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Unformatted text preview: Problem Set 3 Answer Key - Econometrics 120A 1. (a) To find the expectation, recall that under random sampling E [ X i ] = μ for all X i . Therefore: E [ˆ μ 1 ] = E 2 n n/ 2 X i =1 X i = 2 n n/ 2 X i =1 E [ X i ] = 2 n n/ 2 X i =1 μ = μ Similarly, since Var( X + Y ) = Var( X ) + Var( Y ) under independence, Var( aX ) = a 2 Var( X ) and under random sampling Var( X i ) = σ 2 for all X i , we get that: V ar (ˆ μ 1 ) = V ar 2 n n/ 2 X i =1 X i = 4 n 2 n X i =1 V ar ( X i ) = 4 n 2 n 2 σ 2 = 2 n σ 2 (b) The average over the whole sample, ¯ X has variance equal to σ 2 /n . Therefore ¯ X has a smaller variance than ˆ μ 1 and it will allow us to better guess what μ is. 2. (f) You should see a distribution with the mean 0.5 in the middle and the distribution roughly symmetric. It may not, however, be close to normal. The reason is that with the mean in the middle, and a standard deviation that doesn’t hit the bounds of 0 or 1 for the proportion (recall that the normal is not bounded like this) then there is no pile-up at either end of the possible...
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This note was uploaded on 12/05/2010 for the course ECON 120A taught by Professor M.abajian during the Spring '10 term at San Diego.
- Spring '10