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# hw10sol - ISyE 2027B Probability with Applications Fall...

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ISyE 2027B Probability with Applications Fall 2010 Homework 10 Solutions 1. X Ber ( p ), Y Ber ( q ), p = 1 / 2 and q = 1 / 4. P ( X + Y = 0) = P ( X = 0 , Y = 0) = (1 - p )(1 - q ) = 1 2 × 3 4 = 3 8 P ( X + Y = 1) = P ( X = 0 , Y = 1) + P ( X = 1 , Y = 0) = (1 - p )( q ) + p (1 - q ) = 1 2 × 1 4 + 1 2 × 3 4 = 1 2 P ( X + Y = 2) = P ( X = 1 , Y = 1) = pq = 1 2 × 1 4 = 1 8 If X + Y Bin (2 , t ), t is an unknown constant. t should satisfy P ( X + Y = 2) = 2 2 t 2 = 1 8 We get t = 1 2 2 . Check P ( X + Y = 0) = 2 0 (1 - t ) 2 6 = 3 8 So there does not exist t such that X + Y Bin (2 , t ). Therefore X + Y does not have a binomial distribution. 2. Notice R -∞ f X ( x ) dx and R -∞ f Y ( y ) dy are unbounded. We need to check the domain of X and Y . Since R 0 f X ( x ) dx = 1 and R 0 f Y ( y ) dy = 1, both X and Y are defined on [0 , ). According to the addition rule, f Z ( z ) = Z z 0 f X ( z - y ) f Y ( y ) dy = Z z 0 1 4 ( z - y ) e - ( z - y ) / 2 1 4 ye - y/ 2 dy = 1 16 e - z/ 2 Z z 0 ( z - y ) ydy = z 3 96 e - z/ 2 . 3. Put Z = X 1 + X 2 + X 3 . Then Z can be interpreted as the time of the third success in a series of independent experiments with probability p of success. Then Z = k means that the first ( k - 1) experiments contained

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