Prove the following; for every n ≥ 1 and ε > 0. If ε > θ , P ( θ − ε ≤ Y ( n ) ≤ θ + ε ) = 1. If ε < θ , P ( θ − ε ≤ Y ( n ) ≤ θ + ε ) = 1 − [ θ − ε θ ] n .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 9.3, Problem 26E

a.

To determine

Prove the following; for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.

If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

a.

Expert Solution

Answer to Problem 26E

It is proved that for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.
If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

Explanation of Solution

It is given that Y1,Y2,...,Yn denote a random sample from the uniform distribution on the interval (0,θ).

Consider Y(n)=max(Y1,Y2,...,Yn) the cumulative distribution function of Y(n) is given below:

F(n)(y)={0 ; y<θ(yθ)n ; 0≤y≤θ1 ; y>θ

Now, consider n≥1 and ε>0.

P(|Y(n)−θ| ≤ε)=P(θ−ε≤Y(n)≤θ+ε)=F(n)(θ+ε)−F(n)(θ−ε)

If ε>θ then, F(n)(θ+ε)=1 and F(n)(θ−ε)=0. Therefore,

P(|Y(n)−θ| ≤ε)=1−0=1

If ε<θ then, F(n)(θ+ε)=1 and F(n)(θ−ε)=(θ−εθ)n. Therefore,

P(|Y(n)−θ| ≤ε)=1−(θ−εθ)n.

Thus, for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.
If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

b.

To determine

Prove that Y(n) is a consistent estimator of θ by showing that for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

b.

Expert Solution

Answer to Problem 26E

It is proved that Y(n) is a consistent estimator of θ.

Explanation of Solution

From Part (a), if ε>θ, then

P(|Y(n)−θ| ≤ε)=1limn→∞P(|Y(n)−θ| ≤ε)=limn→∞(1)=1

From Part (a), if ε<θ, then

P(|Y(n)−θ| ≤ε)=1−[θ−εθ]nlimn→∞P(|Y(n)−θ| ≤ε)=limn→∞(1−[θ−εθ]n)=1−limn→∞([θ−εθ]n)

Since ε>θ, therefore, 0<θ−εθ<1.

Hence, limn→∞([θ−εθ]n)=0.

Therefore,

limn→∞P(|Y(n)−θ| ≤ε)=1−0=1.

Hence, for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

Thus, Y(n) is a consistent estimator of θ.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

. Let X be a finite random variable on (2, F,P) (i.e., P(X = ∞) = P(X = -x) = 0). Show that its distribution function Fx satisfies the following properties: (a) Fx is non-decreasing. (b) lim,- Fx(x) = 0, and lim, +∞ Fx(x) = 1.

Let x1 ~ N(0,1), X2 ~N(-1,4) and X3~N(1,9) be independent. Derive the distribution of Y=x1-x2-2x3 and compute P(-3<y<1)

Suppose that the random variables X1,...,Xn form a random sample of size n from the uniform distribution on the interval [0, 1]. Let Y1 = min{X1,. . .,Xn}, and let Yn = max{X1,...,Xn}. Find E(Y1) and E(Yn).

Answer 2

Question

Chapter 9.3, Problem 26E

a.

To determine

Prove the following; for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.

If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

a.

Expert Solution

Answer to Problem 26E

It is proved that for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.
If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

Explanation of Solution

It is given that Y1,Y2,...,Yn denote a random sample from the uniform distribution on the interval (0,θ).

Consider Y(n)=max(Y1,Y2,...,Yn) the cumulative distribution function of Y(n) is given below:

F(n)(y)={0 ; y<θ(yθ)n ; 0≤y≤θ1 ; y>θ

Now, consider n≥1 and ε>0.

P(|Y(n)−θ| ≤ε)=P(θ−ε≤Y(n)≤θ+ε)=F(n)(θ+ε)−F(n)(θ−ε)

If ε>θ then, F(n)(θ+ε)=1 and F(n)(θ−ε)=0. Therefore,

P(|Y(n)−θ| ≤ε)=1−0=1

If ε<θ then, F(n)(θ+ε)=1 and F(n)(θ−ε)=(θ−εθ)n. Therefore,

P(|Y(n)−θ| ≤ε)=1−(θ−εθ)n.

Thus, for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.
If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

b.

To determine

Prove that Y(n) is a consistent estimator of θ by showing that for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

b.

Expert Solution

Answer to Problem 26E

It is proved that Y(n) is a consistent estimator of θ.

