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Solution_Assignment1

Solution_Assignment1 - STAT 333 Spring 2009 Solutions to...

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STAT 333 Spring 2009 Solutions to Assignment 1 By Juan Manuel Martinez e-mail:[email protected] June 3, 2009 Type I Problems Problem 1 a) Here T has an exponential distribution with parameter λ = 2. Its cumulative distribution function is therefore given by: F T ( t ) = 1 - e - 2 t We conclude that Pr( T > 1 2 ) = 1 - Pr( T 1 2 ) = 1 - F T ( 1 2 ) = 1 - (1 - e - 1 ) = e - 1 b) We make use of the memoryless property of the exponential distribution. For any positive real numbers s and t: Pr ( T > s + t | T > s ) = Pr ( T > t ) Therefore, Pr( T > 12 . 5 | T > 12 ) = Pr( T > 1 2 ) = e - 1 Problem 2 Let: F = event that fair coin is picked up T = event that two-headed coin is picked up B = event that biased coin is picked up H = event that coin picked up shows head Clearly, Pr ( H | B ) = 0.75, Pr ( H | T ) = 1 and Pr ( H | F ) = 0 . 5 . Since any coin is equally likely to be picked up, Pr(T) = Pr(F) = Pr(B) = 1 3 The problem asks for Pr ( T | H ) . For this we make use of Bayes’ Rule: Pr ( T | H ) = Pr ( T, H ) Pr ( H ) = Pr ( H | T ) Pr ( T ) Pr ( H | T ) Pr ( T ) + Pr ( H | B ) Pr ( B ) + Pr ( H | F ) Pr ( F ) 1
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= (1)( 1 3 ) 1 3 (1 + 0 . 75 + 0 . 5) = 4 9 Problem 3 We have Pr ( Y > n ) = Pr ( FF...F bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright ntimes ) = (1 - p 1 )(1 - p 2 ) ... (1 - p n ) (because the trials are independent) = producttext n i =1 (1 - p i ) Letting n → ∞ , we have Pr ( Y = ) = f( ) = producttext i =1 (1 - p i ) . To determine whether or not Y is a proper random variable, we make use of the Sum-Product lemma, which states that productdisplay i =1 (1 - p i ) > 0 summationdisplay i =1 p i < Therefore, the problem boils down to determining whether or not the infinite series converges or diverges. i) Here, p n = 4 - n . To determine if Y is proper, we must determine if i =1 ( 1 4 ) i converges or not. Observe that: summationdisplay i =1 parenleftbigg 1 4 parenrightbigg i = summationdisplay i =0 parenleftbigg 1 4 parenrightbigg i - 1 = 1 1 - 1 4 - 1 = 1 3 < Thus, the series converges, and by the Sum-Product lemma, Y is not a proper random variable. ii) Here, p n = 1 - e - 1 n . To determine if i =1 (1 - e - 1 i ) converges, we apply the Limit Comparison Test with the harmonic series i =1 ( 1 i ) . Then: lim n →∞ 1 n 1 - e - 1 n = lim n →∞ - 1 n 2 e - 1 n * 1 n 2 = lim n →∞ e 1 n = 1 < Where we used l’Hopital’s rule to go from the first to the second equality. By the Limit Comparison Test, i =1 (1 - e - 1 i ) diverges, and by the Sum-Product lemma, Y is a proper random variable. 2
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Problem 4 Since the X prime i s are independent and identically distributed (i.i.d.) we must have that for any integers i, j, i negationslash = j , Pr ( X i < X j ) = Pr ( X i > X j ) = 1 2 . Therefore, the value of Pr ( X 1 < X 2 < X 4 < X 3 ) is independent of the order of the 4 variables. All we care about is the number of different ways to arrange the random variables in ascending order. There are 24 such ways, and so we conclude that Pr ( X 1 < X 2 < X 4 < X 3 ) = 1 24 Problem 5 a) It is clear that X = n summationdisplay i =1 X i b) To find E [ X ] , we must determine E [ X i ] , which is the same as Pr ( X i ) since X i is an indicator random variable. Since we
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  • Spring '08
  • Chisholm
  • Probability theory, probability density function, proper waiting time, Sum-Product Lemma, Juan Manuel Martinez

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