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# solution06 - Solutions to Homework 06 1 a From the gure...

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Solutions to Homework 06 1. a. From the figure, f X ( x ) = ( 0 | x | > a c 1 - | x | a | x | ≤ a Now, 1 = Z -∞ f X ( x ) dx = Z a - a c 1 - | x | a ! dx = 2 Z a 0 c 1 - x a dx = 2 c " x - x 2 2 a # a 0 = ac c = 1 a b. F X ( x ) = 0 for x < - a and F X ( x ) = 1 for x > a . For - a x 0, | x | = - x . Then F X ( x ) = 1 a Z x - a 1 + x 0 a ! dx 0 = 1 2 + 1 a x + x 2 2 a ! For 0 x a , | x | = x . Then F X ( x ) = Z 0 - a f X ( x 0 ) dx 0 + Z x 0 f X ( x 0 ) dx 0 = 1 2 + 1 a Z x - a 1 + x 0 a ! dx 0 = 1 2 + 1 a x - x 2 2 a ! c. From part b , we know 0 < b < a . 1 2 = P [ | X | < b ] = F X ( b ) - F X ( - b ) = 2 a b - b 2 2 a ! b = a 1 - 1 2 ! 1

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2. For a geometric random variable, P [ X = k ] = (1 - p ) k - 1 p , k = 1 , 2 , · · · . a. A = { X > k } . F X ( x | A ) = P [ { X x } ∩ A ] P [ A ] = P [ { X x } ∩ { X > k } ] P [ { X > k } ] = ( F X ( x ) - F X ( k ) 1 - F X ( k ) for x > k 0 else = ( (1 - p ) k - (1 - p ) x (1 - p ) k for x > k 0 else b. A = { X < k } . F X ( x | A ) = P [ { X x } ∩ A ] P [ A ] = P [ { X x } ] F X ( k - 1) for x k - 1 P [ A ] P [ A ] else = ( 1 - (1 - p ) x 1 - (1 - p ) k - 1 for x k - 1 1 else c. A = { X is even } . P [ A ] = X k =1 p (1 - p ) 2 k - 1 = p ((1 - p ) + (1 - p ) 3 + · · · ) = p 1 - p 1 - (1 - p ) 2 F X ( x | A ) = P [ { X x } ∩ A ] P [ A ] = x/ 2 k =1 p (1 - p ) 2 k - 1 P [ A ] = p (1 - p ) 1 - (1 - p ) x 1 - (1 - p ) 2 p (1 - p ) 1 1 - (1 - p ) 2 = 1 - (1 - p ) x 3.
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