Explanation of Solution

From Part (a), if ε>θ, then

P(|Y(n)−θ| ≤ε)=1limn→∞P(|Y(n)−θ| ≤ε)=limn→∞(1)=1

From Part (a), if ε<θ, then

P(|Y(n)−θ| ≤ε)=1−[θ−εθ]nlimn→∞P(|Y(n)−θ| ≤ε)=limn→∞(1−[θ−εθ]n)=1−limn→∞([θ−εθ]n)

Since ε>θ, therefore, 0<θ−εθ<1.

Hence, limn→∞([θ−εθ]n)=0.

Therefore,

limn→∞P(|Y(n)−θ| ≤ε)=1−0=1.

Hence, for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

Thus, Y(n) is a consistent estimator of θ.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 9.3, Problem 26E

a.

To determine

Prove the following; for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.

If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

a.

Expert Solution

Answer to Problem 26E

It is proved that for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.
If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

Explanation of Solution

It is given that Y1,Y2,...,Yn denote a random sample from the uniform distribution on the interval (0,θ).

Consider Y(n)=max(Y1,Y2,...,Yn) the cumulative distribution function of Y(n) is given below:

F(n)(y)={0 ; y<θ(yθ)n ; 0≤y≤θ1 ; y>θ

Now, consider n≥1 and ε>0.

P(|Y(n)−θ| ≤ε)=P(θ−ε≤Y(n)≤θ+ε)=F(n)(θ+ε)−F(n)(θ−ε)

If ε>θ then, F(n)(θ+ε)=1 and F(n)(θ−ε)=0. Therefore,

P(|Y(n)−θ| ≤ε)=1−0=1

If ε<θ then, F(n)(θ+ε)=1 and F(n)(θ−ε)=(θ−εθ)n. Therefore,

P(|Y(n)−θ| ≤ε)=1−(θ−εθ)n.

Thus, for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.
If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

b.

To determine

Prove that Y(n) is a consistent estimator of θ by showing that for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

b.

Expert Solution

Answer to Problem 26E

It is proved that Y(n) is a consistent estimator of θ.

Explanation of Solution

From Part (a), if ε>θ, then

P(|Y(n)−θ| ≤ε)=1limn→∞P(|Y(n)−θ| ≤ε)=limn→∞(1)=1

From Part (a), if ε<θ, then

P(|Y(n)−θ| ≤ε)=1−[θ−εθ]nlimn→∞P(|Y(n)−θ| ≤ε)=limn→∞(1−[θ−εθ]n)=1−limn→∞([θ−εθ]n)

Since ε>θ, therefore, 0<θ−εθ<1.

Hence, limn→∞([θ−εθ]n)=0.

Therefore,

limn→∞P(|Y(n)−θ| ≤ε)=1−0=1.

Hence, for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

Thus, Y(n) is a consistent estimator of θ.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 9.3, Problem 26E

a.

To determine

Prove the following; for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.

If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

a.

Expert Solution

Answer to Problem 26E

It is proved that for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.
If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

Explanation of Solution

It is given that Y1,Y2,...,Yn denote a random sample from the uniform distribution on the interval (0,θ).

Consider Y(n)=max(Y1,Y2,...,Yn) the cumulative distribution function of Y(n) is given below:

F(n)(y)={0 ; y<θ(yθ)n ; 0≤y≤θ1 ; y>θ

Now, consider n≥1 and ε>0.

P(|Y(n)−θ| ≤ε)=P(θ−ε≤Y(n)≤θ+ε)=F(n)(θ+ε)−F(n)(θ−ε)

If ε>θ then, F(n)(θ+ε)=1 and F(n)(θ−ε)=0. Therefore,

P(|Y(n)−θ| ≤ε)=1−0=1

If ε<θ then, F(n)(θ+ε)=1 and F(n)(θ−ε)=(θ−εθ)n. Therefore,

P(|Y(n)−θ| ≤ε)=1−(θ−εθ)n.

Thus, for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.
If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

b.

To determine

Prove that Y(n) is a consistent estimator of θ by showing that for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

b.

Expert Solution

Answer to Problem 26E

It is proved that Y(n) is a consistent estimator of θ.

