Let Y have probability density function f Y ( y ) = { 2 ( θ − y ) θ 2 , 0 < y < θ , 0 , elsewhere . a Show that Y has distribution function F Y ( y ) = { 0 , y ≤ 0 , 2 y θ − y 2 θ 2 , 0 < y < θ , 1 , y ≥ θ . b Show that Y / θ is a pivotal quantity. c Use the pivotal quantity from part (b) to find a 90% lower confidence limit for θ .

Question

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Answer 1

Textbook Question

Chapter 8.5, Problem 44E

Let Y have probability density function

f Y ( y ) = { 2 ( θ − y ) θ 2 , 0 < y < θ , 0 , elsewhere .

a Show that Y has distribution function

F Y ( y ) = { 0 , y ≤ 0 , 2 y θ − y 2 θ 2 , 0 < y < θ , 1 , y ≥ θ .

b Show that Y/θ is a pivotal quantity.

c Use the pivotal quantity from part (b) to find a 90% lower confidence limit for θ.

a.

Expert Solution

To determine

Verify that the distribution of Y is FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Explanation of Solution

The distribution function for y≤0 is FY(y)=0.

The distribution functions for 0≤y≤θ and FY(y) is given below:

FY(y)=P(Y≤y)=∫−∞yfY(t)dt=∫0y2(θ−t)θ2dt=2θ2(θ∫0ydt−∫0ytdt)=2θ2(θy−t22|0y)=2θ2(θy−y22)=2yθ−y2θ2

The distribution functions for y≥θ and FY(y) is given below:

FY(y)=P(Y≤y)=∫0θ2(θ−t)θ2dt+∫θy0dt=1

Thus, the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Hence, it is verified.

b.

Expert Solution

To determine

Verify that Yθ is a pivotal quantity.

Explanation of Solution

From Part (a), the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

The distribution function for Yθ is obtained as follows:

FU(y)=P(U≤y)=P(Yθ≤y)=P(Y≤θy)=FY(θy)

It can be written as follows:

FU(y)={0, yθ<02yθθ−(θy)2θ2, 0≤yθ≤θ,1, yθ≥θ. ={0, y<02y−y2, 0≤y≤1,1, y≥1.

From the above expression, it is clear that the distribution function for Yθ does not depend on θ. That is, Yθ is a pivotal quantity.

c.

Expert Solution

To determine

Obtain 90% lower confidence limit for θ.

Answer to Problem 44E

90% lower confidence limit for θ is Y0.68.

Explanation of Solution

From Part (b), the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Since 0.90 lies between 0 and θ, it can be written as follows:

0.90=P(Yθ≤a)=2a−a2a2−2a+0.90=0a=2±4−(4×1×0.90)2=2−4−(4×1×0.90)2, 2+4−(4×1×0.90)2=0.68, 1.32a=0.68 [since a∈(0,1)]0.90=P(Yθ≤0.68)0.90=P(Y0.68≤θ)

Thus, 90% lower confidence limit for θ is Y0.68.

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Answer 2

Textbook Question

Chapter 8.5, Problem 44E

Let Y have probability density function

f Y ( y ) = { 2 ( θ − y ) θ 2 , 0 < y < θ , 0 , elsewhere .

a Show that Y has distribution function

F Y ( y ) = { 0 , y ≤ 0 , 2 y θ − y 2 θ 2 , 0 < y < θ , 1 , y ≥ θ .

b Show that Y/θ is a pivotal quantity.

c Use the pivotal quantity from part (b) to find a 90% lower confidence limit for θ.

a.

Expert Solution

To determine

Verify that the distribution of Y is FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Explanation of Solution

The distribution function for y≤0 is FY(y)=0.

The distribution functions for 0≤y≤θ and FY(y) is given below:

FY(y)=P(Y≤y)=∫−∞yfY(t)dt=∫0y2(θ−t)θ2dt=2θ2(θ∫0ydt−∫0ytdt)=2θ2(θy−t22|0y)=2θ2(θy−y22)=2yθ−y2θ2

The distribution functions for y≥θ and FY(y) is given below:

FY(y)=P(Y≤y)=∫0θ2(θ−t)θ2dt+∫θy0dt=1

Thus, the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Hence, it is verified.

b.

