Find the fractions that should be assigned to each value of x to minimize V ( β ^ 2 ) .

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Answer 1

Question

Chapter 12, Problem 37SE

To determine

Find the fractions that should be assigned to each value of x to minimize V(β^2).

Expert Solution & Answer

Answer to Problem 37SE

The fractions of total number of observations that can be assigned at x=−1,0 and 1 to minimize V(β^2) are 0.25, 0.50, and 0.25, respectively.

Explanation of Solution

Calculation:

Consider that k1,k2,k3 are the fractions of total number of observations that can be assigned at x=−1,0 and 1, respectively.

Thus, it can be aid that if total number of observation is n that is large enough, then there are nk1 points at x=−1, there are nk2 points at x=0, and there are nk3 points at x=1.

Now, the Design matrix X can be written as,

X=[1−111−11......1−11100100......100111111..1 ..1 ..1]n×n

Now,

X′X=[nn(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)]=n[1babababa]=nA, where a=k1+k3, b=k3−k1.

Now, it is needed to minimize V(β^2)=σ2c22, where c33 is the (3×3) element of (X′X)−1.

The determination of matrix A is obtained as,

A=[1babababa]|A|=1|abba|−b|bbaa|+a|baab|=a2−b2−b(ab−ba)+a(b2−a2)=a2−b2−0+ab2−a3=(k1+k3)2−(k3−k1)2+(k1+k3)(k3+k1)2−(k1+k3)3=k12+k32+2k1k3−k32−k12+2k1k3+k13+k1k32+2k12k3 +k3k12+k33+2k1k32−k13−3k12k3−3k1k32−k33=4k1k2k3

Now, the inverse of A is obtained as,

A−1=1n|A|[(−1)1+1|abba|(−1)1+2|bbaa|(−1)1+3|baab|(−1)2+1|baba|(−1)2+2|bbaa|(−1)2+3|1bab|(−1)3+1|baab|(−1)3+2|1bab|(−1)3+3|1bba|]=14nk1k2k3[a2−b20b2−a20a−a2ab−bb2−a2ab−ba−b2]

Now, the c22 element of A−1 matrix is a−b24nk1k2k3.

Hence, the V(β^2) can be written as,

V(β^2)=σ2a−b24nk1k2k3=σ2n[k1+k3−(k3−k1)24k1k2k3]=σ2n[k1+k3−[(k3+k1)2−4k1k3]4k1k2k3]=σ2n[(k1+k3)[1−k1−k3]4k1k2k3−4k1k34k1k2k3]=σ2n[(k1+k3)4k1k3−1k2]=σ2n[(k1+k3)4k1k3−11−k1−k3] [as k1+k2+k3=1]

Now, it is needed to partial differentiate V(β^2) with respect to k1 and k3 and equate to .

That is,

∂(V(β^2))∂k1=0∂(σ2n[(k1+k3)4k1k3−11−k1−k3])∂k1=04k12=(1−k1−k3)2............(1),

And

∂(V(β^2))∂k3=0∂(σ2n[(k1+k3)4k1k3−11−k1−k3])∂k3=04k32=(1−k1−k3)2..........(2).

As k1,k2,k3 are positive constants and by symmetry of k1=k3, the equation (1) can be written as,

4k12=(1−2k1)2k1=0.25

Similarly, k3=0.25.

Thus,

1−k1−k3=1−0.25−0.25=0.50.

Thus, the fractions of total number of observations that can be assigned at x=−1,0 and 1 to minimize V(β^2) are 0.25, 0.50, and 0.25, respectively.

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Answer 2

Question

Chapter 12, Problem 37SE

To determine

Find the fractions that should be assigned to each value of x to minimize V(β^2).

Expert Solution & Answer

Answer to Problem 37SE

The fractions of total number of observations that can be assigned at x=−1,0 and 1 to minimize V(β^2) are 0.25, 0.50, and 0.25, respectively.

Explanation of Solution

Calculation:

Consider that k1,k2,k3 are the fractions of total number of observations that can be assigned at x=−1,0 and 1, respectively.

