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Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Fall 2008) Review Problems – Derived, Markov, Chebyshev, Convergence, Bernoulli Process 1. Let X and Y be the distances of Xavier’s and Yolanda’s throws, respectively. The throws are independent, so the joint PDF is f X,Y ( x, y ) = f X ( x ) f Y ( y ) = 1 100 1 60 e - y 60 , 0 x 100 , y 0 0 otherwise (a) E [ X ] = 50 , E [ Y ] = 60. (b) We want to find P ( X > Y ) and compare it to P ( X < Y ). To do this, we look at the joint sample space: Thus, P ( X > Y ) = 100 0 x 0 f X,Y ( x, y ) dydx = 100 0 x 0 1 100 1 60 e - y 60 dydx 0 . 5133 . We also have that P ( Y > X ) = 1 - P ( X > Y ) 0 . 4867. Looking just at the expected values of the individual random variables might suggest that Yolanda is more likely to throw further, but we have found that the probability of Xavier throwing further is slightly higher. (c) Let W = Y - X . First find the CDF of W : F W ( w ) = P ( W w ) = P ( Y - X w ) = P ( Y X + w ) The event is shown below. For - 100 w 0, we have F W ( w ) = 100 - w x + w 0 f X,Y ( x, y ) dydx = 100 - w x + w 0 1 100 1 60 e - y 60 dydx = 1 100 (60 e - 1 60 ( w +100) + w + 40) Compiled November 13, 2008 Page 1 of 5
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Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Fall 2008) For w 0, we have F W ( w ) = 100 0 x + w 0 f X,Y ( x, y ) dxdy = 100 0 x + w 0 1 100 1 60 e - y 60 dydx = 3 5 e - w 60 ( e - 5 3 - 1) + 1 Now we differentiate to obtain the PDF: f W ( w ) = d dw F W ( w ) = 1 100 (1 - e - 1 60 ( w +100) ) , - 100 w 0 1 100 e - w 60 (1 - e - 5 3 ) , w 0 0 otherwise 2. a) The random variables, Y i , are not independent. We can guess this intuitively by observing that consecutive Y i depend on the same value of X i ; we can also prove it using variances.
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