Your Textbook States That an Implication of the Gauss-Markov Theorem$\bar { Y }$

Question

Examlex · Accepted Answer

The Answer is (a)E(

ilde { \mu }

)=

E \left( \sum _ { i = 1 } ^ { n } a _ { i } Y _ { i } ight) = \sum _ { i = 1 } ^ { n } a _ { i } E \left( Y _ { i } ight)

= μ_Y

\sum _ { i = 1 } ^ { n } a _ { i }

Hence for this to be an unbiased estimator, the following condition must hold:

\sum _ { i = 1 } ^ { n } a _ { i } = 1

(b)var(

ilde { \mu }

)= E(

ilde { \mu }

- E(

ilde { \mu }

))² =

E \left( \sum _ { i = 1 } ^ { n } a _ { i } Y _ { i } - \mu _ { y } ight) ^ { 2 } = \left( \sum _ { i = 1 } ^ { n } a _ { i } ^ { 2 } E \left( Y _ { i } - \mu _ { y } ight) ^ { 2 } ight.

=

\sigma \stackrel { 2 } { Y }

\sum _ { i = 1 } ^ { n } a _ { i } ^ { 2 }

(c)Define the Lagrangian L =

\sigma \stackrel { 2 } { Y }

\sum _ { i = 1 } ^ { n } a _ { i } ^ { 2 } - \lambda \left( \sum _ { i = 1 } ^ { n } a _ { i } ^ { 2 } - 1 ight)

, where λ is the Lagrange multiplier. To obtain the first order conditions, minimize L with respect to the n weights and the Lagrange multiplier, and solve the resulting (n+1)equations in the (n+1)unknowns.

\frac { \partial } { \partial a _ { i } }

L = 0 = 2

\sigma \stackrel { 2 } { Y }

a_i - λ; i = 1,..., n

\frac { \partial } { \partial \lambda }

L = 0 =

\sum _ { i = 1 } ^ { n } a _ { i } ^ { 2 } - 1

Summing the first equation 2

\sigma \stackrel { 2 } { Y }

\sum _ { i = 1 } ^ { n } a _ { i }

= nλ and bringing in the second equation subsequently, results in λ =

\frac { 2 \sigma ^ { 2 }_Y } { n }

Substituting this result into the first equation then gives
2

\sigma \stackrel { 2 } { Y }

a_i=

\frac { 2 \sigma ^ { 2 }_Y } { n }

; i = 1,..., or a_i =

\frac { 1 } { n }

; i = 1,..., n. Since these are also the OLS weights, then OLS is BLUE.

Your Textbook States That an Implication of the Gauss-Markov Theorem $\bar { Y }$

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Your Textbook States That an Implication of the Gauss-Markov Theorem Yˉ\bar { Y }Yˉ

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Your Textbook States That an Implication of the Gauss-Markov Theorem $\bar { Y }$

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