Thus, it can be aid that if total number of observation is n that is large enough, then there are nk1 points at x=−1, there are nk2 points at x=0, and there are nk3 points at x=1.

Now, the Design matrix X can be written as,

X=[1−111−11......1−11100100......100111111..1 ..1 ..1]n×n

Now,

X′X=[nn(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)]=n[1babababa]=nA, where a=k1+k3, b=k3−k1.

Now, it is needed to minimize V(β^2)=σ2c22, where c33 is the (3×3) element of (X′X)−1.

The determination of matrix A is obtained as,

A=[1babababa]|A|=1|abba|−b|bbaa|+a|baab|=a2−b2−b(ab−ba)+a(b2−a2)=a2−b2−0+ab2−a3=(k1+k3)2−(k3−k1)2+(k1+k3)(k3+k1)2−(k1+k3)3=k12+k32+2k1k3−k32−k12+2k1k3+k13+k1k32+2k12k3 +k3k12+k33+2k1k32−k13−3k12k3−3k1k32−k33=4k1k2k3

Now, the inverse of A is obtained as,

A−1=1n|A|[(−1)1+1|abba|(−1)1+2|bbaa|(−1)1+3|baab|(−1)2+1|baba|(−1)2+2|bbaa|(−1)2+3|1bab|(−1)3+1|baab|(−1)3+2|1bab|(−1)3+3|1bba|]=14nk1k2k3[a2−b20b2−a20a−a2ab−bb2−a2ab−ba−b2]

Now, the c22 element of A−1 matrix is a−b24nk1k2k3.

Hence, the V(β^2) can be written as,

V(β^2)=σ2a−b24nk1k2k3=σ2n[k1+k3−(k3−k1)24k1k2k3]=σ2n[k1+k3−[(k3+k1)2−4k1k3]4k1k2k3]=σ2n[(k1+k3)[1−k1−k3]4k1k2k3−4k1k34k1k2k3]=σ2n[(k1+k3)4k1k3−1k2]=σ2n[(k1+k3)4k1k3−11−k1−k3] [as k1+k2+k3=1]

Now, it is needed to partial differentiate V(β^2) with respect to k1 and k3 and equate to .

That is,

∂(V(β^2))∂k1=0∂(σ2n[(k1+k3)4k1k3−11−k1−k3])∂k1=04k12=(1−k1−k3)2............(1),

And

∂(V(β^2))∂k3=0∂(σ2n[(k1+k3)4k1k3−11−k1−k3])∂k3=04k32=(1−k1−k3)2..........(2).

As k1,k2,k3 are positive constants and by symmetry of k1=k3, the equation (1) can be written as,

4k12=(1−2k1)2k1=0.25

Similarly, k3=0.25.

Thus,

1−k1−k3=1−0.25−0.25=0.50.

Thus, the fractions of total number of observations that can be assigned at x=−1,0 and 1 to minimize V(β^2) are 0.25, 0.50, and 0.25, respectively.

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Answer 3

Question

Chapter 12, Problem 37SE

To determine

Find the fractions that should be assigned to each value of x to minimize V(β^2).

Expert Solution & Answer

Answer to Problem 37SE

The fractions of total number of observations that can be assigned at x=−1,0 and 1 to minimize V(β^2) are 0.25, 0.50, and 0.25, respectively.

Explanation of Solution

Calculation:

Consider that k1,k2,k3 are the fractions of total number of observations that can be assigned at x=−1,0 and 1, respectively.

Thus, it can be aid that if total number of observation is n that is large enough, then there are nk1 points at x=−1, there are nk2 points at x=0, and there are nk3 points at x=1.

Now, the Design matrix X can be written as,

X=[1−111−11......1−11100100......100111111..1 ..1 ..1]n×n

Now,

X′X=[nn(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)]=n[1babababa]=nA, where a=k1+k3, b=k3−k1.

Now, it is needed to minimize V(β^2)=σ2c22, where c33 is the (3×3) element of (X′X)−1.

