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Unformatted text preview: Probability and Stochastic Processes Problem set 8 Solutions Spring’2010 Problem Solutions : Yates and Goodman, 4.5.2 4.5.5 4.5.6 4.6.4 4.6.6 4.6.8 4.6.10 4.7.2 4.7.5 4.7.6 4.7.7 4.7.9 4.8.1 4.8.2 and 4.8.5 Problem 4.5.2 Solution f X,Y ( x,y ) = braceleftbigg 2 x + y ≤ 1 ,x,y ≥ 0 otherwise (1) Y X Y + X = 1 1 1 Using the figure to the left we can find the marginal PDFs by inte grating over the appropriate regions. f X ( x ) = integraldisplay 1 − x 2 dy = braceleftbigg 2(1 x ) 0 ≤ x ≤ 1 otherwise (2) Likewise for f Y ( y ): f Y ( y ) = integraldisplay 1 − y 2 dx = braceleftbigg 2(1 y ) 0 ≤ y ≤ 1 otherwise (3) Problem 4.5.5 Solution In this problem, the joint PDF is f X,Y ( x,y ) = braceleftbigg 2  xy  /r 4 ≤ x 2 + y 2 ≤ r 2 otherwise (1) (a) Since  xy  =  x  y  , for r ≤ x ≤ r , we can write f X ( x ) = integraldisplay ∞ −∞ f X,Y ( x,y ) dy = 2  x  r 4 integraldisplay √ r 2 − x 2 − √ r 2 − x 2  y  dy (2) Since  y  is symmetric about the origin, we can simplify the integral to f X ( x ) = 4  x  r 4 integraldisplay √ r 2 − x 2 y dy = 2  x  r 4 y 2 vextendsingle vextendsingle vextendsingle vextendsingle √ r 2 − x 2 = 2  x  ( r 2 x 2 ) r 4 (3) Note that for  x  > r , f X ( x ) = 0. Hence the complete expression for the PDF of X is f X ( x ) = braceleftBigg 2  x  ( r 2 − x 2 ) r 4 r ≤ x ≤ r otherwise (4) (b) Note that the joint PDF is symmetric in x and y so that f Y ( y ) = f X ( y ). 1 Problem 4.5.6 Solution (a) The joint PDF of X and Y and the region of nonzero probability are Y X 1 1 f X,Y ( x,y ) = braceleftbigg cy ≤ y ≤ x ≤ 1 otherwise (1) (b) To find the value of the constant, c , we integrate the joint PDF over all x and y . integraldisplay ∞ −∞ integraldisplay ∞ −∞ f X,Y ( x,y ) dxdy = integraldisplay 1 integraldisplay x cy dy dx = integraldisplay 1 cx 2 2 dx = cx 3 6 vextendsingle vextendsingle vextendsingle vextendsingle 1 = c 6 . (2) Thus c = 6. (c) We can find the CDF F X ( x ) = P [ X ≤ x ] by integrating the joint PDF over the event X ≤ x . For x < 0, F X ( x ) = 0. For x > 1, F X ( x ) = 1. For 0 ≤ x ≤ 1, Y X 1 x 1 F X ( x ) = integraldisplayintegraldisplay x ′ ≤ x f X,Y ( x ′ ,y ′ ) dy ′ dx ′ (3) = integraldisplay x integraldisplay x ′ 6 y ′ dy ′ dx ′ (4) = integraldisplay x 3( x ′ ) 2 dx ′ = x 3 . (5) The complete expression for the joint CDF is F X ( x ) = x < x 3 ≤ x ≤ 1 1 x ≥ 1 (6) (d) Similarly, we find the CDF of Y by integrating f X,Y ( x,y ) over the event Y ≤ y . For y < 0, F Y ( y ) = 0 and for y > 1, F Y ( y ) = 1. For 0 ≤ y ≤ 1, Y X 1 y 1 F Y ( y ) = integraldisplayintegraldisplay y ′ ≤ y f X,Y ( x ′ ,y ′ ) dy ′ dx ′ (7) = integraldisplay y integraldisplay 1 y ′ 6 y ′ dx ′ dy ′ (8) = integraldisplay y 6 y ′ (1 y ′ ) dy ′ (9) = 3( y ′ ) 2 2( y ′ ) 3 vextendsingle vextendsingle y = 3 y 2 2 y 3 . (10) 2 The complete expression for the CDF of...
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 Spring '08
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 Variance, CDF, dy dx, joint PDF

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