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20065ee131A_1_EE131AF06HW5Solution

# 20065ee131A_1_EE131AF06HW5Solution - Solution of Assignment...

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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 > 0 , 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 > 0 , 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 σ = 4000 miles. (a) What is the probability that such a tire lasts over 40,000 miles? (b) What is the probability that it lasts between 30,000 and 35,000 miles?

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(c) Given that it has survived 30,000 miles, what is the conditional probability that it survives another 10,000 miles?
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