Expert Solution

To determine

Verify that Yθ is a pivotal quantity.

Explanation of Solution

From Part (a), the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

The distribution function for Yθ is obtained as follows:

FU(y)=P(U≤y)=P(Yθ≤y)=P(Y≤θy)=FY(θy)

It can be written as follows:

FU(y)={0, yθ<02yθθ−(θy)2θ2, 0≤yθ≤θ,1, yθ≥θ. ={0, y<02y−y2, 0≤y≤1,1, y≥1.

From the above expression, it is clear that the distribution function for Yθ does not depend on θ. That is, Yθ is a pivotal quantity.

c.

Expert Solution

To determine

Obtain 90% lower confidence limit for θ.

Answer to Problem 44E

90% lower confidence limit for θ is Y0.68.

Explanation of Solution

From Part (b), the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Since 0.90 lies between 0 and θ, it can be written as follows:

0.90=P(Yθ≤a)=2a−a2a2−2a+0.90=0a=2±4−(4×1×0.90)2=2−4−(4×1×0.90)2, 2+4−(4×1×0.90)2=0.68, 1.32a=0.68 [since a∈(0,1)]0.90=P(Yθ≤0.68)0.90=P(Y0.68≤θ)

Thus, 90% lower confidence limit for θ is Y0.68.

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Answer 3

Textbook Question

Chapter 8.5, Problem 44E

Let Y have probability density function

f Y ( y ) = { 2 ( θ − y ) θ 2 , 0 < y < θ , 0 , elsewhere .

a Show that Y has distribution function

F Y ( y ) = { 0 , y ≤ 0 , 2 y θ − y 2 θ 2 , 0 < y < θ , 1 , y ≥ θ .

b Show that Y/θ is a pivotal quantity.

c Use the pivotal quantity from part (b) to find a 90% lower confidence limit for θ.

a.

Expert Solution

To determine

Verify that the distribution of Y is FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Explanation of Solution

The distribution function for y≤0 is FY(y)=0.

The distribution functions for 0≤y≤θ and FY(y) is given below:

FY(y)=P(Y≤y)=∫−∞yfY(t)dt=∫0y2(θ−t)θ2dt=2θ2(θ∫0ydt−∫0ytdt)=2θ2(θy−t22|0y)=2θ2(θy−y22)=2yθ−y2θ2

The distribution functions for y≥θ and FY(y) is given below:

FY(y)=P(Y≤y)=∫0θ2(θ−t)θ2dt+∫θy0dt=1

Thus, the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Hence, it is verified.

b.

Expert Solution

To determine

Verify that Yθ is a pivotal quantity.

Explanation of Solution

From Part (a), the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

The distribution function for Yθ is obtained as follows:

FU(y)=P(U≤y)=P(Yθ≤y)=P(Y≤θy)=FY(θy)

It can be written as follows:

FU(y)={0, yθ<02yθθ−(θy)2θ2, 0≤yθ≤θ,1, yθ≥θ. ={0, y<02y−y2, 0≤y≤1,1, y≥1.

From the above expression, it is clear that the distribution function for Yθ does not depend on θ. That is, Yθ is a pivotal quantity.

c.

Expert Solution

To determine

Obtain 90% lower confidence limit for θ.

Answer to Problem 44E

90% lower confidence limit for θ is Y0.68.

Explanation of Solution

From Part (b), the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Since 0.90 lies between 0 and θ, it can be written as follows:

0.90=P(Yθ≤a)=2a−a2a2−2a+0.90=0a=2±4−(4×1×0.90)2=2−4−(4×1×0.90)2, 2+4−(4×1×0.90)2=0.68, 1.32a=0.68 [since a∈(0,1)]0.90=P(Yθ≤0.68)0.90=P(Y0.68≤θ)

Thus, 90% lower confidence limit for θ is Y0.68.

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Answer 4

Textbook Question

Chapter 8.5, Problem 44E

Let Y have probability density function

f Y ( y ) = { 2 ( θ − y ) θ 2 , 0 < y < θ , 0 , elsewhere .

a Show that Y has distribution function

F Y ( y ) = { 0 , y ≤ 0 , 2 y θ − y 2 θ 2 , 0 < y < θ , 1 , y ≥ θ .

b Show that Y/θ is a pivotal quantity.

c Use the pivotal quantity from part (b) to find a 90% lower confidence limit for θ.

a.

