Download University of Kansas Math526: Probability and Statistics Instructor

University of Kansas
Instructor: Le Chen
Math526: Probability and Statistics
Class 4172-52503 (2017 Spring)
Homework of Weeks 4 & 5
Due on Feb. 21
3.38. It is helpful to fill the table first.
fX,Y px, yq x “ 0
1
2
3
fY pyq
y“0
0
30
1
30
2
30
3
30
6
30
1
1
30
2
30
3
30
4
30
10
30
2
2
30
3
30
4
30
5
30
14
30
fX pxq
3
30
6
30
9
30
12
30
1
Table 1: The joint distribution function for Ex. 3.38.
(a) P pX ď 2, Y “ 1q “ fX,Y p0, 1q ` fX,Y p1, 1q ` fX,Y p2, 1q “
(b) P pX ą 2, Y ď 1q “ fX,Y p3, 0q ` fX,Y p3, 1q “
3`4
30
“
1`2`3
30
“ 51 .
7
.
30
(c)
P pX ą Y q “ fX,Y p1, 0q ` fX,Y p2, 0q ` fX,Y p3, 0q ` fX,Y p2, 1q ` fX,Y p3, 1q ` fX,Y p3, 2q
1`2`3`3`4`5
“
30
3
“ .
5
(d)P pX ` Y “ 4q “ fX,Y p2, 2q ` fX,Y p3, 1q “
1
4`4
30
“
4
.
15
3.44. (a)
ż 50
ż 50
dx
30
ż 50
2
2
dypx ` y q “
30
ˆ
dx
30
ż 50
“
“
“
“
˙ˇy“50
y 3 ˇˇ
x y`
3 ˇy“30
2
ˆ
503 ´ 303
dx x p50 ´ 30q `
3
30
ˆ 3
˙ˇ50
x
503 ´ 303 ˇˇ
p50 ´ 30q ` x
ˇ
3
3
30
˘
2` 3
50 ´ 303 p50 ´ 30q
3
3920000
.
3
(1)
˙
2
(2)
(3)
(4)
(5)
Hence,
k“
3
.
3920000
(b) By similar calculation as those in (a),
ż 40
P p30 ă X ă 40, 40 ă Y ă 50q “
ż 50
dx
30
dy
40
3
px2 ` y 2 q
3920000
1
“ .
4
(6)
(7)
(c) By similar calculation as those in (a),
ż 40
ż 40
30
“
dy
dx
P p30 ă X ă 40, 30 ă Y ă 40q “
30
3
px2 ` y 2 q
3920000
37
.
196
(8)
(9)
3.49. (a) The marginal distribution of X is
x
fX pxq
1
0.1
2
0.35
3
0.55
3
0.5
5
0.3
(b) The marginal distribution of Y is
y
1
fY pyq 0.2
(d)
P pY “ 3|X “ 2q “
P pY “ 3, X “ 2q
0.1
2
“
“ .
P pX “ 2q
0.35
7
3.56. Note that the domain
0 ă x ă 1,
0ăy ă1´x
(10)
0 ă y ă 1,
0 ă x ă 1 ´ y,
(11)
can be equivalently written as
which is clear from the graph below.
2
y
1
x
o
1
(a) To show that X and Y are not independent, we need to calculate their marginal
distribution functions: If x P p0, 1q, then
ż 1´x
6xdy
(12)
fX pxq “
0
“ 6xp1 ´ xq.
(13)
If x R p0, 1q, then fX pxq “ 0. Hence,
#
6xp1 ´ xq if x P p0, 1q,
fX pxq “
0
otherwise.
(14)
Similarly, if y P p0, 1q, then
ż 1´y
fY pyq “
6xdx
(15)
“ 3p1 ´ yq2 .
(16)
0
If y R p0, 1q, then fY pyq “ 0. Hence,
#
3p1 ´ yq2
fY pyq “
0
if y P p0, 1q,
otherwise.
