FM5002-HW10-4.18.12 (1)

B we first evaluate the left hand side e g d2 n 2n we

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Unformatted text preview: . We con dition on the n umber of heads. Let k be the n umber of heads in 2 n flips, so that 2 n k is the number of tails. Then for an y in terval that corresponds to exactly k flips that were heads, the value of D2 n is k 2n 1 2n 22 n 2n k0 fn x 2k 2 n, so that the value of g D2 n g 2k 2n 2n 2 is g 2 k 2n 2 n . The n umber of , so that summing over all k we obtain 2n 2n . 2n han d side, x by summin g over the bars in H n since f n x a n d p n x a re constant on bars . For the kth bar, n 2n x 1 2n an d p n x g fn x p n x 2 2n k We n ow evaluate the right g fn x p n x 2k 2n an d the length of each such in terval is k ways to have k heads is E e D2 n k 1 22 n k 22 n 2n k0 k 2 , so that 2n 2 2n n . Each bar is of width 2n g 2k 2n 1 2n 22 n k0 2n k 2 2n g 2k 2n 2n ....
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This note was uploaded on 01/19/2014 for the course MATH 5002 taught by Professor Adams during the Spring '08 term at Minnesota.

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