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Consider the likelihood of an i.i.d. sample from a Bernoulli population with parameter ๐‘ ๐ฟ ( ๐‘ฅ 1 , . . . , ๐‘ฅ ๐‘‡ ) = โˆ ๐‘ก = 1 ๐‘‡ ๐‘ ๐‘ฅ ๐‘ก ( 1 โˆ’ ๐‘ ) 1 โˆ’ ๐‘ฅ ๐‘ก . If you estimate the parameter ๐‘ using a Maximum Likelihood estimator, you obtain the point estimate ๐‘ ฬ‚ = 1 ๐‘‡ โˆ‘ ๐‘ก = 1 ๐‘‡ ๐‘ฅ ๐‘ก , which corresponds to the sample mean. We know that for a Bernoulli random variable the expected value and the variance are ๐”ผ ( ๐‘ฅ ๐‘ก ) = ๐‘ , ๐• ( ๐‘ฅ ๐‘ก ) = ๐‘ ( 1 โˆ’ ๐‘ ) . Using this information, what is the variance of the estimator ๐• ( ๐‘ ฬ‚ ) ?

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A.The variance of ๐‘ ฬ‚ is ๐• ( ๐‘ ฬ‚ ) = ๐‘ 2
B.The variance of ๐‘ ฬ‚ is ๐• ( ๐‘ ฬ‚ ) = ๐‘ ( 1 โˆ’ ๐‘ ) ๐‘‡
C.All the answers are incorrect.
D.The variance of ๐‘ ฬ‚ is ๐• ( ๐‘ ฬ‚ ) = ๐”ผ ( ๐‘ 2 )
E.The variance of ๐‘ ฬ‚ is ๐• ( ๐‘ ฬ‚ ) = ๐‘ ( 1 โˆ’ ๐‘ )
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Let's walk through each answer choice carefully and connect them to the underlying statistics of an i.i.d. Bernoulli sample. Option 1: The variance of pฬ‚ is V(pฬ‚) = p^2. This is incorrect. The estimator pฬ‚ = (1/T) โˆ‘ x_t is the sample mean of Bernoulli( p ) draws. Each x_t has variance p(1โˆ’p). The variance of the average of T i.i.d. observations is the individual variance divided by T, not p^2. So V(pฬ‚) = p(......Login to view full explanation

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