Let x1 x2 X9 denote a random sample from a population ha

Let x1, x2. . . . . X9 denote a random sample from a population having mean mu and variance O_2 . Consider the following estimators of mu: Theta 1 = x_1 + x_2 +. . . + x9/9 Theta 2 = 3x_1 - x6 + 2x_4/2 Calculate the bias of both estimators. Calculate the variance of both estimators. Calculate the mean square error of both estimators. What is the relative efficiency of thete1 to thete2? Is one estimator best? If so, which one and why?

Solution

Given that X1,X2,.........,X9 denote a random sample from population having mean µ and variance ² .

Also given the following estimator of µ.

1^ = (X1+X2+........+X9) / 9 and

2^ = (3X1 - X6 + 2X4) / 2

1. Calculate the bias of the both estimator.

Bias = E(^) -

Consider E(1^) = E[ 1/9( X1+X2+.....+X9)]

= 1/9 E[X1+X2+.......+X9]

= 1/9 { E(X1)+E(X2)+......+E(X9) }

But we have given that , X1,X2,.....X9 have mean µ.

= 1/9 [ µ+µ+µ+......+µ] = 9µ / 9 = µ

E(1^) = µ

Bias(1^) =  E(1^) -   = µ - µ = 0

Now we can calculate variance of 1^.

Var(1^) = Var [ 1/9 (X1+X2+X3+......+X9] ]

= (1/9)² Var(X1+X2+X3+........+X9)

= 1/81 [Var(X1)+Var(X2)+......+Var(X9)]

But here also the variance of X1,X2,.......X9 is ².

= 1/81 [ ² + ² +.......+ ² ]

Var(1^) = 1/81 * (9² ) = ² / 9

Now we calculate bias of 2^.

Bias(2^) = E(2^) -

E(2^) = E [(3X1 - X6 + 2X4) / 2]

= 1/2 E[3X1 - X6 + 2X4]

= 1/2 [3*E(X1) - E(X6) + 2*E(X4) ]

= 1/2 [ 3*µ - µ + 2*µ ]

= 4*µ / 2

E(2^) =2*µ

Bias(2^) = 2*µ - µ = µ

Again we find variance(2^).

Var(2^) = Var [ (3X1 - X6 + 2X4) / 2]

= 1/4 * Var (3X1 - X6 + 2X4)

= 1/4 * [ 9*Var(X1) - Var(X6) +4*Var(X4) ]

= 1/4 * [ 9*² - ² + 4*² ]

= 12² / 4 = 3²

standard deviation(1^) = sd(1^)=( sqrt(1^) = sqrt(² / 9) = / 3

standard deviation(2^) = sd(2^) = sqrt(2^) = sqrt(3² ) = sqrt(3)*  

standard error(1^) =se1 = sd(1^) / sqrt(n) Here n is 9.

= ( / 3 ) / sqrt(9) = / 9

standard error(2^) =se2 = sd( 2^) / sqrt(n)

= sqrt(3)* / sqrt(9)

= sqrt(3) * / 3

= / sqrt(3)

MSE( 1^) = [se(1^)]² + bias² = (   / 9)² + 0² =  ² / 81

MSE( 2^) = [se(2^)]² + bias² = [ / sqrt(3) ]² + µ² =  ² / 3 + µ²

Relative efficiency of 1^ and 2^,

Relative efficiency = MSE( 1^) / MSE( 2^) = [² / 81] / [² / 3 + µ²]

= ² / 27(² + 3µ² )

Variance of 2^ is larger than the variance of 1^.

Thus 1^ is best estimator than 2^.