STAT 400 Discussion 07 Solutions Spring 2018 1. Suppose that number of accidents at a construction site follows a Poisson process with the average rate of 0.80 accidents per month. Assume all months are independent of each other. “Hint”: If T a has a Gamma ( a , q = 1 /l ) distribution, where a is an integer, then F T a ( t ) = P ( T a £ t ) = P ( X t ³ a ) and P ( T a > t ) = P ( X t £ a – 1 ), where X t has a Poisson ( l t ) distribution. a) Find the probability that the first accident of a calendar year would occur during March. T 1 has Exponential distribution with l = 0.80 or q = 1 / 0.80 = 1.25. P ( 2 < T 1 < 3 ) = = e – 1.60 – e – 2.40 » 0.1112. OR P ( 2 < T 1 < 3 ) = P ( T 1 > 2 ) – P ( T 1 > 3 ) = P ( X 2 = 0 ) – P ( X 3 = 0 ) = P ( Poisson ( 1.60 ) = 0 ) – P ( Poisson ( 2.40 ) = 0 ) = 0.202 – 0.091 = 0.111. OR = P ( X 2 = 0 ) ´ P ( X 1 ³ 1 ) = 0.202 ´ ( 1 – 0.449 ) » 0.111. ò - 3 2 80 . 0 80 . 0 dt t e ÷ ÷ ø ö ç ç è æ month third the during accident one least at months first two the during accidents no P AND
Transcript
STAT 400 Discussion 07 Solutions Spring 2018
1. Suppose that number of accidents at a construction site follows a Poisson process with the average rate of 0.80 accidents per month. Assume all months are independent of each other.
“Hint”: If T a has a Gamma ( a , q = 1/l ) distribution, where a is an integer, then
F T a ( t ) = P ( T a £ t ) = P ( X t ³ a ) and P ( T a > t ) = P ( X t £ a – 1 ),
where X t has a Poisson ( l t ) distribution.
a) Find the probability that the first accident of a calendar year would occur during March. T 1 has Exponential distribution with l = 0.80 or q = 1/0.80 = 1.25.
P ( 2 < T 1 < 3 ) = = e – 1.60 – e – 2.40 » 0.1112.
OR
P ( 2 < T 1 < 3 ) = P ( T 1 > 2 ) – P ( T 1 > 3 ) = P ( X 2 = 0 ) – P ( X 3 = 0 )
= P ( X 2 = 0 ) ´ P ( X 1 ³ 1 ) = 0.202 ´ ( 1 – 0.449 ) » 0.111.
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0.449 ´ 0.449 ´ 0.551 » 0.111. b) Find the probability that the third accident of a calendar year would occur during April.
T 3 has Gamma distribution with a = 3 and l = 0.80 or q = 1/0.80 = 1.25. P ( 3 < T 3 < 4 ) = P ( T 3 > 3 ) – P ( T 3 > 4 ) = P ( X 3 £ 2 ) – P ( X 4 £ 2 )
c) Find the probability that the third accident of a calendar year would occur during spring (March, April, or May).
T 3 has Gamma distribution with a = 3 and l = 0.80 or q = 1/0.80 = 1.25. P ( 2 < T 3 < 5 ) = P ( T 3 > 2 ) – P ( T 3 > 5 ) = P ( X 2 £ 2 ) – P ( X 5 £ 2 )
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2. As Alex is leaving for college, his parents give him a car, but warn him that they would take the car away if Alex gets 6 speeding tickets. Suppose that Alex receives speeding tickets according to Poisson process with the average rate of one ticket per six months.
X t = number of speeding tickets in t years. Poisson ( l t )
T k = time of the k th speeding ticket. Gamma, a = k. one ticket per six months Þ l = 2.
If T a has a Gamma ( a , q = 1/l ) distribution, where a is an integer, then
P ( T a £ t ) = P ( X t ³ a ) and P ( T a > t ) = P ( X t £ a – 1 ), where X t has a Poisson ( l t ) distribution.
a) Find the probability that it would take Alex longer than two years to get his sixth speeding ticket.
P ( T 6 > 2 ) = P ( X 2 £ 5 ) = P ( Poisson ( 4 ) £ 5 ) = 0.785.