The determination of matrix A is obtained as,

A=[1babababa]|A|=1|abba|−b|bbaa|+a|baab|=a2−b2−b(ab−ba)+a(b2−a2)=a2−b2−0+ab2−a3=(k1+k3)2−(k3−k1)2+(k1+k3)(k3+k1)2−(k1+k3)3=k12+k32+2k1k3−k32−k12+2k1k3+k13+k1k32+2k12k3 +k3k12+k33+2k1k32−k13−3k12k3−3k1k32−k33=4k1k2k3

Now, the inverse of A is obtained as,

A−1=1n|A|[(−1)1+1|abba|(−1)1+2|bbaa|(−1)1+3|baab|(−1)2+1|baba|(−1)2+2|bbaa|(−1)2+3|1bab|(−1)3+1|baab|(−1)3+2|1bab|(−1)3+3|1bba|]=14nk1k2k3[a2−b20b2−a20a−a2ab−bb2−a2ab−ba−b2]

Now, the c22 element of A−1 matrix is a−b24nk1k2k3.

Hence, the V(β^2) can be written as,

V(β^2)=σ2a−b24nk1k2k3=σ2n[k1+k3−(k3−k1)24k1k2k3]=σ2n[k1+k3−[(k3+k1)2−4k1k3]4k1k2k3]=σ2n[(k1+k3)[1−k1−k3]4k1k2k3−4k1k34k1k2k3]=σ2n[(k1+k3)4k1k3−1k2]=σ2n[(k1+k3)4k1k3−11−k1−k3] [as k1+k2+k3=1]

Now, it is needed to partial differentiate V(β^2) with respect to k1 and k3 and equate to .

That is,

∂(V(β^2))∂k1=0∂(σ2n[(k1+k3)4k1k3−11−k1−k3])∂k1=04k12=(1−k1−k3)2............(1),

And

∂(V(β^2))∂k3=0∂(σ2n[(k1+k3)4k1k3−11−k1−k3])∂k3=04k32=(1−k1−k3)2..........(2).

As k1,k2,k3 are positive constants and by symmetry of k1=k3, the equation (1) can be written as,

4k12=(1−2k1)2k1=0.25

Similarly, k3=0.25.

Thus,

1−k1−k3=1−0.25−0.25=0.50.

Thus, the fractions of total number of observations that can be assigned at x=−1,0 and 1 to minimize V(β^2) are 0.25, 0.50, and 0.25, respectively.

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Answer 4

Question

Chapter 12, Problem 37SE

To determine

Find the fractions that should be assigned to each value of x to minimize V(β^2).

Expert Solution & Answer

Answer to Problem 37SE

The fractions of total number of observations that can be assigned at x=−1,0 and 1 to minimize V(β^2) are 0.25, 0.50, and 0.25, respectively.

Explanation of Solution

Calculation:

Consider that k1,k2,k3 are the fractions of total number of observations that can be assigned at x=−1,0 and 1, respectively.

Thus, it can be aid that if total number of observation is n that is large enough, then there are nk1 points at x=−1, there are nk2 points at x=0, and there are nk3 points at x=1.

Now, the Design matrix X can be written as,

X=[1−111−11......1−11100100......100111111..1 ..1 ..1]n×n

Now,

X′X=[nn(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)]=n[1babababa]=nA, where a=k1+k3, b=k3−k1.

Now, it is needed to minimize V(β^2)=σ2c22, where c33 is the (3×3) element of (X′X)−1.

The determination of matrix A is obtained as,

A=[1babababa]|A|=1|abba|−b|bbaa|+a|baab|=a2−b2−b(ab−ba)+a(b2−a2)=a2−b2−0+ab2−a3=(k1+k3)2−(k3−k1)2+(k1+k3)(k3+k1)2−(k1+k3)3=k12+k32+2k1k3−k32−k12+2k1k3+k13+k1k32+2k12k3 +k3k12+k33+2k1k32−k13−3k12k3−3k1k32−k33=4k1k2k3

Now, the inverse of A is obtained as,

A−1=1n|A|[(−1)1+1|abba|(−1)1+2|bbaa|(−1)1+3|baab|(−1)2+1|baba|(−1)2+2|bbaa|(−1)2+3|1bab|(−1)3+1|baab|(−1)3+2|1bab|(−1)3+3|1bba|]=14nk1k2k3[a2−b20b2−a20a−a2ab−bb2−a2ab−ba−b2]

Now, the c22 element of A−1 matrix is a−b24nk1k2k3.

