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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 2010) Problem Set 4: Solutions 1. (a) From the joint PMF, there are six ( x , y ) coordinate pairs with nonzero probabilities of occurring. These pairs are (1, 1), (1, 3), (2, 1), (2, 3), (4, 1), and (4, 3). The probability of a pair is proportional to the sum of the squares of the coordinates of the pair, x 2 + y 2 . Because the probability of the entire sample space must equal 1, we have: (1 + 1) c + (1 + 9) c + (4 + 1) c + (4 + 9) c + (16 + 1) c + (16 + 9) c = 1 . 1 Solving for c , we get c = 72 . (b) There are three sample points for which y < x : P ( Y < X ) = P ( { (2 , 1) } ) + P ( { (4 , 1) } ) + P ( { (4 , 3) } ) = 5 + 17 + 25 = 47 . 72 72 72 72 (c) There are two sample points for which y > x : P ( Y > X ) = P ( { (1 , 3) } ) + P ( { (2 , 3) } ) = 10 + 13 = 23 . 72 72 72 (d) There is only one sample point for which y = x : P ( Y = X ) = P ( { (1 , 1) } ) = 2 . 72 Notice that, using the above two parts, 47 23 2 P ( Y < X ) + P ( Y > X ) + P ( Y = X ) = + + = 1 72 72 72 as expected. (e) There are three sample points for which y = 3: 10 13 25 48 P ( Y = 3) = P ( { (1 , 3) } ) + P ( { (2 , 3) } ) + P ( { (4 , 3) } ) = + + = 72 . 72 72 72 (f) In general, for two discrete random variable X and Y for which a joint PMF is defined, we have p X ( x ) = p X,Y ( x, y ) and p Y ( y ) = p X,Y ( x, y ) . y = x = In this problem the ranges of X and Y are quite restricted so we can determine the marginal PMFs by enumeration. For example, 18 p X (2) = P ( { (2 , 1) } ) + P ( { (2 , 3) } ) = . 72 Overall, we get: 12 / 72 , if x = 1 , 24 / 72 , if y = 1 , 18 / 72 , if x = 2 , p X ( x ) = and p Y ( y ) = 48 / 72 , if y = 3 , 42 / 72 , if x = 4 , , otherwise . , otherwise Page 1 of 7 Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Fall 2010) (g) In general, the expected value of any discrete random variable X equals E [ X ] = xp X ( x ) . x = For this problem, 12 18 42 E [ X ] = 1 + 2 + 4 = 3 72 72 72 and 24 48 7 E [ Y ] = 1 + 3 = 3 . 72 72 To compute E [ XY ], note that p X,Y ( x, y ) = negationslash p X ( x ) p Y ( y ). Therefore, X and Y are not inde pendent and we cannot assume E [ XY ] = E [ X ] E [ Y ]. Thus, we have E [ XY ] = xyp X,Y ( x, y ) x y 2 5 17 10 13 25 61 = 1 + 2 + 4 + 3 + 6 + 12 = . 72 72 72 72 72 72 9 (h) The variance of a random variable X can be computed as E [ X 2 ] E [ X ] 2 or as E [( X E [ X ]) 2 ]....
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 Fall '10
 Prof.DimitriBertsekas
 Electrical Engineering

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