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Unformatted text preview: Math 213 Exam 2 (Solutions) Prof. I.Kapovich April 10, 2007 Problem 1 A binary string x of length 20 is chosen at random. Let X be the random variable where X ( x ) is the number of 0s in x plus twice the number of 1s in x . Find E ( X ) and V ( X ). Solution. For i = 1 , 2 , . . . , 20 let X i = 1 , if the ith bit is 0 2 , if the ibit is 1 Then X = X 1 + · · · + X 20 . Clearly, the random variables X 1 , . . . , X 20 are independent. Therefore EX = ∑ 20 i =1 EX i = 20 EX 1 and V X = ∑ 20 i =1 V X i = 20 V X 1 . We have EX 1 = 1 · 1 2 + 2 · 1 2 = 3 2 and E ( X 2 1 ) = 1 2 · 1 2 + 2 2 · 1 2 = 5 2 Therefore EX = 20 X i =1 EX i = 20 EX 1 = 20 · 3 2 = 30 , and V ( X ) = 20 V X 1 = 20 ( E ( X 2 1 ) ( EX 1 ) 2 ) = 20 5 2 9 4 = 20 · 1 4 = 5 . Problem 2 Find the general solution of the recurrence relation ( † ) a n = 4 a n 1 4 a n 2 + n. Solution. The associated homogeneous relation is a n = 4 a n 1 4 a n 2 . It has characteristic equation r 2 = 4 r 4, that is r 2 4 r...
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 Spring '08
 JACK
 Math, Equivalence relation, Transitive relation, α, random variables X1, ﬁrst ﬂip

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