OR
P ( T 6 > 2 ) = = = …
b) Find the probability that it would take Alex less than four years to get his sixth speeding ticket.
P ( T 6 < 4 ) = P ( X 4 ³ 6 ) = P ( X 4 ³ 6 ) = 1 – P ( X 4 £ 5 )
3. Consider two continuous random variables X and Y with joint p.d.f.
f X, Y ( x, y ) = C ( x + 2 y ), 0 < x < 2, 0 < y < 3, zero elsewhere.
a) Sketch the support of ( X , Y ). That is, sketch { 0 < x < 2, 0 < y < 3 }. b) Find the value of C so that f X, Y ( x, y ) is a valid joint p.d.f.
1 =
=
=
= = 24 C. Þ C = .
c) Find the marginal probability density function of X, f X ( x ).
f X ( x ) = = = = , 0 < x < 2.
d) Find the marginal probability density function of Y, f Y ( y ).
f Y ( y ) = =
= = , 0 < y < 3.
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4. Let X and Y have the joint p.d.f. f X , Y ( x, y ) = C x 2 y, 0 < x < 4, 0 < y < , zero elsewhere. a) Sketch the support of ( X , Y ). That is, sketch { 0 < x < 4, 0 < y < }.
b) Find the value of C so that f X, Y ( x, y ) is a valid joint p.d.f.
1 = = = = = 32 C.
Þ C = .
c) Find the marginal probability density function of X, f X ( x ).
f X ( x ) = = = , 0 < x < 4.
x
x
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d) Find the marginal probability density function of Y, f Y ( y ).
f Y ( y ) = = = = , 0 < y < 2.
e) Are X and Y independent? If X and Y are not independent, find Cov ( X, Y ).
f ( x, y ) ≠ f X ( x ) × f Y ( y ). Þ X and Y are NOT independent. The support of ( X, Y ) is NOT a rectangle. Þ X and Y are NOT independent.
E ( X ) = = = = = 3.2.
E ( Y ) = = =
= = » 1.1852.
E ( X Y ) = = =
= = » 3.8788.
Cov ( X, Y ) = E ( X Y ) – E ( X ) × E ( Y ) = = » 0.0862.
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5. Let the joint probability density
function for ( X , Y ) be f ( x, y ) = x + y,
x > 0, y > 0, x + 2 y < 2, zero otherwise.
a) Find the probability P ( Y > X ).
intersection point:
y = x and x + 2 y = 2
x = and y =
P ( Y > X ) = = = .
OR
P ( Y > X ) = = = .
b) Find the marginal p.d.f. of X, f X ( x ).
f X ( x ) = = , 0 < x < 2.
32
32
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c) Find the marginal p.d.f. of Y, f Y ( y ).
f Y ( y ) = = 2 – 2 y, 0 < y < 1.
d)* Are X and Y independent? If not, find Cov ( X, Y ).
The support of ( X, Y ) is NOT a rectangle. Þ X and Y are NOT independent.
OR f X, Y ( x, y ) ≠ f X ( x ) ´ f Y ( y ). Þ X and Y are NOT independent.
E ( X ) = = =
= = = .
E ( Y ) = = =
= = = .
E ( X Y ) = =
= = = .
Cov ( X, Y ) = E ( X Y ) – E ( X ) ´ E ( Y ) = = » – 0.077778.
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6 – 9. Let the joint probability density function for ( X , Y ) be
f ( x, y ) = , 0 < y < 1, y < x < 2, zero otherwise.
Do NOT use a computer. You may only use +, –, ´, ÷, and on a calculator. Show all work. Example:
= =
= = = 1. Þ f ( x, y ) is a valid joint p.d.f.
6. a) Sketch the support of ( X , Y ). That is, sketch { 0 < y < 1, y < x < 2 }.
b) Find the marginal probability density function of X, f X ( x ).
For 0 < x < 1, f X ( x ) = = = .
For 1 < x < 2, f X ( x ) = = = .
3
512 yx
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1
0
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53
Check: + = + = = 1.
c) Find the marginal probability density function of Y, f Y ( y ).
For 0 < y < 1, f Y ( y ) = = = .
