EE 503 Problem Set #7 Due Thursday, October 15, 2015 Fall 2015 Sawchuk ANNOUNCEMENT Friday October 16 - Lecture 16 will be recorded in advance - 3:00 pm-4:50 pm - OHE 132 - please attend live recording if possible - playback at usual time Tuesday October 20 from 10:00 am-11:50 am; TA will be present 7-1 (45) Two random variables X and Y have the joint probability density $ 2, x 0 and y 0 and x + y 1 & f XY ( x, y ) = % &0, elsewhere ' Answer each of these independent questions about X, Y, carefully indicating the domain of all functions where needed. Parts a). - i). are 4 points; part j). is 9 points. a). Find E(Z), where Z is a new random variable defined by Z = XY % 3" b). A is the event & X > # . Find P(A). ' 4$ c). B is the event {2X>Y}. Find P(B). d). Find f X ( x ) . e). Find fY |X ( y | x ) . f). Are X and Y independent? Why or why not? g). Find E(Y | X = x) . % 3" h). Find and sketch the density f X |A ( x | A ) where A is the event & X > # as in part b). ' 4$ % 3" i). Find and sketch the density fY |A ( y | A ) where A is the event & X > # as in part b). ' 4$ j). A new random variable W is defined by W = X + Y. Find and sketch FW ( w) and fW (w ) . 7-2 (15) Text problem 4.85. (Linear function - Gaussian) 7-3 (15) Text problem 4.86. (Function of uniform RV) 7-4 (15) Text problem 5.80. (Conditional probability - problem 5.26)) 7-5 (15) Random variable X has probability density f X ( x ) = k(2 x), 0 x 2, with zero value elsewhere. a). Find the value of k and the distribution function FX (x) . 2 b). A new RV Y is defined by Y = X . Find E(Y ) and probability density of Y. 7-6 (15) Random variable X has a Gaussian distribution with mean = 1 and standard deviation = 0.6. Find the following in terms of Q functions: a). P( X > 0) b). P(0.2 < X < 1.8) c). f X |A ( x | A) where A = {0.5 < X < 1.5} ----------------------------Useful series expansions: x2 x3 x 4 e x = 1+ x + + + ..............for all real x 2! 3! 4! 1 = 1+ x + x 2 + x 3 ................for x < 1 1 x Derivative of the quotient of two variables: " u % vdu udv d$ ' = #v& v2 1 2 3 2 = 1+ 2x + 3x + 4x + .................for x < 1 (1 x) x2 x 3 x 4 ln(1 x) = x + + + .................for x < 1 2 3 4