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hw4sol - APPM 4/5520 Solutions to Problem Set Four 1(a 2...

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APPM 4/5520 Solutions to Problem Set Four 1. (a) E [ S 2 2 ] = 1 n 1 E [ summationdisplay ( X i X ) 2 ] Option One: Square ( X i X ) and take the sum through: summationdisplay ( X i X ) 2 = summationdisplay ( X 2 i 2 X + X 2 ) = X 2 i 2 X X i + n X 2 = X 2 i 2 X · n · 1 n X i + n X 2 = X 2 i 2 X · n X + n X 2 = X 2 i 2 n X 2 + n X 2 = X 2 i n X 2 So E [ ( X i X ) 2 ] = E [ X 2 X n X 2 ] = E [ X 2 i ] n E [ X 2 ] = n · E [ X 2 1 ] n E [ X 2 ] Since E [ X 2 1 ] = V ar [ X 1 ] + ( E [ X 1 ]) 2 = σ 2 + μ 2 and E [ X 2 ] = V ar [ X ] + parenleftBig E [ X ] parenrightBig 2 = σ 2 n + μ 2 we have E [ ( X i X ) 2 ] = n ( σ 2 + μ 2 ) n parenleftBig σ 2 n + μ 2 parenrightBig = 2 σ 2 = ( n 1) σ 2 . Putting it all together: E [ S 2 2 ] = 1 n 1 E [ summationdisplay ( X i X ) 2 ] = 1 n 1 · ( n 1) σ 2 = σ 2
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So S 2 2 is an unbiased estimator of σ 2 . Option Two: Square ( X i X ) and find the expectation but leave the sum outside of the expected value: E [( X i X ) 2 ] = E [ X 2 i 2 XX i + X 2 ] = E [ X 2 i ] 2 E [ XX i ] + E [ X 2 ] As before, we have E [ X 2 i ] = V ar [ X i ] + ( E [ X i ]) 2 = σ 2 + μ 2 and E [ X 2 ] = V ar [ X ] + parenleftBig E [ X ] parenrightBig 2 = σ 2 n + μ 2 . As for the middle term, E [ XX i ] = E [ X i X ] E bracketleftBig X i 1 n n j =1 X j bracketrightBig = 1 n n j =1 E [ X i X j ] When j negationslash = i , this expectation is E [ X i X j ] indep = E [ X
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