STA3007
Applied probability (0809)
Solution to Assignment 8
Chapter 5
Problem 3.1
If
W
1
> w
1
, W
2
> w
2
, then there is no event occurring in the interval [0
, w
1
], and there
may 0 or 1 event occurring in the interval (
w
1
, w
2
], which imply
X
(
w
1
) = 0
, X
(
w
2
)

X
(
w
1
) = 0 or 1.
It is obvious that
X
(
w
1
) = 0
, X
(
w
2
)

X
(
w
1
) = 0 or 1 also imply
W
1
> w
1
, W
2
> w
2
.
Pr(
W
1
> w
1
, W
2
> w
2
)
=
Pr(
X
(
W
1
) = 0
, X
(
w
2
)

X
(
w
1
) = 0 or 1)
=
Pr(
X
(
W
1
) = 0)Pr(
X
(
w
2
)

X
(
w
1
) = 0 or 1)
=
exp(

λw
1
)[1 +
λ
(
w
2

w
1
)] exp(

λ
(
w
2

w
1
))
=
[1 +
λ
(
w
2

w
1
)] exp(

λw
2
)
.
Since
Pr(
W
1
> w
1
, W
2
> w
2
) =
∞
w
1
∞
w
2
f
(
x
1
, x
2
)
dx
1
dx
2
,
where
f
(
x
1
, x
2
) is the joint density function of
W
1
, W
2
, thus
f
(
w
1
, w
2
) =
d
dw
1
d
dw
2
∞
w
1
∞
w
2
f
(
x
1
, x
2
)
dx
1
dx
2
=
λ
2
exp(

λw
2
)
for 0
< w
1
< w
2
.
Problem 3.3
By changing variables according to
S
0
=
W
1
,
S
1
=
W
2

W
1
,
we have
W
1
=
S
0
,
W
2
=
S
0
+
S
1
,
and the Jacobean matrix of this transformation
∂
(
w
1
, w
2
)
∂
(
s
0
, s
1
)
=
1
1
0
1
Therefore, the joint distribution of the first two sojourn times is
f
(
s
0
, s
1
)
=
f
(
w
1
(
s
0
, s
1
)
, w
2
(
s
0
, s
1
)) det(
∂
(
w
1
, w
2
)
∂
(
s
0
, s
1
)
)
=
λ
2
exp(

λ
(
s
0
+
s
1
))
=
λ
exp(

λs
0
)
λ
exp(

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 Spring '11
 afdsa
 Probability theory, WI, 1 1 wk

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