Hence, the V(β^2) can be written as,

V(β^2)=σ2a−b24nk1k2k3=σ2n[k1+k3−(k3−k1)24k1k2k3]=σ2n[k1+k3−[(k3+k1)2−4k1k3]4k1k2k3]=σ2n[(k1+k3)[1−k1−k3]4k1k2k3−4k1k34k1k2k3]=σ2n[(k1+k3)4k1k3−1k2]=σ2n[(k1+k3)4k1k3−11−k1−k3] [as k1+k2+k3=1]

Now, it is needed to partial differentiate V(β^2) with respect to k1 and k3 and equate to .

That is,

∂(V(β^2))∂k1=0∂(σ2n[(k1+k3)4k1k3−11−k1−k3])∂k1=04k12=(1−k1−k3)2............(1),

And

∂(V(β^2))∂k3=0∂(σ2n[(k1+k3)4k1k3−11−k1−k3])∂k3=04k32=(1−k1−k3)2..........(2).

As k1,k2,k3 are positive constants and by symmetry of k1=k3, the equation (1) can be written as,

4k12=(1−2k1)2k1=0.25

Similarly, k3=0.25.

Thus,

1−k1−k3=1−0.25−0.25=0.50.

Thus, the fractions of total number of observations that can be assigned at x=−1,0 and 1 to minimize V(β^2) are 0.25, 0.50, and 0.25, respectively.

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Answer 5

Chapter 12, Problem 37SE

To determine

Find the fractions that should be assigned to each value of x to minimize V(β^2).

Expert Solution & Answer

Answer to Problem 37SE

The fractions of total number of observations that can be assigned at x=−1,0 and 1 to minimize V(β^2) are 0.25, 0.50, and 0.25, respectively.

Explanation of Solution

Calculation:

Consider that k1,k2,k3 are the fractions of total number of observations that can be assigned at x=−1,0 and 1, respectively.

Thus, it can be aid that if total number of observation is n that is large enough, then there are nk1 points at x=−1, there are nk2 points at x=0, and there are nk3 points at x=1.

Now, the Design matrix X can be written as,

X=[1−111−11......1−11100100......100111111..1 ..1 ..1]n×n

Now,

X′X=[nn(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)]=n[1babababa]=nA, where a=k1+k3, b=k3−k1.

Now, it is needed to minimize V(β^2)=σ2c22, where c33 is the (3×3) element of (X′X)−1.

The determination of matrix A is obtained as,

A=[1babababa]|A|=1|abba|−b|bbaa|+a|baab|=a2−b2−b(ab−ba)+a(b2−a2)=a2−b2−0+ab2−a3=(k1+k3)2−(k3−k1)2+(k1+k3)(k3+k1)2−(k1+k3)3=k12+k32+2k1k3−k32−k12+2k1k3+k13+k1k32+2k12k3 +k3k12+k33+2k1k32−k13−3k12k3−3k1k32−k33=4k1k2k3

Now, the inverse of A is obtained as,

A−1=1n|A|[(−1)1+1|abba|(−1)1+2|bbaa|(−1)1+3|baab|(−1)2+1|baba|(−1)2+2|bbaa|(−1)2+3|1bab|(−1)3+1|baab|(−1)3+2|1bab|(−1)3+3|1bba|]=14nk1k2k3[a2−b20b2−a20a−a2ab−bb2−a2ab−ba−b2]

Now, the c22 element of A−1 matrix is a−b24nk1k2k3.

Hence, the V(β^2) can be written as,

V(β^2)=σ2a−b24nk1k2k3=σ2n[k1+k3−(k3−k1)24k1k2k3]=σ2n[k1+k3−[(k3+k1)2−4k1k3]4k1k2k3]=σ2n[(k1+k3)[1−k1−k3]4k1k2k3−4k1k34k1k2k3]=σ2n[(k1+k3)4k1k3−1k2]=σ2n[(k1+k3)4k1k3−11−k1−k3] [as k1+k2+k3=1]

Now, it is needed to partial differentiate V(β^2) with respect to k1 and k3 and equate to .

