Question c)
We have
E
(
Z
) = 1 (see e.g. the mean of an exponential variable page
279 or the distribution summary page 477 or page 480).
1
IMM  DTU
02405 Probability
20031016
BFN/bfn
Solution for review exercise 23 (chapter 4) in Pitman
We introduce
Y
=
M
−
3 such that
Y
has the exponential distribution with mean 2.
Question a)
E
(
M
) =
E
(
Y
+ 3) =
E
(
Y
) + 3 = 5
V ar
(
M
) =
V ar
(
Y
+ 3) =
V ar
(
Y
) = 4
where we have used standard rules for mean and variance see eg. page 249, and
the result page 279 for the variance of the exponential distribution.
Question b)
We get the density
f
M
(
m
) of the random variable
M
is
f
M
(
m
) =
1
2
e
−
1
2
(
m
−
3)
m >
3
.
from the stated assumptions. We can apply the box page 304 to get
f
X
(
x
) =
f
M
(
m
)
d
x
d
m
=
1
2
e
−
1
2
(log (
x
)
−
3)
x
=
e
3
2
2
x
√
x
,
x > e
3
where
X
=
g
(
M
) =
e
M
. Alternatively
F
X
(
x
) =
P
(
X
≤
x
) =
P
(log (
X
)
≤
log (
x
) =
P
(log (
X
)
−
3
≤
log (
x
)
−
3)
=
P
(
Y
≤
log (
x
)
−
3) = 1
−
e
−
(log (
x
)
−
3)
2
= 1
−
e
3
2
√
x
x > e
3
taking derivative we get
f
X
(
x
) =
d
F
X
(
x
)
d
x
==
e
3
2
2
x
√
x
,
x > e
3
Question c)
We do the calculations in terms of the random variables
Y
i
=
M
i
−
3,
M
i
= log (
X
i
).
Here
X
i
denotes the magnitude of the
i
’th earthquake.
From
Example 3 page 317 we know that the minimum
Z
of the
Y
i
’s,
Z
= min (
Y
1
, Y
2
)
is exponentially distributed with mean 1.
P
(
M >
4) =
P
(
Z >
1) =
e
−
1
1
IMM  DTU
02405 Probability
2003111
BFN/bfn
Solution for review exercise 25 (chapter 4) in Pitman
Question a)
We first note that the range of
Y
is 0
< Y
≤
1
2
.
P
(
Y
≤
y
) =
P
U
≤
1
2
P
(
Y
≤
y

U
≤
1
2
+
P
1
2
< U
P
(
Y
≤
y

1
2
< U
= 2
P
(
U
≤
y
)
The density is 2 for 0
< y <
1
2
0 elsewhere.
Question b)
The standard uniform density
f
(
y
) = 1 for 0
< y <
1, 0 elsewhere.
Question c)
E
(
Y
) =
1
2
−
0
2
=
1
4
, V ar
(
Y
) =
(
1
2
−
0
)
2
12
=
1
48
1
IMM  DTU
02405 Probability
20031112
BFN/bfn
Solution for review exercise 26 (chapter 4) in Pitman
Question a)
E
(
W
t
) =
E
(
Xe
tY
)
=
E
(
X
)
E
(
e
tY
)
by the independence of
X
and
Y
.
We find
E
(
e
tY
)
from the definition of the
mean.
E
(
e
tY
)
=
3
2
1
e
ty
·
2d
y
=
2
e
t
t
e
t
2
−
1
Inserting this result and
E
(
X
) = 2 we get
E
(
W
t
) = 2
2
e
t
t
e
t
2
−
1
Alternatively we could derive the joint density of
X
and
Y
to
f
(
x, y
) = 2(2
x
)
3
e
−
2
x
,
0
< x,
0
< y <
1
where we have used that
X
has Gamma (4,2) density, and apply the formula for
E
(
g
(
X, Y
)) page 349.
Question b)
Since
X
and
Y
are independent we find
E
(
W
2
t
)
E
(
W
2
t
) =
E
(
X
2
)
E
(
e
tY
)
2
where
E
(
X
2
) =
V ar
(
X
) + (
E
(
X
))
2
= 5, see eg. page 481. Next we derive
E
(
e
tY
)
2
=
e
2
t
t
(
e
t
−
1
)
and apply the computational formula for the variance page 261
SD
(
W
t
) =
5
e
2
t
t
(
e
t
−
1)
−
2
2
e
t
t
e
t
2
−
1
2
=
1
IMM  DTU
02405 Probability
2003111
BFN/bfn
Solution for review exercise 1 (chapter 5) in Pitman
First apply the definition of conditional probability page 36
P
Y
≥
1
2

Y
≥
X
2
=
P
(
Y
≥
1
2
∩
Y
≥
X
2
)
P
(
Y
≥
X
2
)
The joint density of
X
and
Y
is the product of the marginal densities since
X
and
Y
are independent (page 349). We calculate the denominator using the formula for the
probability of a set
B
page 349
P
(
Y
≥
X
2
) =
1
0
1
x
2
1
·
1
·
d
y
d
x
=
1
0
(1
−
x
2
)d
x
= 1
−
1
3
=
2
3
and the numerator
P
Y
≥
1
2
∩
Y
≥
X
2
=
P
(
Y
≥
X
2
)
−
P
Y <
1
2
∩
Y
≥
X
2
Now for the last term
P
Y <
1
2
∩
Y
≥
X
2
=
1
√
2
0
1
2
x
2
1
·
d
y
d
x
=
1
√
2
0
(
1
2
−
x
2
)d
x
=
1
2
1
√
2
−
1
3
1
2
1
√
2
=
1
3
√
2
Finally we get
P
Y
≥
1
2

Y
≥
X
2
=
2
3
−
1
3
√
2
2
3
= 1
−
√
2
4
1
IMM  DTU
02405 Probability
20031017
BFN/bfn
Solution for review exercise 20 (chapter 5) in Pitman
Question a)
This is example 3 page 317. A rederivation gives us
P
(
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 Spring '13
 Mattingly
 Normal Distribution, Probability, Probability theory, Binomial distribution, pitman