Quiz6Spring08 - V X n = 9 n 4 Alternative Solution Let Y =...

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Math 213, Quiz 6 (Solutions); Friday, February 29, 2008 1. Let n 2 be an integer. A bit-string of length n is chosen at random. Let X = 3 j where j is the number of 1-bits in the string chosen. Find EX and V X . Provide a detailed explanation of your answer. Solution. For i = 1 , 2 . . . n let X i = ± 3 , if i -th bit is 1 0 , if i -th bit is 0 . Then X = X 1 + ··· + X n and hence EX = EX 1 + . . . EX n . Moreover, since X 1 , . . . , X n are clearly independent, we also have V X = V X 1 + ··· + V X n . We have EX i = 3 · 1 2 + 0 · 1 2 = 3 2 , E ( X 2 i ) = 3 2 · 1 2 + 0 2 · 1 2 = 9 2 , and hence V X i = E ( X 2 i ) - ( EX i ) 2 = 9 2 - 9 4 = 9 4 . Therefore EX = EX 1 + . . . EX n = 3 n 2 and V X = V X 1 + ···
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Unformatted text preview: + V X n = 9 n 4 Alternative Solution. Let Y = j be the number of 1-bits in the string chosen. Then X = 3 Y and the random variable Y has the binomial distribution with the parameters p = q = 1 / 2 on the set { , 1 , . . . , n } . Therefore, by the results of Example 5 and Example 18 in Ch 6.4 we have EX = np = n/ 2 and V X = npq = n/ 4. Since X = 3 Y , we have EX = 3 EY = 3 n/ 2 and V X = 3 2 V Y = 9 n/ 4. 1...
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This note was uploaded on 01/27/2010 for the course MATH 213 taught by Professor Jack during the Spring '08 term at University of Illinois at Urbana–Champaign.

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