Expert Solution

To determine

Verify that the distribution of Y is FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Explanation of Solution

The distribution function for y≤0 is FY(y)=0.

The distribution functions for 0≤y≤θ and FY(y) is given below:

FY(y)=P(Y≤y)=∫−∞yfY(t)dt=∫0y2(θ−t)θ2dt=2θ2(θ∫0ydt−∫0ytdt)=2θ2(θy−t22|0y)=2θ2(θy−y22)=2yθ−y2θ2

The distribution functions for y≥θ and FY(y) is given below:

FY(y)=P(Y≤y)=∫0θ2(θ−t)θ2dt+∫θy0dt=1

Thus, the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Hence, it is verified.

b.

Expert Solution

To determine

Verify that Yθ is a pivotal quantity.

Explanation of Solution

From Part (a), the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

The distribution function for Yθ is obtained as follows:

FU(y)=P(U≤y)=P(Yθ≤y)=P(Y≤θy)=FY(θy)

It can be written as follows:

FU(y)={0, yθ<02yθθ−(θy)2θ2, 0≤yθ≤θ,1, yθ≥θ. ={0, y<02y−y2, 0≤y≤1,1, y≥1.

From the above expression, it is clear that the distribution function for Yθ does not depend on θ. That is, Yθ is a pivotal quantity.

c.

Expert Solution

To determine

Obtain 90% lower confidence limit for θ.

Answer to Problem 44E

90% lower confidence limit for θ is Y0.68.

Explanation of Solution

From Part (b), the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Since 0.90 lies between 0 and θ, it can be written as follows:

0.90=P(Yθ≤a)=2a−a2a2−2a+0.90=0a=2±4−(4×1×0.90)2=2−4−(4×1×0.90)2, 2+4−(4×1×0.90)2=0.68, 1.32a=0.68 [since a∈(0,1)]0.90=P(Yθ≤0.68)0.90=P(Y0.68≤θ)

Thus, 90% lower confidence limit for θ is Y0.68.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

Verify that the distribution of Y is FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Explanation of Solution

The distribution function for y≤0 is FY(y)=0.

The distribution functions for 0≤y≤θ and FY(y) is given below:

FY(y)=P(Y≤y)=∫−∞yfY(t)dt=∫0y2(θ−t)θ2dt=2θ2(θ∫0ydt−∫0ytdt)=2θ2(θy−t22|0y)=2θ2(θy−y22)=2yθ−y2θ2

The distribution functions for y≥θ and FY(y) is given below:

FY(y)=P(Y≤y)=∫0θ2(θ−t)θ2dt+∫θy0dt=1

Thus, the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Hence, it is verified.

b.

Expert Solution

To determine

Verify that Yθ is a pivotal quantity.

Explanation of Solution

From Part (a), the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

The distribution function for Yθ is obtained as follows:

FU(y)=P(U≤y)=P(Yθ≤y)=P(Y≤θy)=FY(θy)

It can be written as follows:

FU(y)={0, yθ<02yθθ−(θy)2θ2, 0≤yθ≤θ,1, yθ≥θ. ={0, y<02y−y2, 0≤y≤1,1, y≥1.

From the above expression, it is clear that the distribution function for Yθ does not depend on θ. That is, Yθ is a pivotal quantity.

c.

Expert Solution

To determine

Obtain 90% lower confidence limit for θ.

Answer to Problem 44E

90% lower confidence limit for θ is Y0.68.

Explanation of Solution

From Part (b), the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Since 0.90 lies between 0 and θ, it can be written as follows:

0.90=P(Yθ≤a)=2a−a2a2−2a+0.90=0a=2±4−(4×1×0.90)2=2−4−(4×1×0.90)2, 2+4−(4×1×0.90)2=0.68, 1.32a=0.68 [since a∈(0,1)]0.90=P(Yθ≤0.68)0.90=P(Y0.68≤θ)

Thus, 90% lower confidence limit for θ is Y0.68.

Answer 8

a.

Expert Solution

To determine

Verify that the distribution of Y is FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Explanation of Solution

The distribution function for y≤0 is FY(y)=0.

