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Statistics 341
Fall 2008 - Homework Assignment #8 Answers
We will go over some of these problems during the last week of
classes
- Let Y 1 and Y 2 have the joint probability density function given by
f (y 1 , y 2 ) =
{ 6(1 − y 2 ) 0 ≤ y 1 ≤ y 2 ≤ 1 0 elsewhere
(a) Find P (Y 1 ≤ 3 / 4 , Y 2 ≥ 1 /2). The region where Y 1 ≤ 0 .75 and Y 2 ≥ 0 .5 within the support of Y 1 and Y 2 is shown in Figure 1. This region must be split into two parts. The two parts I chose are shown in the graph. You could choose to break this region up in a different manner.
0.0 0.2 0.4 0.6 0.8 1.
0.^ 0.^ 0.^ 0.^ 0.^
First region ∫ (^1)
- 5
∫ (^0). 5 0
6(1 − y 2 )dy 1 dy 2 =
∫ (^1)
- 5
6(1 − y 2 )
( y 1 |^00.^5
) dy 2
=
∫ (^1)
- 5 3(1 − y 2 )dy 2 =
( 3 y 2 − 1. 5 y 22
) |^10. 5 = 1. 5 − 1 .125 = 0. 375
Second region ∫ (^0). 75
- 5
∫ (^1) y 1
6(1 − y 2 )dy 2 dy 1 =
∫ (^0). 75
- 5
( 6 y 2 − 3 y 22 |^1 y 1
) dy 1
=
∫ (^0). 75
- 5
3 − 6 y 1 + 3y 12 dy 1
( 3 y 1 − 3 y^21 + y^31
) |^00 ..^755 = 0. 984375 − 0 .875 = 0. 109375
Adding the two regions together gives 0.375 + 0.109375 = 0.484375. (b) Find the marginal distributions for both Y 1 and Y 2. The marginal distribution of Y 1 is
f 1 (y 1 ) =
∫ (^1) y 1
6(1 − y 2 )dy 2
=
( 6 y 2 − 3 y^22 |^1 y 1
)
= 3 y 12 − 6 y 1 + 3 = 3(1 − y 1 )^2 for 0 ≤ y 1 ≤ 1 and 0 otherwise. The marginal distribution of Y 2 is
f 2 (y 2 ) =
∫ (^) y 2 0
6(1 − y 2 )dy 1 = (6(1 − y 2 )y 1 |y 02 ) = 6 y 2 (1 − y 2 ) for 0 ≤ y 2 ≤ 1 and 0 otherwise. (c) Find the conditional density function of Y 1 when Y 2 = y 2.
f (y 1 |Y 2 = y 2 ) = f (y 1 , y 2 ) f 2 (y 2 ) =^
6(1 − y 2 ) 6 y 2 (1 − y 2 ) =
y 2 for 0 ≤ y 1 ≤ y 2 and 0 elsewhere. (d) Find the conditional density function of Y 2 when Y 1 = y 1.
f (y 2 |Y 1 = y 1 ) = f^ (y^1 , y^2 ) f 1 (y 1 ) = 6(1^ −^ y^2 ) 3(1 − y 1 )^2 = 2(1^ −^ y^2 ) (1 − y 1 )^2 for y 1 ≤ y 2 ≤ 1 and 0 elsewhere. (e) Find P (Y 2 ≥ 3 / 4 |Y 1 = 1/2).
P (Y 2 ≥ 0. 75 |Y 1 = 0.5) =
∫ (^1)
- 75
2(1 − y 2 ) (1 − 0 .5)^2 dy 2
=
∫ (^1)
- 75
8(1 − y 2 )dy 2
=
( 8 y 2 − 4 y^22 |^10. 75
)
= 4 − (8(3/4) − 4(3/4)^2 ) = 0. 25 (f) Are Y 1 and Y 2 independent random variables? Explain your answer. We have found the marginal distributions as f 1 (y 1 ) = 3(1 − y 1 )^2 for 0 ≤ y 1 ≤ 1 f 2 (y 2 ) = 6 y 2 (1 − y 2 ) for 0 ≤ y 2 ≤ 1
E(Y 12 ) = 3
∫ (^1) 0
y 12 − 2 y^31 + y 14 dy 1
= 3
( y^31 3 −^
y^41 2 +^
y 15 5
) |^10
= 3
+^1
)
V (Y 1 ) = E(Y 12 ) − (E(Y 1 ))^2
E(Y 22 ) = 6
∫ (^1) 0 y 23 − y 24 dy 2
= 6
( y^42 4 − y
(^52) 5
) |^10
= 6
4 −^
)
V (Y 2 ) = E(Y 22 ) − (E(Y 2 ))^2
Corr(Y 1 , Y 2 ) = √Cov(Y^1 , Y^2 ) V (Y 1 )V (Y 2 ) = √^0.^025 0 .0375(0.05) = 0. 5774
- An environmental engineer measures the amount (by weight) of particulate pollution in air samples of a certain volume collected over two smokestacks at a coal-operated power plant. One of the stacks is equipped with a cleaning device. Let Y 1 denote the amount of pollutant per sample collected above the stack that has no cleaning device and let Y 2 denote the amount of pollutant per sample collected above the stack that is equipped with the cleaning device. Suppose that the relative frequency behavior of Y 1 and Y 2 can be modeled by
f (y 1 , y 2 ) =
{ 1 0 ≤ y 1 ≤ 2 , 0 ≤ y 2 ≤ 1 , 2 y 2 ≤ y 1 0 elsewhere
(a) Find P (Y 1 ≥ 3 Y 2 ). This is the probability the cleaning device reduces the amount of pollutant by one-third or more. The region where Y 1 ≥ 3 Y 2 within the support of Y 1 and Y 2 is given in Figure 2.