Explanation of Solution

From Part (a), if ε>θ, then

P(|Y(n)−θ| ≤ε)=1limn→∞P(|Y(n)−θ| ≤ε)=limn→∞(1)=1

From Part (a), if ε<θ, then

P(|Y(n)−θ| ≤ε)=1−[θ−εθ]nlimn→∞P(|Y(n)−θ| ≤ε)=limn→∞(1−[θ−εθ]n)=1−limn→∞([θ−εθ]n)

Since ε>θ, therefore, 0<θ−εθ<1.

Hence, limn→∞([θ−εθ]n)=0.

Therefore,

limn→∞P(|Y(n)−θ| ≤ε)=1−0=1.

Hence, for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

Thus, Y(n) is a consistent estimator of θ.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 9.3, Problem 26E

a.

To determine

Prove the following; for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.

If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

a.

Expert Solution

Answer to Problem 26E

It is proved that for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.
If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

Explanation of Solution

It is given that Y1,Y2,...,Yn denote a random sample from the uniform distribution on the interval (0,θ).

Consider Y(n)=max(Y1,Y2,...,Yn) the cumulative distribution function of Y(n) is given below:

F(n)(y)={0 ; y<θ(yθ)n ; 0≤y≤θ1 ; y>θ

Now, consider n≥1 and ε>0.

P(|Y(n)−θ| ≤ε)=P(θ−ε≤Y(n)≤θ+ε)=F(n)(θ+ε)−F(n)(θ−ε)

If ε>θ then, F(n)(θ+ε)=1 and F(n)(θ−ε)=0. Therefore,

P(|Y(n)−θ| ≤ε)=1−0=1

If ε<θ then, F(n)(θ+ε)=1 and F(n)(θ−ε)=(θ−εθ)n. Therefore,

P(|Y(n)−θ| ≤ε)=1−(θ−εθ)n.

Thus, for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.
If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

b.

To determine

Prove that Y(n) is a consistent estimator of θ by showing that for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

b.

Expert Solution

Answer to Problem 26E

It is proved that Y(n) is a consistent estimator of θ.

Explanation of Solution

From Part (a), if ε>θ, then

P(|Y(n)−θ| ≤ε)=1limn→∞P(|Y(n)−θ| ≤ε)=limn→∞(1)=1

From Part (a), if ε<θ, then

P(|Y(n)−θ| ≤ε)=1−[θ−εθ]nlimn→∞P(|Y(n)−θ| ≤ε)=limn→∞(1−[θ−εθ]n)=1−limn→∞([θ−εθ]n)

Since ε>θ, therefore, 0<θ−εθ<1.

Hence, limn→∞([θ−εθ]n)=0.

Therefore,

limn→∞P(|Y(n)−θ| ≤ε)=1−0=1.

Hence, for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

Thus, Y(n) is a consistent estimator of θ.

Answer 6

Chapter 9.3, Problem 26E

a.

To determine

Prove the following; for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.

If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

a.

Expert Solution

Answer to Problem 26E

It is proved that for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.
If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

Explanation of Solution

It is given that Y1,Y2,...,Yn denote a random sample from the uniform distribution on the interval (0,θ).

Consider Y(n)=max(Y1,Y2,...,Yn) the cumulative distribution function of Y(n) is given below:

F(n)(y)={0 ; y<θ(yθ)n ; 0≤y≤θ1 ; y>θ

Now, consider n≥1 and ε>0.

P(|Y(n)−θ| ≤ε)=P(θ−ε≤Y(n)≤θ+ε)=F(n)(θ+ε)−F(n)(θ−ε)

If ε>θ then, F(n)(θ+ε)=1 and F(n)(θ−ε)=0. Therefore,

P(|Y(n)−θ| ≤ε)=1−0=1

If ε<θ then, F(n)(θ+ε)=1 and F(n)(θ−ε)=(θ−εθ)n. Therefore,

P(|Y(n)−θ| ≤ε)=1−(θ−εθ)n.

Thus, for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.
If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

Answer 7

Chapter 9.3, Problem 26E

a.

To determine

Prove the following; for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.

If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

Answer 8

a.

Expert Solution

Answer to Problem 26E

It is proved that for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.
If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

It is given that Y1,Y2,...,Yn denote a random sample from the uniform distribution on the interval (0,θ).

Consider Y(n)=max(Y1,Y2,...,Yn) the cumulative distribution function of Y(n) is given below:

F(n)(y)={0 ; y<θ(yθ)n ; 0≤y≤θ1 ; y>θ

Now, consider n≥1 and ε>0.

