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MIT6_041F10_assn06_sol

# MIT6_041F10_assn06_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 6: Solutions 1. Let us draw the region where f X,Y ( x, y ) is nonzero: y 2 1 y-x=z 0 1 2 x x =2 y = x The joint PDF has to integrate to 1. From x =1 y =0 ax dy dx = 3 7 a = 1, we get a = 7 3 . 2 3 9 1 7 x dx, if 0 y 1 , , if 0 y 1 , 14 (a) b f Y ( y ) = f X,Y ( x, y ) dy = 2 3 x dx, if 1 < y 2 , = 3 (4 y 2 ) , if 1 < y 2 , y 7 14 0 , otherwise 0 , otherwise . (c) f X | Y ( x | 3 ) = f X,Y ( x, 2 3 ) = 8 x, for 3 2 x 2 and 0 otherwise . 2 f Y ( 3 2 ) 7 Then, 2 1 3 1 8 4 E | Y = = x dx = . X 2 3 / 2 x 7 7 (d) We use the technique of first finding the CDF and differentiating it to get the PDF. F Z ( z ) = P ( Z z ) = P ( Y X z ) 0 , if z < 2 ,  � x =2 y = x + z 3 8 6 3 x dy dx = + z 1 z , if 2 z ≤ − 1 , 7 7 14 7 x = z y =0 = x =2 y = x + z 3 x dy dx = 1 + 9 z, if 1 < z 0 , 14 7 x =1 y =0 1 , if 0 < z. 6 3 z 2 , if 2 z ≤ − 1 , d 7 14 9 f Z ( z ) = F Z ( z ) = 14 , if 1 < z 0 , dz 0 , otherwise . Page 1 of 4

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Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Fall 2010) 2. The PDF of Z , f Z ( z ), can be readily computed using the convolution integral: f Z ( z ) = f X ( t ) f Y ( z t ) dt.
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MIT6_041F10_assn06_sol - Massachusetts Institute of...

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