soln8 - Probability and Stochastic Processes A Friendly...

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Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman,5.1.2 5.2.1 5.3.1 5.3.2 5.5.1 5.5.2 5.6.1 5.6.2 5.7.1 and 5.8.1 Problem 5.1.2 (a) Because the probability that any random variable is less than - is zero, we have F X , Y ( x , - ) = P [ X x , Y ≤ - ] P [ Y ≤ - ] = 0 (b) The probability that any random variable is less than infinity is always one. F X , Y ( x , ) = P [ X x , Y ] = P [ X x ] = F X ( x ) (c) Although P [ Y ] = 1, P [ X ≤ - ] = 0. Therefore the following is true. F X , Y ( - , ) = P [ X ≤ - , Y ] P [ X ≤ - ] = 0 (d) Part (d) follows the same logic as that of part (a). F X , Y ( - , y ) = P [ X ≤ - , Y y ] P [ X ≤ - ] = 0 (e) Analogous to Part (b), we find that F X , Y ( , y ) = P [ X , Y y ] = P [ Y y ] = F Y ( y ) Problem 5.2.1 (a) The joint PDF of X and Y is f X , Y ( x , y ) = c x + y 1 , x , y 0 0 otherwise Y X Y + X = 1 1 1 To find the constant c we integrate over the region shown. This gives 1 0 1 - x 0 cdydx = cx - cx 2 1 0 = c 2 = 1 Therefore c = 2. 1
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(b) To find the P [ X Y ] we look to integrate over the area indicated by the graph P [ X Y ] = 1 / 2 0 1 - x x dydx = 1 / 2 0 ( 2 - 4 x ) dx = 1 / 2 Y X X=Y 1 1 X Y £ (c) The probability P [ X + Y 1 / 2 ] can be seen in the figure at right. Here we can set up the fol- lowing integrals P [ X + Y 1 / 2 ] = 1 / 2 0 1 / 2 - x 0 2 dydx = 1 / 2 0 ( 1 - 2 x ) dx = 1 / 2 - 1 / 4 = 1 / 4 Y X Y + X = 1 Y + X = ½ 1 1 Problem 5.3.1 (a) The joint PDF (and the corresponding region of nonzero probability) are f X , Y ( x , y ) = 1 / 2 - 1 x y 1 0 otherwise Y X 1 -1 (b) P [ X > 0 ] = 1 0 1 x 1 2 dydx = 1 0 1 - x 2 dx = 1 / 4 This result can be deduced by geometry. The shaded triangle of the
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