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Unformatted text preview: Solutions to Homework 06 1. a. From the figure, f X ( x ) = (  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 c 1 x a dx = 2 c " x x 2 2 a # a = 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 a ! dx = 1 2 + 1 a x + x 2 2 a ! For 0 ≤ x ≤ a ,  x  = x . Then F X ( x ) = Z a f X ( x ) dx + Z x f X ( x ) dx = 1 2 + 1 a Z x a 1 + x a ! dx = 1 2 + 1 a x x 2 2 a ! c. 1 4 = P [  X  < b ] = F X ( b ) F X ( b ) = 2 a b b 2 2 a ! ⇒ b = a 1 s 3 4 1 2. F X ( x ) = n X k =0 n k ! p k (1 p ) n k u ( x k ) f X ( x ) = n X k =0 n k ! p k (1 p ) n k δ ( x k ) a. Following the example done in class (refer to page 99, Example 3.10 of textbook) F X ( x  A ) = ( x < 4 F X ( x ) F X (4) 1 F X (4) for x > 4 f X ( x  A ) = ( x < 4 f X ( x ) 1 F X (4) for x > 4 b....
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This note was uploaded on 09/07/2009 for the course ECE 302 taught by Professor Gelfand during the Fall '08 term at Purdue University.
 Fall '08
 GELFAND

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