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BU.232.630.W6.SP25 sample_quiz_3
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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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The question presents an i.i.d. Bernoulli(p) sample X1, X2, ..., XT and defines the MLE for p as the sample mean p_hat = (1/T) sum_t X_t. To find Var(p_hat), we can use the properties of independent Bernoulli trials. Each X_t has Var(X_t) = p(1 โ p). Since the X_t are independent, the variance of the sum is the sum......Login to view full explanation็ปๅฝๅณๅฏๆฅ็ๅฎๆด็ญๆก
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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
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