Check: = = = 1.
d) Are X and Y independent? Justify your answer. The support of ( X, Y ) is NOT a rectangle. X and Y are NOT independent.
OR Since f ( x, y ) ≠ f X ( x ) × f Y ( y ), X and Y are NOT independent.
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7. Find the probability P ( X > 2 Y ).
a) Set up the double integral(s) over the region that “we want” with the outside integral w.r.t. x and the inside integral w.r.t. y.
b) Set up the double integral(s) over the region that “we want” with the outside integral w.r.t. y and the inside integral w.r.t. x.
c) Set up the double integral(s) over the region that “we do not want” with the outside integral w.r.t. x and the inside integral w.r.t. y.
+
d) Set up the double integral(s) over the region that “we do not want” with the outside integral w.r.t. y and the inside integral w.r.t. x.
dxdyyxx
2
0
2
0 3
512
ò ò ÷÷÷
ø
ö
ççç
è
æ
dydxyxy
1
0
2
2
3
512
ò ò÷÷÷
ø
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ççç
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æ
dxdyyxx
x
1
0 2 3
512
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ø
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è
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x
2
1
1
2 3
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ò ò ÷÷÷
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y
1
0
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512
ò ò÷÷÷
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ççç
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æ
e) Use one of (a) – (d) to find the desired probability.
(a) = = = = 0.40.
(b) = = = .
(c) 1 – –
= 1 – –
= 1 – –
= = .
(d) 1 – = 1 – = 1 –
= 1 – = = .
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2
0
2
0 3
512
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1
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8. Find the probability P ( X + Y < 2 ).
a) Set up the double integral(s) over the region that “we want” with the outside integral w.r.t. x and the inside integral w.r.t. y.
+
b) Set up the double integral(s) over the region that “we want” with the outside integral w.r.t. y and the inside integral w.r.t. x.
c) Set up the double integral(s) over the region that “we do not want” with the outside integral w.r.t. x and the inside integral w.r.t. y.
d) Set up the double integral(s) over the region that “we do not want” with the outside integral w.r.t. y and the inside integral w.r.t. x.
dxdyyxx
1
0 0 3
512
ò ò ÷÷÷
ø
ö
ççç
è
ædxdyyx
x
2
1
2
0 3
512
ò ò ÷÷
ø
ö
çç
è
æ -
dydxyxy
y
1
0
2 3
512
ò ò÷÷÷
ø
ö
ççç
è
æ -
dxdyyxx
2
1
1
2 3
512
ò ò ÷÷
ø
ö
çç
è
æ
-
dydxyxy
1
0
2
2 3
512
ò ò ÷÷÷
ø
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ççç
è
æ
-
e) Use one of (a) – (d) to find the desired probability.
(b) = =
= = = 0.24.
(d) 1 – = 1 –
= 1 – = 1 – = .
(a) +
= + = + = …
(c) 1 – = 1 – = …
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1
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9. Find the probability P ( X Y < 1 ).
a) Set up the double integral(s) over the region that “we want” with the outside integral w.r.t. x and the inside integral w.r.t. y.
+
b) Set up the double integral(s) over the region that “we want” with the outside integral w.r.t. y and the inside integral w.r.t. x.
+
c) Set up the double integral(s) over the region that “we do not want” with the outside integral w.r.t. x and the inside integral w.r.t. y.
d) Set up the double integral(s) over the region that “we do not want” with the outside integral w.r.t. y and the inside integral w.r.t. x.
dxdyyxx
1
0 0 3
512
ò ò ÷÷÷
ø
ö
ççç
è
ædxdyyx
x
2
1
1
0 3
512
ò ò ÷÷÷
ø
ö
ççç
è
æ
dydxyxy
21
0
2 3
512
ò ò÷÷÷
ø
ö
ççç
è
ædydxyx
y
y
1
21
1 3
512
ò ò÷÷÷
ø
ö
ççç
è
æ
dxdyyxx
2
1
1
1 3
512
ò ò ÷÷÷
ø
ö
ççç
è
æ
dydxyxy
1
21
2
1 3
512
ò ò ÷÷÷
ø
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ççç
è
æ
e) Use one of (a) – (d) to find the desired probability.
(a) + = +
= + = = = 0.325.
(b) +
= +
= +
= = = .
