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HW 2 - Solutions - EC489.02 Summer 2011 Homework 2...

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Unformatted text preview: EC489.02 Summer 2011 Homework 2 - Solutions 1. The left hand side is equal to n X i =1 ( X i & & ) 2 = n X i =1 & X 2 i & 2 X i & + & 2 ¡ = n X i =1 X 2 i & n X i =1 2 X i & + n X i =1 & 2 = n X i =1 X 2 i & 2 & n X i =1 X i + n X i =1 & 2 = n X i =1 X 2 i & 2 &n & X + n& 2 : (1) The right hand side is equal to n X i =1 ( X i & & X ) 2 + n ( & X & & ) 2 = " n X i =1 & X 2 i & 2 X i & X + & X 2 ¡ # + n & & X 2 & 2 & X& + & 2 ¡ = n X i =1 X 2 i & n X i =1 2 X i & X + n X i =1 & X 2 + n & & X 2 & 2 & X& + & 2 ¡ = n X i =1 X 2 i & 2 & X n X i =1 X i + n & X 2 + n & X 2 & 2 n & X& + n& 2 = n X i =1 X 2 i & 2 & Xn & X + n & X 2 + n & X 2 & 2 n & X& + n& 2 = n X i =1 X 2 i & 2 n & X 2 + n & X 2 + n & X 2 & 2 n & X& + n& 2 = n X i =1 X 2 i & 2 n & X& + n& 2 : (2) Therefore, (2) is equal to (1), which completes the answer. 2. Suppose that we are given a random sample X 1 ; :::; X n of size n from a continuous Uniform population. Remember that for a random variable X i ; distributed with the...
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HW 2 - Solutions - EC489.02 Summer 2011 Homework 2...

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