P(|Y(n)−θ| ≤ε)=P(θ−ε≤Y(n)≤θ+ε)=F(n)(θ+ε)−F(n)(θ−ε)

If ε>θ then, F(n)(θ+ε)=1 and F(n)(θ−ε)=0. Therefore,

P(|Y(n)−θ| ≤ε)=1−0=1

If ε<θ then, F(n)(θ+ε)=1 and F(n)(θ−ε)=(θ−εθ)n. Therefore,

P(|Y(n)−θ| ≤ε)=1−(θ−εθ)n.

Thus, for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.
If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

Answer 13

b.

To determine

Prove that Y(n) is a consistent estimator of θ by showing that for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

b.

Expert Solution

Answer to Problem 26E

It is proved that Y(n) is a consistent estimator of θ.

Explanation of Solution

From Part (a), if ε>θ, then

P(|Y(n)−θ| ≤ε)=1limn→∞P(|Y(n)−θ| ≤ε)=limn→∞(1)=1

From Part (a), if ε<θ, then

P(|Y(n)−θ| ≤ε)=1−[θ−εθ]nlimn→∞P(|Y(n)−θ| ≤ε)=limn→∞(1−[θ−εθ]n)=1−limn→∞([θ−εθ]n)

Since ε>θ, therefore, 0<θ−εθ<1.

Hence, limn→∞([θ−εθ]n)=0.

Therefore,

limn→∞P(|Y(n)−θ| ≤ε)=1−0=1.

Hence, for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

Thus, Y(n) is a consistent estimator of θ.

Answer 14

b.

To determine

Prove that Y(n) is a consistent estimator of θ by showing that for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

Answer 15

b.

Expert Solution

Answer to Problem 26E

It is proved that Y(n) is a consistent estimator of θ.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

From Part (a), if ε>θ, then

P(|Y(n)−θ| ≤ε)=1limn→∞P(|Y(n)−θ| ≤ε)=limn→∞(1)=1

From Part (a), if ε<θ, then

P(|Y(n)−θ| ≤ε)=1−[θ−εθ]nlimn→∞P(|Y(n)−θ| ≤ε)=limn→∞(1−[θ−εθ]n)=1−limn→∞([θ−εθ]n)

Since ε>θ, therefore, 0<θ−εθ<1.

Hence, limn→∞([θ−εθ]n)=0.

Therefore,

limn→∞P(|Y(n)−θ| ≤ε)=1−0=1.

Hence, for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

Thus, Y(n) is a consistent estimator of θ.

Answer 20

Ch. 9.2 - Prob. 1E Ch. 9.2 - Let Y1, Y2,, Yn denote a random sample from a...Ch. 9.2 - Let Y1, Y2, , Yn denote a random sample from the...Ch. 9.2 - Let Y1, Y2, , Yn denote a random sample of size n...Ch. 9.2 - Suppose that Y1, Y2, , Yn is a random sample from...Ch. 9.2 - Prob. 6E Ch. 9.2 - Prob. 7E Ch. 9.2 - Let Y1, Y2, , Yn denote a random sample from a...Ch. 9.3 - Applet Exercise How was Figure 9.1 obtained?...Ch. 9.3 - Applet Exercise Refer to Exercise 9.9. Scroll down...

Prove the following; for every n ≥ 1 and ε &gt; 0. If ε &gt; θ , P ( θ − ε ≤ Y ( n ) ≤ θ + ε ) = 1. If ε &lt; θ , P ( θ − ε ≤ Y ( n ) ≤ θ + ε ) = 1 − [ θ − ε θ ] n .

Prove the following; for every n≥1 and ε>0. If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1. If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.

Answer to Problem 26E

Explanation of Solution

Prove that Y(n) is a consistent estimator of θ by showing that for every ε>0, limn→∞P(|Y(n)−θ| ≤ε)=1.

Answer to Problem 26E

Explanation of Solution

Want to see more full solutions like this?

Chapter 9 Solutions

Prove the following; for every n ≥ 1 and ε > 0. If ε > θ , P ( θ − ε ≤ Y ( n ) ≤ θ + ε ) = 1. If ε < θ , P ( θ − ε ≤ Y ( n ) ≤ θ + ε ) = 1 − [ θ − ε θ ] n .

Prove the following; for every n≥1 and ε>0.

If ε>θ, P(θ−ε≤Y(n)≤θ+ε)=1.

If ε<θ, P(θ−ε≤Y(n)≤θ+ε)=1−[θ−εθ]n.