(c) 1 – = 1 – = 1 –
= 1 – = 1 – = = .
(d) 1 – = 1 – = 1 –
= 1 – = 1 – = = .
dxdyyxx
1
0 0 3
512
ò ò ÷÷÷
ø
ö
ççç
è
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x
2
1
1
0 3
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101
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1
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10. Let the joint probability density
function for ( X , Y ) be
f ( x , y ) = ,
x > 0, y > 0,
x 2 + ( y + 3 )
2 < 25,
zero elsewhere. a) Find the value of C so that f ( x, y ) is a valid joint p.d.f.
Must have 1 = =
=
= = 6 C.
Þ C = .
b) Find P ( 2 X + Y > 2 ).
=
= = = = .
yxC
( )
ò òúúú
û
ù
êêê
ë
é +-2
0
325
0
2
dydxyxyC ( )[ ]ò +-
2
0 2 325
2 dyyyC
[ ]ò --2
0 32 616
2 dyyyyC
02
4128
2 43 2
úûù
êëé -- yyyC
61
ò ò ÷÷
ø
ö
çç
è
æ-
-1
0
22
0
61 1
dxdyyxx
( )ò --1
0 2 22
121 1 dxxx
( )ò +--1
0 32 2
31 1 dxxxx ÷
øö
çèæ +--
41
32
21
311
3611-
3635
OR
= … = …
OR
= …
c) Find P ( X – 3 Y > 0 ). P ( X – 3 Y > 0 ) = P ( X > 3 Y )
=
= =
= = = » 0.2916667.
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11. Suppose that ( X, Y ) is uniformly distributed over the region defined by
x ³ 0, y ³ 0, x 2 + y
2 £ 1. That is, f ( x, y ) = C, x ³ 0, y ³ 0, x
2 + y 2 £ 1, zero elsewhere.
a) What is the joint probability density function of X and Y ? That is, find the value of C so that f ( x, y ) is a valid joint p.d.f. The area of a circle is p r
2.
Þ The area of the support of
( X, Y ) is .
Þ C = » 1.27324.
b) Find P ( X + Y < 1 ). Since uniform,
= =
» 0.63662.
4π
π4
areatotalareawant
4
21
π π 2
c) Find P ( Y > 2 X ). Since uniform,
=
» 0.295167.
OR
=
» 0.295167.
d)* Are X and Y independent? The support of ( X, Y ) is not a rectangle. Þ X and Y are NOT independent.
( )
2
2arctan1 π-( )
π2arctan21 ´
-
2
21arctan
π÷øö
çèæ
π21arctan2 ÷øö
çèæ´
12. Consider two continuous random variables X and Y with joint p.d.f.
f X, Y ( x, y ) = , y > 1, 0 < x < y, zero elsewhere.
a) Sketch the support of ( X , Y ). That is, sketch { y > 1, 0 < x < y }.
( )
3 2
C
x y+
b) Find the value of C so that f X, Y ( x, y ) is a valid joint p.d.f.
1 = =
= = = = .
Þ C = = 4.5.
c) Find the marginal probability density function of X, f X ( x ). For 0 < x < 1,
f X ( x ) = = = , 0 < x < 1.
For 1 < x < ∞,
f X ( x ) = = = , 1 < x < ∞.
d) Find the marginal probability density function of Y, f Y ( y ).
f Y ( y ) = =
= = , 1 < y < ∞.
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yx03
22
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9
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y
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1
y
e) Find P ( X + Y < 2 ).
P ( X + Y < 2 )
=
=
=
=
= = = 0.375.
f) Find P ( X + Y > 5 ).
P ( X + Y > 5 )
=
+
= +
= – –
= – + – 0 + = 0.25.
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9
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yxx
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54
9
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x ò¥
5.22
4
1 dx
x
( ) 0
5.2
549+x 5.2
41 ¥x
309
209
101
g) Find P ( Y > 3 X ).
P ( Y > 3 X )
=
=
=
= = 0.72.
h)* Are X and Y independent? f X, Y ( x, y ) ≠ f X ( x ) × f Y ( y ). Þ X and Y are NOT independent.
OR
The support of ( X, Y ) is NOT a rectangle. Þ X and Y are NOT independent.
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