That is,

∂(V(β^2))∂k1=0∂(σ2n[(k1+k3)4k1k3−11−k1−k3])∂k1=04k12=(1−k1−k3)2............(1),

And

∂(V(β^2))∂k3=0∂(σ2n[(k1+k3)4k1k3−11−k1−k3])∂k3=04k32=(1−k1−k3)2..........(2).

As k1,k2,k3 are positive constants and by symmetry of k1=k3, the equation (1) can be written as,

4k12=(1−2k1)2k1=0.25

Similarly, k3=0.25.

Thus,

1−k1−k3=1−0.25−0.25=0.50.

Thus, the fractions of total number of observations that can be assigned at x=−1,0 and 1 to minimize V(β^2) are 0.25, 0.50, and 0.25, respectively.

Answer 6

Chapter 12, Problem 37SE

To determine

Find the fractions that should be assigned to each value of x to minimize V(β^2).

Expert Solution & Answer

Answer to Problem 37SE

The fractions of total number of observations that can be assigned at x=−1,0 and 1 to minimize V(β^2) are 0.25, 0.50, and 0.25, respectively.

Explanation of Solution

Calculation:

Consider that k1,k2,k3 are the fractions of total number of observations that can be assigned at x=−1,0 and 1, respectively.

Thus, it can be aid that if total number of observation is n that is large enough, then there are nk1 points at x=−1, there are nk2 points at x=0, and there are nk3 points at x=1.

Now, the Design matrix X can be written as,

X=[1−111−11......1−11100100......100111111..1 ..1 ..1]n×n

Now,

X′X=[nn(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)]=n[1babababa]=nA, where a=k1+k3, b=k3−k1.

Now, it is needed to minimize V(β^2)=σ2c22, where c33 is the (3×3) element of (X′X)−1.

The determination of matrix A is obtained as,

A=[1babababa]|A|=1|abba|−b|bbaa|+a|baab|=a2−b2−b(ab−ba)+a(b2−a2)=a2−b2−0+ab2−a3=(k1+k3)2−(k3−k1)2+(k1+k3)(k3+k1)2−(k1+k3)3=k12+k32+2k1k3−k32−k12+2k1k3+k13+k1k32+2k12k3 +k3k12+k33+2k1k32−k13−3k12k3−3k1k32−k33=4k1k2k3

Now, the inverse of A is obtained as,

A−1=1n|A|[(−1)1+1|abba|(−1)1+2|bbaa|(−1)1+3|baab|(−1)2+1|baba|(−1)2+2|bbaa|(−1)2+3|1bab|(−1)3+1|baab|(−1)3+2|1bab|(−1)3+3|1bba|]=14nk1k2k3[a2−b20b2−a20a−a2ab−bb2−a2ab−ba−b2]

Now, the c22 element of A−1 matrix is a−b24nk1k2k3.

Hence, the V(β^2) can be written as,

V(β^2)=σ2a−b24nk1k2k3=σ2n[k1+k3−(k3−k1)24k1k2k3]=σ2n[k1+k3−[(k3+k1)2−4k1k3]4k1k2k3]=σ2n[(k1+k3)[1−k1−k3]4k1k2k3−4k1k34k1k2k3]=σ2n[(k1+k3)4k1k3−1k2]=σ2n[(k1+k3)4k1k3−11−k1−k3] [as k1+k2+k3=1]

Now, it is needed to partial differentiate V(β^2) with respect to k1 and k3 and equate to .

That is,

∂(V(β^2))∂k1=0∂(σ2n[(k1+k3)4k1k3−11−k1−k3])∂k1=04k12=(1−k1−k3)2............(1),

And

∂(V(β^2))∂k3=0∂(σ2n[(k1+k3)4k1k3−11−k1−k3])∂k3=04k32=(1−k1−k3)2..........(2).

