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MIT6_041F10_assn05_sol

# MIT6_041F10_assn05_sol - Massachusetts Institute of...

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Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Fall 2010) Problem Set 5: Solutions 1. (a) Because of the required normalization property of any joint PDF, 2 2 2 2 3 1 3 2 1 = ax dy dx = ax (2 x ) dx = a 2 2 1 2 + = a 3 3 3 x =1 y = x x =1 so a = 3 / 2. (b) For 1 y 2, y a 3 f Y ( y ) = ax dx = ( y 2 1) = ( y 2 1) , 2 4 1 and f Y ( y ) = 0 otherwise. (c) First notice that for 1 x 3 / 2, f X,Y ( x, 3 / 2) (3 / 2) x 8 x f X | Y ( x | 3 / 2) = = � � 2 = . f Y (3 / 2) 3 3 1 2 5 4 2 Therefore, 3 / 2 1 8 x E [1 /X | Y = 3 / 2] = dx = 4 / 5 . x 5 1 2. (a) By definition f X,Y ( x, y ) = f X ( x ) f Y | X ( y | x ). f X ( x ) = ax as shown in the graph. We have that 40 1 = ax dx = 800 a. 0 So f X ( x ) = x/ 800. From the problem statement f Y | X ( y | x ) = 2 1 x for y [0 , 2 x ]. Therefore, 1 / 1600 , if 0 x 4 and 0 < y < 2 x, f X,Y ( x, y ) = 0 , otherwise . (b) Paul makes a positive profit if Y > X . This occurs with probability � � 40 2 x 1 1 P ( Y > X ) = f X,Y ( x, y ) dy dx = dy dx = . 1600 2 y>x 0 x We could have also arrived at this answer by realizing that for each possible value of X , there is a 1 / 2 probability that Y > X . (c) The joint density function satisfies f X,Z ( x, z ) = f X ( x ) f Z | X ( z | x ). Since Z is conditionally uniformly distributed given X , f Z | X ( z | x ) = 2 1 x for x z x . Therefore, f X,Z ( x, z ) = 1 / 1600 for 0 x 40 and x z x . The marginal density f z ( z ) is calculated as 40 1 40 −| z | , if | z | < 40 , f Z ( z ) = f X,Z ( x, z ) dx = dx = 1600 x x = | z | 1600 0 , otherwise. Page 1 of 7

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Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Fall 2010) 3. (a) In order for X and Y to be independent, any observation of X should not give any information on Y . If X is observed to be equal to 0, then Y must be 0. In other words, f Y |{ X =0 } ( y | 0) negationslash f Y ( y ). Therefore, X and Y are not independent. = 1.5 1 0 , otherwise. 0.5 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x 2 1.8 1.6 1.4 x/ 2 , if 0 x 1 , (b) f X ( x ) = 3 x/ 2 + 3 , if 1 < x 2 , f X | Y ( x | 0 . 5) f Y | X ( y | 0 .
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MIT6_041F10_assn05_sol - Massachusetts Institute of...

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