Mathematical Statistics with Applications
Mathematical Statistics with Applications
7th Edition
ISBN: 9780495110811
Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer
Publisher: Cengage Learning
Question
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Chapter 6, Problem 76SE

a.

To determine

Find the density function of Y(k), kth-order statistic.

a.

Expert Solution
Check Mark

Answer to Problem 76SE

The density function of Y(k), kth-order statistic is,

g(k)(y)=n!(k1)!(nk)!yk1(θy)nkθn,0ykθ.

Explanation of Solution

Calculation:

From Theorem 6.5, the density function for kth order statistic, Y(k) is,

g(k)(yk)=n!(k1)!(nk)![F(yk)]k1×[1F(yk)]nkf(yk),<yk<.

It is known that, the density function of Uniform distribution over the interval [0,θ] is f(y)=1θ,0yθ and the distribution function is F(y)=yθ.

The density function for Y(k) is,

g(k)(y)=n!(k1)!(nk)![yθ]k1×[1yθ]nk1θ,<yk<=n!(k1)!(nk)!yk1(θy)nkθn,0ykθ

b.

To determine

Find the value of E(Y(k)).

b.

Expert Solution
Check Mark

Answer to Problem 76SE

The value of E(Y(k)) is kn+1θ.

Explanation of Solution

Calculation:

Consider,

E(Y(k))=0θyg(k)(y)dy=0θyn!(k1)!(nk)!yk1(θy)nkθndy=kn+1Γ(n+2)Γ(k+1)Γ(nk+1)0θ(yθ)k(1yθ)nkdy=kn+1Γ(n+2)Γ(k+1)Γ(nk+1)0θ(z)k(1z)nkθdz    [Let, yθ=zdy=θdz]

This function  0θ(yθ)k(1yθ)nkdy representing the beta density function with α=k+1 and β=nk+1.

Therefore,

E(Y(k))=kn+1θ

c.

To determine

Find the value of V(Y(k)).

c.

Expert Solution
Check Mark

Answer to Problem 76SE

The value of V(Y(k)) is k(nk+1)(n+1)2(n+2)θ2.

Explanation of Solution

Calculation:

Consider,

E(Y2(k))=0θy2g(k)(y)dy=0θy2n!(k1)!(nk)!yk1(θy)nkθndy=kθn+1Γ(n+2)Γ(k+1)Γ(nk+1)0θ(yθ)k+1(1yθ)nkdy=kθn+1Γ(n+2)Γ(k+1)Γ(nk+1)0θ(z)k+1(1z)nkθdz    [Let, yθ=zdy=θdz]

                   =k(k+1)(n+1)(n+2)θ2

Therefore,

V(Y(k))=E(Y(k)2)[E(Y(k))]2=k(k+1)(n+1)(n+2)θ2(kn+1θ)2=k(nk+1)(n+1)2(n+2)θ2

d.

To determine

Find the mean difference between two successive order statistics, E(Y(k)Y(k1)).

d.

Expert Solution
Check Mark

Answer to Problem 76SE

The mean difference between two successive order statistics, E(Y(k)Y(k1)) is 1n+1θ.

The expected order statistics are equally spaced.

Explanation of Solution

Calculation:

Consider,

E(Y(k)Y(k1))=kn+1θk1n+1θ=(kk+1n+1)θ=1n+1θ

This function is a constant for all k. Thus, the expected order statistics are equally spaced.

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Chapter 6 Solutions

Mathematical Statistics with Applications

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