(17)
Therefore, X and Y are not independent because
fX,Y px, yq ‰ fX pxqfY pyq.
(b) To calculate the conditional probability, we need to find the conditional distribution
function first. For x and y in the right domain (i.e., x and y satisfy the conditions (10)),
fX,Y px, yq
fY pyq
2x
“
.
p1 ´ yq2
fX px|Y “ yq “
(18)
(19)
Hence, if y P p0, 1q,
2x
fX px|Y “ yq “ p1 ´ yq2
%
0
$
&
3
if 0 ă x ă 1 ´ y,
otherwise.
(20)
Therefore, for y P p0, 1q,
ż `8
fX px|Y “ yqdx
P pX ą 0.3|Y “ yq “
(21)
ż0.3
1´y
2x
dx
2
0.3 p1 ´ yq
ˇx“1´y
x2 ˇˇ
“
p1 ´ yq2 ˇ
“
(22)
(23)
x“0.3
“1´
0.09
.
p1 ´ yq2
(24)
In particular, if y “ 0.5, we have
P pX ą 0.3|Y “ 0.5q “ 1 ´
0.09
16
“ .
2
p1 ´ 0.5q
25
4.2. We first calculate all the probability distribution function
ˆ ˙ ˆ ˙0 ˆ ˙3
3
1
3
33
f p0q “
“ 1 ˆ 1 ˆ 3,
0
4
4
4
ˆ ˙ ˆ ˙1 ˆ ˙2
3!
3
1
3
1 32
33
f p1q “
“
ˆ ˆ 2 “ 3,
1
4
4
2!1! 4 4
4
ˆ ˙ ˆ ˙2 ˆ ˙1
1
3
32
3
3!
3
1
ˆ 2 ˆ “ 3,
f p2q “
“
4
4
1!2! 4
4
4
2
ˆ ˙ ˆ ˙3 ˆ ˙0
1
3
3
3!
1
1
f p3q “
“
ˆ 3 ˆ 1 “ 3.
3
4
4
3!0! 4
4
Therefore,
EpXq “ f p0q ˆ 0 ` f p1q ˆ 1 ` f p2q ˆ 2 ` f p3q ˆ 3 “
0 ˆ 33 ` 33 ` 2 ˆ 32 ` 3
3
“ .
3
4
4
4.4. Suppose the probability of tail is p, then the probability of head is 3p. Because
p ` 3p “ 1, we see that p “ 1{4. The sample space is
S “ tpH, Hq, pH, T q, pT, Hq, pT, T qu.
Let X be the number of tails in the two tosses. The two tosses are independent. Hence,
ˆ ˙2
3
fX p0q “ P pX “ 0q “ P ptpH, Hquq “
,
4
fX p1q “ P pX “ 1q “ P ptpT, Hq, pH, T quq “ P ptpT, Hquq ` P ptpH, T quq
1 3 3 1
3
“ ˆ ` ˆ “
4 4 4 4
8
ˆ ˙2
1
fX p2q “ P pX “ 2q “ P ptpT, T quq “ P ptpT, T quq “
.
4
4
Therefore,
ˆ ˙2
ˆ ˙2
3
3
1
1
EpXq “ 0 ˆ
`1ˆ `2ˆ
“ .
4
8
4
2
4.12.
ż8
xf pxqdx
EpXq “
´8
ż1
2xp1 ´ xqdx
ˆ
˙ˇx“1
ˇ
2
1
“ x2 ´ x ˇˇ
“ .
3
3
x“0
“
0
Hence, the average profit per automobile is 5000{3 « 1667 USD.
4.17.
µgpXq “ EpgpXqq “ gp´3qf p´3q ` gp6qf p6q ` gp9qf p9q
1
1
1
“ 25 ˆ ` 169 ˆ ` 361 ˆ “ 209.
