1.
Suppose that people arrive at a bus stop in accordance with a Poisson process with rate
λ
. The bus departs at time
t
. Let
X
denote the total amount of waiting time of the person
who ﬁrst arrives at the bus stop. Let
Y
denote the total amount of waiting time of all those
who get on the bus at time
t
.
a) Find
E
[
X
].
b) Find
E
[
Y
].
c) Find
V ar
(
Y
).
Solution to Problem 1:
The number of passengers who get on the bus at time
t
is given by
N
(
t
). If
N
(
t
) = 0, no
customers arrive before the bus departs and we deﬁne the waiting time to be zero.
a)
Method 1:
Given
N
(
t
) =
n >
0,
S
1
has the same distribution as
U
(1)
, the smallest of
U
1
,...,U
n
, where
U
1
,U
2
,
···
,U
n
are independent uniform (0
,t
) random variables.
E
[
X

N
(
t
) =
n
] =
E
[
t

U
(1)
]
.
Since
P
(
U
(1)
> s
) =
P
(
U
i
> s
for all
i
) =
±
t

s
t
²
n
,
we have the probability density function of
U
(1)
,
f
U
(1)
(
s
) =
n
(
t

s
)
n

1
/t
n
.
Consequently
E
[
t

U
(1)
] =
Z
t
0
(
t

s
)
n
(
t

s
)
n

1
t
n
ds
=
nt
n
+ 1
.
So
E
[
X
] =
∞
X
n
=0
nt
n
+ 1
e

λt
(
λt
)
n
n
!
=
te

λt
(
∞
X
n
=0
(
λt
)
n
n
!

1
λt
∞
X
n
=0
(
λt
)
n
+1
(
n
+ 1)!
)
=
te

λt
(
e
λt

1
λt
(
e
λt

1))
=
t
(1

1
λt
(1

e

λt
))
=
t

1
λ
(1

e

λt
)
.
Method 2:
Let
T
1
be the arrival time of the ﬁrst passenger. We know that
T
1
follows exponential
distribution with parameter
λ
. Then we have
X
=
³
t

T
1
T
1
< t
0
otherwise
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThen
E
[
X
] =
E
[
t

T
1

T
1
< t
]
P
(
T
1
< t
) + 0
=
t
·
P
(
T
1
< t
)

E
[
T
1

T
1
< t
]
P
(
T
1
< t
)
=
t
(1

e

λt
)

Z
t
0
sλe

λs
ds
=
t

1
λ
(1

e

λt
)
.
b) Since, given
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Han
 Probability, Exponential distribution, X1, wA, xa

Click to edit the document details