As k1,k2,k3 are positive constants and by symmetry of k1=k3, the equation (1) can be written as,

4k12=(1−2k1)2k1=0.25

Similarly, k3=0.25.

Thus,

1−k1−k3=1−0.25−0.25=0.50.

Thus, the fractions of total number of observations that can be assigned at x=−1,0 and 1 to minimize V(β^2) are 0.25, 0.50, and 0.25, respectively.

Answer 7

Chapter 12, Problem 37SE

To determine

Find the fractions that should be assigned to each value of x to minimize V(β^2).

Answer 8

Expert Solution & Answer

Answer to Problem 37SE

The fractions of total number of observations that can be assigned at x=−1,0 and 1 to minimize V(β^2) are 0.25, 0.50, and 0.25, respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Calculation:

Consider that k1,k2,k3 are the fractions of total number of observations that can be assigned at x=−1,0 and 1, respectively.

Thus, it can be aid that if total number of observation is n that is large enough, then there are nk1 points at x=−1, there are nk2 points at x=0, and there are nk3 points at x=1.

Now, the Design matrix X can be written as,

X=[1−111−11......1−11100100......100111111..1 ..1 ..1]n×n

Now,

X′X=[nn(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)n(k3−k1)n(k1+k3)]=n[1babababa]=nA, where a=k1+k3, b=k3−k1.

Now, it is needed to minimize V(β^2)=σ2c22, where c33 is the (3×3) element of (X′X)−1.

The determination of matrix A is obtained as,

A=[1babababa]|A|=1|abba|−b|bbaa|+a|baab|=a2−b2−b(ab−ba)+a(b2−a2)=a2−b2−0+ab2−a3=(k1+k3)2−(k3−k1)2+(k1+k3)(k3+k1)2−(k1+k3)3=k12+k32+2k1k3−k32−k12+2k1k3+k13+k1k32+2k12k3 +k3k12+k33+2k1k32−k13−3k12k3−3k1k32−k33=4k1k2k3

Now, the inverse of A is obtained as,

A−1=1n|A|[(−1)1+1|abba|(−1)1+2|bbaa|(−1)1+3|baab|(−1)2+1|baba|(−1)2+2|bbaa|(−1)2+3|1bab|(−1)3+1|baab|(−1)3+2|1bab|(−1)3+3|1bba|]=14nk1k2k3[a2−b20b2−a20a−a2ab−bb2−a2ab−ba−b2]

Now, the c22 element of A−1 matrix is a−b24nk1k2k3.

Hence, the V(β^2) can be written as,

V(β^2)=σ2a−b24nk1k2k3=σ2n[k1+k3−(k3−k1)24k1k2k3]=σ2n[k1+k3−[(k3+k1)2−4k1k3]4k1k2k3]=σ2n[(k1+k3)[1−k1−k3]4k1k2k3−4k1k34k1k2k3]=σ2n[(k1+k3)4k1k3−1k2]=σ2n[(k1+k3)4k1k3−11−k1−k3] [as k1+k2+k3=1]

Now, it is needed to partial differentiate V(β^2) with respect to k1 and k3 and equate to .

That is,

∂(V(β^2))∂k1=0∂(σ2n[(k1+k3)4k1k3−11−k1−k3])∂k1=04k12=(1−k1−k3)2............(1),

And

∂(V(β^2))∂k3=0∂(σ2n[(k1+k3)4k1k3−11−k1−k3])∂k3=04k32=(1−k1−k3)2..........(2).

As k1,k2,k3 are positive constants and by symmetry of k1=k3, the equation (1) can be written as,

4k12=(1−2k1)2k1=0.25

Similarly, k3=0.25.

Thus,

1−k1−k3=1−0.25−0.25=0.50.

Thus, the fractions of total number of observations that can be assigned at x=−1,0 and 1 to minimize V(β^2) are 0.25, 0.50, and 0.25, respectively.

Answer 12

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Find the fractions that should be assigned to each value of x to minimize V ( β ^ 2 ) .

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Find the fractions that should be assigned to each value of x to minimize V(β^2).

Answer to Problem 37SE

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