# hw2sol - UCSD ECE 153 Handout #8 Prof. Young-Han Kim...

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Unformatted text preview: UCSD ECE 153 Handout #8 Prof. Young-Han Kim Thursday, October 16, 2008 Solutions to Homework Set #2 (Prepared by TA Halyun Jeong) 1. Read Sections 3.1–3.2, 3.4–3.6, 4.1–4.2, 4.4–4.5, 4.7, 4.9, 5.1-5.5, 5.7-5.8 in the text. Try to work on all examples. 2. Juror’s fallacy. Suppose that P ( A | B ) ≥ P ( A ) and P ( A | C ) ≥ P ( A ). Is it always true that P ( A | B,C ) ≥ P ( A ) ? Prove or provide a counterexample. Solution: The answer is no. There are many counterexamples that can be given. For example, suppose a fair die is thrown and let X denote the number of dots. Let A be the event that X = 3 or 6; let B be the event that X = 3 or 5; and let C be the event that X = 5 or 6. Then, we have P ( A ) = 1 / 3 , P ( A | B ) = P ( A | C ) = 1 / 2 , but P ( A | B,C ) = 0 . Apparently, having two positive evidences does not necessarily lead to a stronger evidence. 3. Let X be a geometric random variable with pmf p X ( k ) = p (1- p ) k- 1 , k = 1 , 2 ,... . Find and plot the conditional pmf p X ( k | A ) = P { X = k | X ∈ A } if: (a) A = { X &gt; m } where m is a positive integer. (b) A = { X &lt; m } . (c) A = { X is an even number } . Comment on the shape of the conditional pmf of part (a). Solution: (a) We have P ( A ) = ∞ summationdisplay n = m +1 p (1- p ) n- 1 = ∞ summationdisplay n =0 p (1- p ) n + m = p (1- p ) m ∞ summationdisplay n =0 (1- p ) n = (1- p ) m . 1 For k ≤ m , p X ( k | A ) = 0. For k &gt; m , p X ( k | A ) = P { X = k | X &gt; m } = P { X = k } P { X &gt; m } = p (1- p ) k- 1 (1- p ) m = p (1- p ) k- m- 1 . (b) P ( A ) = m- 2 summationdisplay n =0 p (1- p ) n = p 1- (1- p ) m- 1 1- (1- p ) = 1- (1- p ) m- 1 . For k ≤ m or k ≤ 0, p X ( k | A ) = 0. For k &lt; m p X ( k | A ) = P { X = k | X &lt; m } = P { X = k } P { X &lt; m } = p (1- p ) k- 1 1- (1- p ) m- 1 . (c) We have P ( A ) = ∞ summationdisplay n =0 p (1- p )((1- p ) 2 ) n = p (1- p ) 1- (1- p ) 2 = 1- p 2- p . Thus for k even p X ( k | A ) = P { X = k | X is even } = P { X = k } P { X is even } = p (1- p ) k- 1 P ( A ) = p (2- p )(1- p ) k- 2 . 2 2 4 6 8 10 12 0.1 0.2 0.3 0.4 (a) p X (x|A) 2 4 6 8 10 12 0.1 0.2 0.3 0.4 (b) p X (x|A) 2 4 6 8 10 12 0.1 0.2 0.3 0.4 (c) p X (x|A) x Figure 1: Plots of the conditional pmf’s using p = 1 4 and m = 5. Plots are shown in Figure 1. The shape of the conditional pmf in part (a) shows that the geometric random variable is memoryless: p X ( x | X &gt; k ) = p X ( x- k ) , for x ≥ k....
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## This note was uploaded on 10/26/2011 for the course MATH 180C taught by Professor Eggers during the Winter '09 term at Aarhus Universitet.

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hw2sol - UCSD ECE 153 Handout #8 Prof. Young-Han Kim...

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