald’s Equation)
4. (a) Here, A and B leave together after 10 minutes, and then C enters service. Therefore
the answer is zero.
(b) The only way A can still be in service after B and C leave is if A’s service time is 3,
and B and C’s both are 1. The probability of this event is (1
/
3)
3
= 1
/
27
.
(c) First of all, A’s service time
S
A
must be greater than B’s ,
S
B
, and this happens with
probability 1
/
2 =
P
(
S
A
> S
B
), and then this must be followed by A’s remaining
service time being greater than C’s service time, which by the memoryless property
also has probability 1
/
2, and is independent of the past. Therefore (by independence)
the answer is the product, 1
/
4
.
37. She is uninjured during time of length
s
if and only if
N
(
s
) = 0;
P
(
N
(
s
) = 0) =
e

λs
,
with
λ
= 3. For
s
=
2
60
,
5
60
,
10
60
,
20
60
minutes, we see that
e

3
s
=
e

1
10
,e

1
4
,e

1
2
,e

1
44. (METHOD I:) Let
X
n
denote the iid exponential
λ
interarrival times for the Poisson
process. Let
τ
= min
{
n
≥
1 :
X
n
> T
}
, a stopping time. Then the waiting time
W
can
be written as
W
=
τ

1
X
n
=1
X
n
=
±
τ
X
n
=1
X
n
²

X
τ
,
where
X
τ
has the conditional distribution of (
X

X > T
).
Clearly,
τ
has a geometric distribution with “success” probability
p
=
P
(
X > T
) =
e

λT
;
and
E
(
τ
) = 1
/p
=
e
λT
.
W
= 0 if and only if
τ
= 1;
P
(
W
= 0) =
p
=
e

λT
. From Wald’s
equation
E
(
W
) =
E
(
τ
)
E
(
X
)

E
(
X

X > T
) =
e