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BU.232.630.F3.SP25 QUIZ 1 2025

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Consider a population with mean ๐œ‡ and variance ๐œŽ 2 < โˆž . You are comparing two estimators ๐œ‡ ฬ‚ 1 and ๐œ‡ ฬ‚ 2 for the mean of the population ๐œ‡ , with the following expected values and variances ๐ธ ( ๐œ‡ ฬ‚ 1 ) = ๐œ‡ ; ๐‘‰ ( ๐œ‡ ฬ‚ 1 ) = 4 ; ๐ธ ( ๐œ‡ ฬ‚ 1 ) = ๐œ‡ + 1 ; ๐‘‰ ( ๐œ‡ ฬ‚ 2 ) = 1 . We also know that the covariance between the two estimators is ๐ถ ๐‘‚ ๐‘‰ ( ๐œ‡ ฬ‚ 1 , ๐œ‡ ฬ‚ 2 ) = โˆ’ 2 . Now consider a new estimator that combines the two previous ones ๐œ‡ ฬ‚ 3 = 2 5 ๐œ‡ ฬ‚ 1 + 3 5 ๐œ‡ ฬ‚ 2 . Then the variance ๐‘‰ ( ๐œ‡ ฬ‚ 3 ) of ๐œ‡ ฬ‚ 3 is

้€‰้กน
A.0.04
B.2.2
C.๐œŽ 2 + 1
D.1
E.๐œŽ 2 -2
F.1.24
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We start by restating the problem setup to ensure clarity. The goal is to find Var(mu_hat_3) where mu_hat_3 = (2/5) mu_hat_1 + (3/5) mu_hat_2, given Var(mu_hat_1) = 4, Var(mu_hat_2) = 1, Cov(mu_hat_1, mu_hat_2) = -2, and the means are mu_hat_1 ~ E = mu, E(mu_hat_2) = mu as well (the means are not directly needed for the variance calculation here). Now, letโ€™s compute Var(mu_hat_3) using the variance of a linear combination formula: Var(aX + bY) = a^2 Var(X) + b^2 Var(Y) + 2ab Cov......Login to view full explanation

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