6
2
3
4.23. (a)
EpXY 2 q “
ÿ
ÿ
xy 2 f px, yq
x“2,4 y“1,3,5
2
“2 ˆ 1 ˆ 0.10 ` 2 ˆ 32 ˆ 0.20 ` 2 ˆ 52 ˆ 0.10
4 ˆ 12 ˆ 0.15 ` 4 ˆ 32 ˆ 0.30 ` 4 ˆ 52 ˆ 0.15
“35.20.
(b)The marginal law of X and Y are
x
2
fX pxq 0.4
4
0.6
and
y
1
fY pyq 0.25
3
0.5
5
0.25
Hence,
µX “ 2 ˆ 0.4 ` 4 ˆ 0.6 “ 3.2
and
µY “ 1 ˆ 0.25 ` 3 ˆ 0.5 ` 5 ˆ 0.25 “ 3.
5
4.26.
ż1
ż1
dy 4xy
dx
EpZq “
a
x2 ` y 2
ż01
ż01
a
dpx2 ` y 2 q 2x x2 ` y 2
0
0
ˇy“1
ż1
1
px2 ` y 2 q 2 `1 ˇˇ
“
dx 2x
ˇ
ˇ
1 ` 12
0
y“0
ż1
2 32
3
p1 ` x q ´ x
dx 4x
“
3
0
ż1
ż1 4
2 32
4x
2 2p1 ` x q
dp1 ` x q
“
´
dx
3
3
0
0
ˇ1
ˇ1
3
4x5 ˇˇ
2p1 ` x2 q 2 `1 ˇˇ
“
ˇ ´
3 ˆ 5 ˇ0
3p1 ` 32 q ˇ
dx
“
0
9{2
4
´4
´
15
15
29{2 ´ 8
“
.
15
“
2
4.30.
ż8
xf pxqdx
(25)
x ´x{4
e
dx
4
(26)
ErXs “
ż´8
8
“
0
ż8
xde´x{4
“´
(27)
0
ż8
ˇx“8
“ ´xe
´
e´x{4 dp´xq
ˇ
x“0
0
ż8
“0´0`
e´x{4 dx
0
ˇx“8
ˇ
“ ´4e´x{4 ˇ
“ ´p0 ´ 4q “ 4.
´x{4 ˇ
(Integration by parts)
x“0
4.34. We first find the mean of X
µX “ ´2 ˆ 0.3 ` 3 ˆ 0.2 ` 5 ˆ 0.5 “ 2.5.
Then we calculate
EpX 2 q “ p´2q2 ˆ 0.3 ` 32 ˆ 0.2 ` 52 ˆ 0.5 “ 15.5.
Hence, the variance of X is equal to
2
σX
“ EpX 2 q ´ µ2X “ 9.25.
6
(28)
(29)
Finally,
σX “
?
9.25 “ 3.041.
4.57.
1
1
1
11
`6ˆ `9ˆ “ ,
6
2
3
2
1
1
1
93
ErX 2 s “ p´3q2 ˆ ` 62 ˆ ` 92 ˆ “ ,
6
2
3
2
ErXs “ p´3q ˆ
(30)
(31)
Therefore,
Erp2X ` 1q2 s “ Er4X 2 ` 4X ` 1s
“ 4ErX 2 s ` 4ErXs ` 1
93
11
“4ˆ
`4ˆ
` 1 “ 209.
2
2
4.62. Because X and Y are independent, σXY “ 0. Hence,
2
σZ2 “ p´2q2 σX
` 42 σY2 ` 2 ˆ p´2q ˆ 4 ˆ σXY “ 4 ˆ 5 ` 16 ˆ 3 ` 0 “ 68.
4.63.
2
σZ2 “ p´2q2 σX
` 42 σY2 ` 2 ˆ p´2q ˆ 4 ˆ σXY “ 4 ˆ 5 ` 16 ˆ 3 ´ 16 “ 52.
7
(32)
(33)
(34)