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# soln6 - f X x = ½ 1 4-1 ≤ x ≤ 3 otherwise We recognize...

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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,4.1.2 4.2.1 4.3.1 and 4.3.4 Problem 4.1.2 F V ( v ) = 0 v < - 5 c ( v + 5 ) 2 - 5 v < 7 1 v 7 (a) For V to be a continuous random variable, F V ( v ) must be a continuous function. This occurs if we choose c such that F V ( v ) doesn’t have a discontinuity at v = 7. We meet this requirement if c ( 7 + 5 ) 2 = 1. This implies c = 1 / 144. (b) P [ V > 4 ] = 1 - P [ V 4 ] = 1 - F V ( 4 ) = 1 - 81 / 144 = 63 / 144 (c) P [ - 3 < V 0 ] = F V ( 0 ) - F V ( - 3 ) = 25 / 144 - 4 / 144 = 21 / 144 (d) Since 0 F V ( v ) 1 and since F V ( v ) is a nondecreasing function, it must be that - 5 a 7. In this range, P [ V > a ] = 1 - F V ( a ) = 1 - ( a + 5 ) 2 / 144 = 2 / 3 The unique solution in the range - 5 a 7 is a = 4 3 - 5 = 1 . 928. Problem 4.2.1 f X ( x ) = cx 0 x 2 0 otherwise (a) From the above PDF we can determine the value of c by integrating the PDF and setting it equal to 1. 2 0 cxdx = 2 c = 1 Therefore c = 1 / 2. (b) P [ 0 X 1 ] = 1 0 x 2 dx = 1 / 4 (c) P [ - 1 / 2 X 1 / 2 ] = 1 / 2 0 x 2 dx = 1 / 16 (d) The CDF of X is found by integrating the PDF from 0 to x . F X ( x ) = x 0 f X ( x 0 ) dx 0 =

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Unformatted text preview: f X ( x ) = ½ 1 / 4-1 ≤ x ≤ 3 otherwise We recognize that X is a uniform random variable from [-1,3]. (a) E [ X ] = 1 and Var [ X ] = ( 3 + 1 ) 2 12 = 4 / 3. (b) The new random variable Y is defined as Y = h ( X ) = X 2 . Therefore h ( E [ X ]) = h ( 1 ) = 1 and E [ h ( X )] = E £ X 2 ¤ = Var [ X ]+ E [ X ] 2 = 4 / 3 + 1 = 7 / 3 Finally E [ Y ] = E [ h ( X )] = E £ X 2 ¤ = 7 / 3 Var [ Y ] = E £ X 4 ¤-E £ X 2 ¤ 2 = 3-1 x 4 4 dx-49 9 = 61 5-49 9 Problem 4.3.4 (a) We can find the expected value of X by direct integration of the given PDF. f Y ( y ) = ½ y / 2 ≤ y ≤ 2 otherwise The expectation is E [ Y ] = 2 y 2 2 dy = 4 / 3 (b) E £ Y 2 ¤ = 2 y 3 2 dy = 2 Var [ Y ] = E £ Y 2 ¤-E [ Y ] 2 = 2-( 4 / 3 ) 2 = 2 / 9 2...
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soln6 - f X x = ½ 1 4-1 ≤ x ≤ 3 otherwise We recognize...

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