HW7-solution

HW7-solution - Probability and Stochastic Processes: A...

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Unformatted text preview: Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Edition 2 Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman, 4.1.1 4.1.5 4.2.2 4.2.6 4.2.7 4.3.3 4.3.4 4.4.4 4.5.3 4.6.7 and 4.6.8 Problem 4.1.1 Solution (a) The probability P [ X 2 , Y 3] can be found be evaluating the joint CDF F X,Y ( x, y ) at x = 2 and y = 3. This yields P [ X 2 , Y 3] = F X,Y (2 , 3) = (1- e- 2 )(1- e- 3 ) (1) (b) To find the marginal CDF of X, F X ( x ), we simply evaluate the joint CDF at y = . F X ( x ) = F X,Y ( x, ) = 1- e- x x otherwise (2) (c) Likewise for the marginal CDF of Y , we evaluate the joint CDF at X = . F Y ( y ) = F X,Y ( , y ) = 1- e- y y otherwise (3) Problem 4.1.5 Solution In this problem, we prove Theorem 4.5 which states P [ x 1 < X x 2 , y 1 < Y y 2 ] = F X,Y ( x 2 , y 2 )- F X,Y ( x 2 , y 1 ) (1)- F X,Y ( x 1 , y 2 ) + F X,Y ( x 1 , y 1 ) (2) (a) The events A , B , and C are Y X x 1 y 1 y 2 Y X x 1 x 2 y 1 y 2 Y X x 1 x 2 y 1 y 2 A B C (3) (b) In terms of the joint CDF F X,Y ( x, y ), we can write P [ A ] = F X,Y ( x 1 , y 2 )- F X,Y ( x 1 , y 1 ) (4) P [ B ] = F X,Y ( x 2 , y 1 )- F X,Y ( x 1 , y 1 ) (5) P [ A B C ] = F X,Y ( x 2 , y 2 )- F X,Y ( x 1 , y 1 ) (6) 1 (c) Since A , B , and C are mutually exclusive, P [ A B C ] = P [ A ] + P [ B ] + P [ C ] (7) However, since we want to express P [ C ] = P [ x 1 < X x 2 , y 1 < Y y 2 ] (8) in terms of the joint CDF F X,Y ( x, y ), we write P [ C ] = P [ A B C ]- P [ A ]- P [ B ] (9) = F X,Y ( x 2 , y 2 )- F X,Y ( x 1 , y 2 )- F X,Y ( x 2 , y 1 ) + F X,Y ( x 1 , y 1 ) (10) which completes the proof of the theorem. Problem 4.2.2 Solution On the X, Y plane, the joint PMF is- 6 ?...
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HW7-solution - Probability and Stochastic Processes: A...

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