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Unformatted text preview: CS237. Problem Set 8: Variance, Geometric Distribution and Coupon Collector Problem Graded Problems due Thurs March 31 11:59PM. April 9, 2011 Reading. Schaum’s Chapter 5.4, 5.5. M&U 2.4 and 3.2 and Lemma 3.8 on page 51. 1 Practice Problems. Exercise 1. In class we proved that if X 1 ,X 2 are independent random variables, then VAR[ X 1 + X 2 ] = VAR[ X 1 ] + VAR[ X 2 ] (1) 1. Show that equation (1) does not hold if X 1 and X 2 are not independent. (To do this, give a specific counterexample, i.e., give example dependent RV’s X 1 and X 2 that show that (1) does not hold.) Ilir notes: Consider the following example. Roll a die n times. Let X be the number of 6 and Y be the number of 1 . X i is indicator if on the i th run 6 appeared, and Y i is the indicator if on the i th trial 1 appeared. We have X = X 1 + ··· + X n and Y = Y 1 + ··· + Y n . Now, EX = EY = n/ 6 and VAR X = VAR Y = 5 / 36 n . However, VAR( X + Y ) VAR X VAR Y 6 = 0 . 2. Prove the general version of this theorem. This is, for independent random variables X 1 ,X 2 ,...,X n , prove that VAR[ n X i =1 X i ] = n X i =1 VAR[ X i ] (2) Ilir notes: Let’s let μ i = E [ X i ] . We have VAR( n X i =1 X i ) = E " n X i =1 ( X i μ i ) ! 2 # = n X i =1 E ( X i μ i )2 + X i 6 = j E (( X i μ i )( X j μ j )) = n X i =1 VAR[ X i ] + X i 6 = j E [( X i μ i )( X j μ j )] To complete the proof we need only show that for any independent pair X j ,X i where i 6 = j we have E [( X i μ i )( X j μ j )] = 0 We do this as follows: E [( X i μ i )( X j μ j )] = E [ X i X j μ i X j μ j X i + μ i μ j ] = E [ X i X j ] μ i E [ X j ] μ j [ X i ] + μ i μ j ] = E [ X i X j ] μ i μ j = E [ X i ] E [ X j ] μ i μ j = μ i μ j μ i μ j = 0 1 were we used linearity of expectation to go from the first line to the second line, the definition of μ i ,μ j to go from the second to the third line, and the independence of X i ,X j to go from the third line to the fourth. This completes our proof....
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 Spring '11
 Goldberg
 Probability theory, Ilir

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