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Unformatted text preview: Probability and Stochastic Processes Homework 10 Solutions Spring2010 Problem Solutions : Yates and Goodman, 6.1.1 6.1.2 6.1.4 6.2.1 6.2.4 6.2.5 6.3.1 6.3.3 6.4.1 6.4.4 6.4.5 6.4.6 6.5.1 6.5.2 and 6.5.3 Problem 6.1.1 Solution The random variable X 33 is a Bernoulli random variable that indicates the result of flip 33. The PMF of X 33 is P X 33 ( x ) = 1 p x = 0 p x = 1 otherwise (1) Note that each X i has expected value E [ X ] = p and variance Var[ X ] = p (1 p ). The random variable Y = X 1 + + X 100 is the number of heads in 100 coin flips. Hence, Y has the binomial PMF P Y ( y ) = braceleftbigg ( 100 y ) p y (1 p ) 100 y y = 0 , 1 ,... , 100 otherwise (2) Since the X i are independent, by Theorems 6.1 and 6.3, the mean and variance of Y are E [ Y ] = 100 E [ X ] = 100 p Var[ Y ] = 100Var[ X ] = 100 p (1 p ) (3) Problem 6.1.2 Solution Let Y = X 1 X 2 . (a) Since Y = X 1 + ( X 2 ), Theorem 6.1 says that the expected value of the difference is E [ Y ] = E [ X 1 ] + E [ X 2 ] = E [ X ] E [ X ] = 0 (1) (b) By Theorem 6.2, the variance of the difference is Var[ Y ] = Var[ X 1 ] + Var[ X 2 ] = 2Var[ X ] (2) Problem 6.1.4 Solution We can solve this problem using Theorem 6.2 which says that Var[ W ] = Var[ X ] + Var[ Y ] + 2Cov [ X,Y ] (1) The first two moments of X are E [ X ] = integraldisplay 1 integraldisplay 1 x 2 xdy dx = integraldisplay 1 2 x (1 x ) dx = 1 / 3 (2) E bracketleftbig X 2 bracketrightbig = integraldisplay 1 integraldisplay 1 x 2 x 2 dy dx = integraldisplay 1 2 x 2 (1 x ) dx = 1 / 6 (3) (4) 1 Thus the variance of X is Var[ X ] = E [ X 2 ] ( E [ X ]) 2 = 1 / 18. By symmetry, it should be apparent that E [ Y ] = E [ X ] = 1 / 3 and Var[ Y ] = Var[ X ] = 1 / 18. To find the covariance, we first find the correlation E [ XY ] = integraldisplay 1 integraldisplay 1 x 2 xy dy dx = integraldisplay 1 x (1 x ) 2 dx = 1 / 12 (5) The covariance is Cov [ X,Y ] = E [ XY ] E [ X ] E [ Y ] = 1 / 12 (1 / 3) 2 = 1 / 36 (6) Finally, the variance of the sum W = X + Y is Var[ W ] = Var[ X ] + Var[ Y ] 2Cov [ X,Y ] = 2 / 18 2 / 36 = 1 / 18 (7) For this specific problem, its arguable whether it would easier to find Var[ W ] by first deriving the CDF and PDF of W . In particular, for 0 w 1, F W ( w ) = P [ X + Y w ] = integraldisplay w integraldisplay w x 2 dy dx = integraldisplay w 2( w x ) dx = w 2 (8) Hence, by taking the derivative of the CDF, the PDF of W is f W ( w ) = braceleftbigg 2 w w 1 otherwise (9) From the PDF, the first and second moments of W are E [ W ] = integraldisplay 1 2 w 2 dw = 2 / 3 E bracketleftbig W 2 bracketrightbig = integraldisplay 1 2 w 3 dw = 1 / 2 (10) The variance of W is Var[ W ] = E [ W 2 ] ( E [ W ]) 2 = 1 / 18. Not surprisingly, we get the same answer both ways....
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This note was uploaded on 02/13/2011 for the course 332 226 taught by Professor Staff during the Spring '08 term at Rutgers.
 Spring '08
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