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Unformatted text preview: 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 , ≤ x ≤ 100 ,y ≥ 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 ) = Z 100 Z x f X,Y ( x,y ) dydx = Z 100 Z x 1 100 1 60 e- y 60 dydx ≈ . 5133 . We also have that P ( Y > X ) = 1- P ( X > Y ) ≈ . 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 ) = Z 100- w Z x + w f X,Y ( x,y ) dydx = Z 100- w Z x + w 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 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 ) = Z 100 Z x + w f X,Y ( x,y ) dxdy = Z 100 Z x + w 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 ≤ 1 100 e- w 60 (1- e- 5 3 ) , w ≥ otherwise 2. a) The random variables,2....
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This note was uploaded on 06/10/2010 for the course ELECTRONIC C1002900 taught by Professor Hyungdongshin during the Spring '10 term at Kyung Hee.

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solutions - Massachusetts Institute of Technology...

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