P (Y 1 ≥ 3 Y 2 ) =
∫ (^2) / 3 0
∫ (^2) 3 y 2
dy 1 dy 2
y
y
0.0 0.5 1.0 1.5 2.
0.^ 0.^ 0.^ 0.^ 0.^
∫ (^2) / 3 0
(2 − 3 y 2 )dy 2
=
( 2 y 2 − 3 2 y^22 )|^20 /^3
)
(b) Find the marginal distributions for both Y 1 and Y 2. The marginal distribution for Y 1 is
f 1 (y 1 ) =
∫ (^0). 5 y 1 0 dy 2 = 0. 5 y 1 for 0 ≤ y 1 ≤ 2
The marginal distribution for Y 2 is
f 2 (y 2 ) =
∫ (^2) 2 y 2
dy 1 = 2(1 − y 2 ) for 0 ≤ y 2 ≤ 1
(c) Find the conditional density function of Y 1 when Y 2 = y 2.
f (y 1 |Y 2 = y 2 ) = f (y 1 , y 2 ) f 2 (y 2 ) =^
2(1 − y 2 ) for 2y 2 ≤ y 1 ≤ 2 and 0 elsewhere. (d) Find the conditional density function of Y 2 when Y 1 = y 1.
f (y 2 |Y 1 = y 1 ) = f^ (y^1 , y^2 ) f 1 (y 1 )
- 5 y 1
=^2
y 1 for 0 ≤ y 2 ≤ 0. 5 y 1 and 0 elsewhere. (e) Find the probability that the amount of pollutant in a sample taken above the stack without the cleaner (Y 1 ) is 1.5 or greater given the amount of pollutant in a sample taken above the stack with the cleaner (Y 2 ) is 0.5. We need to use f (y 1 |Y 2 = 0.5) = (^) 2(1−^1 y 2 ) = (^) 2(1−^10 .5) = 1.
P (Y 1 ≥ 1. 5 |Y 2 = 0.5) =
∫ (^2)
- 5
dy 1 = 2 − 1 .5 = 0. 5
)
(h) Find the correlation between Y 1 and Y 2. For the correlation, we need V (Y 1 ) and V (Y 2 ).
E(Y 12 ) =
∫ (^2) 0
y^21 ∗ 0. 5 y 1 dy 1
=
∫ (^2) 0
- 5 y^31 dy 1
= 18 y^41 |^20
= 18 (16 − 0) = 2
V (Y 1 ) = E(Y 12 ) − (E(Y 1 ))^2
) 2
E(Y 22 ) =
∫ (^1) 0
y^22 ∗ 2(1 − y 2 )dy 2
=
∫ (^1) 0
2 y 22 − 2 y 23 dy 2
= 2 3 y^32 − 1 2 y^42 |^10
= 23 − (^12)
=
V (Y 2 ) = E(Y 22 ) − (E(Y 2 ))^2
) 2
Corr(Y 1 , Y 2 ) = √Cov(Y^1 , Y^2 ) V (Y 1 )V (Y 2 ) =
1 √( 18 2 9
) ( (^1) 18
)
= 0. 5
(i) Find the expected value and variance of Y 1 − Y 2 , the amount of pollutant removed by the cleaner.
E(Y 1 − Y 2 ) = E(Y 1 ) − E(Y 2 )
3 −^
V (Y 1 − Y 2 ) = V (Y 1 ) + V (Y 2 ) + 2(1)(−1)Cov(Y 1 , Y 2 )
18 −^2
)
- Let Y 1 , Y 2 ,... , Yn be independent Poisson random variables with means λi for i = 1, 2 ,... , n. Define the mean of these n random variables to be
Y =
∑n i=1 Yi n
Find the expected value and variance of Y.
E(Y ) = E
( ∑^ Y
i n
)
n E
(∑ Yi
)
n
∑ E(Yi)
= 1 n
∑^ n i=
λi
V (Y ) = V
( ∑ Y
i n
)
n^2
V
(∑ Yi
)
n^2
( (^) n ∑ i=
V (Yi) + 2
∑ ∑ Cov(Yi, Yj )
)
n^2
∑^ n i=
V (Yi) since the Yi variables are all independent
n^2
∑^ n i=
λi
