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Let X1,X2......X7 denote a random sample from a population having mean μ and variance σ. Consider the following estimators of μ:

θ1=X1+X2+......+X7 / 7
θ2= (2X1-X3+X5) / 2

a. Is either estimator unbiased?
b. Which estimator is best? In what sense is it best? Calculate the relative efficiency of the 2 estimtors.

User Leecbaker
by
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1 Answer

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Answer:

a) In order to check if an estimator is unbiased we need to check this condition:


E(\theta) = \mu

And we can find the expected value of each estimator like this:


E(\theta_1 ) = (1)/(7) E(X_1 +X_2 +... +X_7) = (1)/(7) [E(X_1) +E(X_2) +....+E(X_7)]= (1)/(7) 7\mu= \mu

So then we conclude that
\theta_1 is unbiased.

For the second estimator we have this:


E(\theta_2) = (1)/(2) [2E(X_1) -E(X_3) +E(X_5)]=(1)/(2) [2\mu -\mu +\mu] = (1)/(2) [2\mu]= \mu

And then we conclude that
\theta_2 is unbiaed too.

b) For this case first we need to find the variance of each estimator:


Var(\theta_1) = (1)/(49) (Var(X_1) +...+Var(X_7))= (1)/(49) (7\sigma^2) = (\sigma^2)/(7)

And for the second estimator we have this:


Var(\theta_2) = (1)/(4) (4\sigma^2 -\sigma^2 +\sigma^2)= (1)/(4) (4\sigma^2)= \sigma^2

And the relative efficiency is given by:


RE= (Var(\theta_1))/(Var(\theta_2))=((\sigma^2)/(7))/(\sigma^2)= (1)/(7)

Explanation:

For this case we assume that we have a random sample given by:
X_1, X_2,....,X_7 and each
X_i \sim N (\mu, \sigma)

Part a

In order to check if an estimator is unbiased we need to check this condition:


E(\theta) = \mu

And we can find the expected value of each estimator like this:


E(\theta_1 ) = (1)/(7) E(X_1 +X_2 +... +X_7) = (1)/(7) [E(X_1) +E(X_2) +....+E(X_7)]= (1)/(7) 7\mu= \mu

So then we conclude that
\theta_1 is unbiased.

For the second estimator we have this:


E(\theta_2) = (1)/(2) [2E(X_1) -E(X_3) +E(X_5)]=(1)/(2) [2\mu -\mu +\mu] = (1)/(2) [2\mu]= \mu

And then we conclude that
\theta_2 is unbiaed too.

Part b

For this case first we need to find the variance of each estimator:


Var(\theta_1) = (1)/(49) (Var(X_1) +...+Var(X_7))= (1)/(49) (7\sigma^2) = (\sigma^2)/(7)

And for the second estimator we have this:


Var(\theta_2) = (1)/(4) (4\sigma^2 -\sigma^2 +\sigma^2)= (1)/(4) (4\sigma^2)= \sigma^2

And the relative efficiency is given by:


RE= (Var(\theta_1))/(Var(\theta_2))=((\sigma^2)/(7))/(\sigma^2)= (1)/(7)

User Truong Hua
by
6.5k points