20065ee131A_1_EE131AF06HW5Solution

20065ee131A_1_EE131AF06HW5Solution - Solution of Assignment...

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Unformatted text preview: Solution of Assignment 5, EE 131A Due: Monday November 20, 2006 Problem 1. (a) First prove that the memoryless property of a random variable X P [ X > k + j vextendsingle vextendsingle X > j ] = P [ X > k ] , k, j > , is equivalent to the following relation: P [ X > k + j ] = P [ X > k ] P [ X > j ] . Prove that the geometric random variable satisfies this property. (b) Suppose that X is a discrete random variable that satisfies the memoryless property. Show that this implies that X is a geometric random variable. Solution: (a) Since, for k, j > , P [ X > k + j vextendsingle vextendsingle X > j ] = P [ { X > k + j } { X > j } ] P [ X > j ] = P [ X > k + j ] P [ X > j ] it follows that the memoryless property is equivalent to P [ X > k + j ] = P [ X > k ] P [ X > j ] . (b) Let p = P [ X = 1] and q = 1- p . By induction, assume that P [ X k ] = 1- q k , then the memoryless property implies that P [ X k + 1] = 1- P [ X > k + 1] = 1- P [ X > k ] P [ X > 1] = 1- q k q = 1- q k +1 . Now, we have P [ X = k ] = P [ X k ]- P [ X k- 1] = 1- q k- 1 + q k- 1 = q k- 1 (1- q ) = q k- 1 p, which shows that X is a geometric random variable. Problem 2. The life of a certain type of automobile tire is normally distributed with mean m = 34 , 000 miles and standard deviation = 4000 miles....
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This note was uploaded on 06/09/2010 for the course EE 131A taught by Professor Lorenzelli during the Spring '08 term at UCLA.

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20065ee131A_1_EE131AF06HW5Solution - Solution of Assignment...

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