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Probability, Third Edition
By David J Carr & Michael A Gauger
Published by BPP Professional Education
Solution 6.1
Let the losses due to storm, fire, and theft be denoted X1, X2 and X3 respectively.
({
23})
({13})
Prmax 1, , >2.5 =1 Pr max XXX , 2, =2.5
XXX
1 Pr (X=2.5 )(
Pr X=2.5 )
Pr (X=2.5 )
#NAME?
123
The distribution function for a uniform distribution with range [, ]
ab is:
xa

() =
Fx
ba

So, we can calculate the probability as:
{
,, }>2.5 )=1 Pr X=2.5 )
Pr (X=2.5 )
Pr (X=2.5 )
Pr max (
XXX (
123 123
.2.5 .
.2.5 .
Err:509
..
.=1 0.521
Solutions to practice questions h
Chapter 6
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1 =0.479
.3 .
.4 .
Solution 6.2
We have:
+
[]=
ab =1,879
EX
2
( )2
var ()=
ba =507 .ba =
12
X 
507 =78
12
Hence:
ab+=
1,879 =
2 3,758
.
a+(78 +a) =3,758
.
a=1,840
.
b=1,918
b
Solution 6.3
Since we have formulas for [ and EX2], it is convenient to first calculate the second moment:
EX][
2
22
EX.
Probability Solutions to practice questions h
Chapter 6
Sheet1
Page 3
.=var ()+([]=18.75 +2.5
..
X EX )=25
Hence:
2.5 =[]
.
EX =
a1
2
2 .2!
..
.
.
(a1)(a2)
We solve these simultaneous equations to find the parameters:
22 2
25 =EX =
25 EX .a
2 a
[] .2! 1 .
1
Err:520
=
..
2 ([]2
2.5 EX )
(a1)(a2).
..
a2
.a=3, .=5
Now it is easy to calculate the 90th percentile:
a
3
...
.
5
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0.9 =Fx =
1
()
0.9 1 .
.=
.
.
x +.
x +5
.0.9 .
.0.9 .
1
.
.
3
.x =50.1 1 =5.772
0.9 .
.
..
Solution 6.4
If we notice that X follows a Pareto distribution with parameters a=3 and .=2,000 , the expected return is
easily calculated using the general result as:
.
2000
[]==
1000
EX
a13 1
Solution 6.5
We want to calculate the following:
Pr (XX 1.5 )
Pr 1.5 X 2)
2 <<
F 1.5
<n>
(
Page 5
F (2)
()
Pr (X <2
X >1.5 )=
Err:520
Pr (X >1.5 )
Pr (X >1.5 )
1 (
)
F 1.5
Note that the pdf is that of a oneparameter Pareto distribution with a=3 and .=1 , so the cdf is:
()=x3 for x >1
Fx 1
Finally:
3 3
12 
1

1.5
)F ()
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This note was uploaded on 09/08/2008 for the course STT 490 taught by Professor Dudziak during the Spring '08 term at Michigan State University.
 Spring '08
 Dudziak

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