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BU.232.630.F3.SP25 QUIZ 1 2025
ๅ้กน้ๆฉ้ข
Consider a population with mean ๐ and variance ๐ 2 < โ . You are comparing two estimators ๐ ฬ 1 and ๐ ฬ 2 for the mean of the population ๐ , with the following expected values and variances ๐ธ ( ๐ ฬ 1 ) = ๐ ; ๐ ( ๐ ฬ 1 ) = 4 ; ๐ธ ( ๐ ฬ 1 ) = ๐ + 1 ; ๐ ( ๐ ฬ 2 ) = 1 . We also know that the covariance between the two estimators is ๐ถ ๐ ๐ ( ๐ ฬ 1 , ๐ ฬ 2 ) = โ 2 . Now consider a new estimator that combines the two previous ones ๐ ฬ 3 = 2 5 ๐ ฬ 1 + 3 5 ๐ ฬ 2 . Then the variance ๐ ( ๐ ฬ 3 ) of ๐ ฬ 3 is
้้กน
A.0.04
B.2.2
C.๐
2
+ 1
D.1
E.๐
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, letโs 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 explanation็ปๅฝๅณๅฏๆฅ็ๅฎๆด็ญๆก
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็ฑปไผผ้ฎ้ข
Consider the likelihood of an i.i.d. sample from a Bernoulli population with parameter ๐ ๐ฟ ( ๐ฅ 1 , . . . , ๐ฅ ๐ ) = โ ๐ก = 1 ๐ ๐ ๐ฅ ๐ก ( 1 โ ๐ ) 1 โ ๐ฅ ๐ก . If you estimate the parameter ๐ using a Maximum Likelihood estimator, you obtain the point estimate ๐ ฬ = 1 ๐ โ ๐ก = 1 ๐ ๐ฅ ๐ก , which corresponds to the sample mean. We know that for a Bernoulli random variable the expected value and the variance are ๐ผ ( ๐ฅ ๐ก ) = ๐ , ๐ ( ๐ฅ ๐ก ) = ๐ ( 1 โ ๐ ) . Using this information, what is the variance of the estimator ๐ ( ๐ ฬ ) ?
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