The distribution functions for 0≤y≤θ and FY(y) is given below:

FY(y)=P(Y≤y)=∫−∞yfY(t)dt=∫0y2(θ−t)θ2dt=2θ2(θ∫0ydt−∫0ytdt)=2θ2(θy−t22|0y)=2θ2(θy−y22)=2yθ−y2θ2

The distribution functions for y≥θ and FY(y) is given below:

FY(y)=P(Y≤y)=∫0θ2(θ−t)θ2dt+∫θy0dt=1

Thus, the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Hence, it is verified.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Verify that the distribution of Y is FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Answer 13

Explanation of Solution

The distribution function for y≤0 is FY(y)=0.

The distribution functions for 0≤y≤θ and FY(y) is given below:

FY(y)=P(Y≤y)=∫−∞yfY(t)dt=∫0y2(θ−t)θ2dt=2θ2(θ∫0ydt−∫0ytdt)=2θ2(θy−t22|0y)=2θ2(θy−y22)=2yθ−y2θ2

The distribution functions for y≥θ and FY(y) is given below:

FY(y)=P(Y≤y)=∫0θ2(θ−t)θ2dt+∫θy0dt=1

Thus, the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Hence, it is verified.

Answer 14

b.

Expert Solution

To determine

Verify that Yθ is a pivotal quantity.

Explanation of Solution

From Part (a), the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

The distribution function for Yθ is obtained as follows:

FU(y)=P(U≤y)=P(Yθ≤y)=P(Y≤θy)=FY(θy)

It can be written as follows:

FU(y)={0, yθ<02yθθ−(θy)2θ2, 0≤yθ≤θ,1, yθ≥θ. ={0, y<02y−y2, 0≤y≤1,1, y≥1.

From the above expression, it is clear that the distribution function for Yθ does not depend on θ. That is, Yθ is a pivotal quantity.

Answer 15

b.

Expert Solution

Answer 16

b.

Expert Solution

Answer 17

Expert Solution

Answer 18

To determine

Verify that Yθ is a pivotal quantity.

Answer 19

Explanation of Solution

From Part (a), the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

The distribution function for Yθ is obtained as follows:

FU(y)=P(U≤y)=P(Yθ≤y)=P(Y≤θy)=FY(θy)

It can be written as follows:

FU(y)={0, yθ<02yθθ−(θy)2θ2, 0≤yθ≤θ,1, yθ≥θ. ={0, y<02y−y2, 0≤y≤1,1, y≥1.

From the above expression, it is clear that the distribution function for Yθ does not depend on θ. That is, Yθ is a pivotal quantity.

Answer 20

c.

Expert Solution

To determine

Obtain 90% lower confidence limit for θ.

Answer to Problem 44E

90% lower confidence limit for θ is Y0.68.

Explanation of Solution

From Part (b), the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Since 0.90 lies between 0 and θ, it can be written as follows:

0.90=P(Yθ≤a)=2a−a2a2−2a+0.90=0a=2±4−(4×1×0.90)2=2−4−(4×1×0.90)2, 2+4−(4×1×0.90)2=0.68, 1.32a=0.68 [since a∈(0,1)]0.90=P(Yθ≤0.68)0.90=P(Y0.68≤θ)

Thus, 90% lower confidence limit for θ is Y0.68.

Answer 21

c.

Expert Solution

Answer 22

c.

Expert Solution

Answer 23

Expert Solution

Answer 24

To determine

Obtain 90% lower confidence limit for θ.

Answer 25

Answer to Problem 44E

90% lower confidence limit for θ is Y0.68.

Answer 26

Explanation of Solution

From Part (b), the distribution of Y is obtained as FY(y)={0, y<02yθ−y2θ2, 0≤y≤θ1, y≥θ. .

Since 0.90 lies between 0 and θ, it can be written as follows:

0.90=P(Yθ≤a)=2a−a2a2−2a+0.90=0a=2±4−(4×1×0.90)2=2−4−(4×1×0.90)2, 2+4−(4×1×0.90)2=0.68, 1.32a=0.68 [since a∈(0,1)]0.90=P(Yθ≤0.68)0.90=P(Y0.68≤θ)

Thus, 90% lower confidence limit for θ is Y0.68.

Answer 27

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