( Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) Recitation 08 Answers March 09, 2006 1. (a) The marginal distributions are obtained by integrating the joint distribution along the X and Y axes and is sh own in the f ollowing figure. y y0 0 2.0 f x,y(x 0,y0) = 0.11.0 x 0 -1.0 -2.0 0.2 -1.0 1.0 2.0 0.4 0.2 0.2 f(x) yf (y) 0.3 x 1.0 2.0-1.0 -1.0 1.0 2.0 -2.0 x 0 Figure 1: Marginal probabilities fX(x) and fY (y) obtained by integration along the y and x axes respectively The conditional PDFs are as shown in the figure below. (b) X and Y are NOT indepenent since fXY (x, y) 6= fX(x)fY (y). Also, fr om the fi gures we have fX|Y (x|y) 6 fX(x). = (c) fX,Y ((x,y) (x, y) ∈ A fX,Y |A(x, y) = P(A) 0 otherwise = ( 0.1 π0.1 (x, y) ∈ A 0 otherwise (d) E[X|Y = y] = 0 1 2 0 −2.0 ≤ y ≤ −1.0 −1.0 ≤ y ≤ 1.0 1.0 ≤ y ≤ 2.0 Page 1 of 3Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) f (x|y) {1 < y <= 2} x|y f (x|y) x|y -1.0 2.0 1/3 {-1<y<=1} f (x|y) x|y -1.0 1.0 1/2 {-2<y<=2} f (y|x) y|x f (y|x) y|x y 0 y 0 1.0-1.0 1/2 2.0 -2.0 1/4 1.0 -1.0 1/2 y0 x 01.0 2.0-1.0 -1.0 1.0 2.0 f (x ,y ) = 0.10x,y 0 -2.0 {-1<x<=1} {1<x<=2} Figure 2: Conditional Probabilities Page 2 of 3Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) The conditional variance var(X|Y = y) is given by var(X|Y = y) = 4 12 −2.0 ≤ y ≤ −1.0 9 12 −1.0 ≤ y ≤ 1.0 4 12 1.0 ≤ y ≤ 2.0 2. (a) We have a = 1/800, so that fXY (x, y) = ( 1/1600 if 0 ≤ x ≤ 40 and 0 ≤ y ≤ 2x 0, otherwise. ) (b) P(Y > X) = 1/2 (c) Let Z = Y − X. We have fZ(z) = 1 1600 z + 1 40 , if − 40 ≤ z ≤ 0, − 1 1600 + 1 40 , if 0 ≤ z ≤ 40, 0, otherwise. E[Z] = 0. Page 3 of
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