Dr. Hackney STA Solutions pg 162

# Dr. Hackney STA Solutions pg 162 - Second Edition 10-5 The...

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Second Edition 10-5 The first column is n n - 1 x 1 s followed by n n - 1 x 2 s followed by, . . . , followed by n n - 1 x n s The second column is n n - 2 x 1 s followed by n n - 2 x 2 s followed by, . . . , followed by n n - 2 x n s, repeated n times The third column is n n - 3 x 1 s followed by n n - 3 x 2 s followed by, . . . , followed by n n - 3 x n s, repeated n 2 times . . . The n th column is 1 x 1 followed by 1 x 2 followed by, . . . , followed by 1 x n , repeated n n - 1 times So now it is easy to see that each column in the data array has mean ¯ x , hence the entire bootstrap data set has mean ¯ x . Appealing to the 3 3 × 3 data array, we can write the numerator of the variance of the bootstrap means as 3 i =1 3 j =1 3 k =1 1 3 ( x i + x j + x k ) - ¯ x 2 = 1 3 2 3 i =1 3 j =1 3 k =1 [( x i - ¯ x ) + ( x j - ¯ x ) + ( x k - ¯ x )] 2 = 1 3 2 3 i =1 3 j =1 3 k =1 ( x i - ¯ x ) 2 + ( x j - ¯ x ) 2 + ( x k - ¯ x ) 2 , because all of the cross terms are zero (since they are the sum of deviations from the mean). Summing up and collecting terms shows that 1 3 2 3 i =1 3 j =1 3 k =1 ( x i - ¯ x ) 2 + ( x j - ¯ x ) 2 + ( x k - ¯ x ) 2 = 3 3 i =1 ( x i - ¯ x ) 2 , and thus the average of the variance of the bootstrap means is 3 3 i =1 ( x i - ¯ x ) 2 3 3 which is the usual estimate of the variance of ¯ X if we divide by
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