Probset10sol

# Probset10sol - Probability and Stochastic Processes...

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Unformatted text preview: Probability and Stochastic Processes Homework 10 Solutions Spring’2010 Problem Solutions : Yates and Goodman, 5.1.2 5.1.3 5.2.2 5.3.2 5.3.6 5.4.3 5.4.4 5.4.5 5.5.3 and 5.5.4 Problem 5.1.2 Solution Whether a computer has feature i is a Bernoulli trial with success probability p i = 2 − i . Given that n computers were sold, the number of computers sold with feature i has the binomial PMF P N i ( n i ) = braceleftbigg ( n n i ) p n i i (1 − p i ) n i n i = 0 , 1 ,... ,n otherwise (1) Since a computer has feature i with probability p i independent of whether any other feature is on the computer, the number N i of computers with feature i is independent of the number of computers with any other features. That is, N 1 ,... ,N 4 are mutually independent and have joint PMF P N 1 ,...,N 4 ( n 1 ,... ,n 4 ) = P N 1 ( n 1 ) P N 2 ( n 2 ) P N 3 ( n 3 ) P N 4 ( n 4 ) (2) Problem 5.1.3 Solution (a) In terms of the joint PDF, we can write joint CDF as F X 1 ,...,X n ( x 1 ,... ,x n ) = integraldisplay x 1 −∞ ··· integraldisplay x n −∞ f X 1 ,...,X n ( y 1 ,... ,y n ) dy 1 ··· dy n (1) However, simplifying the above integral depends on the values of each x i . In particular, f X 1 ,...,X n ( y 1 ,... ,y n ) = 1 if and only if 0 ≤ y i ≤ 1 for each i . Since F X 1 ,...,X n ( x 1 ,... ,x n ) = 0 if any x i < 0, we limit, for the moment, our attention to the case where x i ≥ 0 for all i . In this case, some thought will show that we can write the limits in the following way: F X 1 ,...,X n ( x 1 ,... ,x n ) = integraldisplay max(1 ,x 1 ) ··· integraldisplay min(1 ,x n ) dy 1 ··· dy n (2) = min(1 ,x 1 )min(1 ,x 2 ) ··· min(1 ,x n ) (3) A complete expression for the CDF of X 1 ,... ,X n is F X 1 ,...,X n ( x 1 ,... ,x n ) = braceleftbigg producttext n i =1 min(1 ,x i ) 0 ≤ x i ,i = 1 , 2 ,... ,n otherwise (4) 1 (b) For n = 3, 1 − P bracketleftbigg min i X i ≤ 3 / 4 bracketrightbigg = P bracketleftbigg min i X i > 3 / 4 bracketrightbigg (5) = P [ X 1 > 3 / 4 ,X 2 > 3 / 4 ,X 3 > 3 / 4] (6) = integraldisplay 1 3 / 4 integraldisplay 1 3 / 4 integraldisplay 1 3 / 4 dx 1 dx 2 dx 3 (7) = (1 − 3 / 4) 3 = 1 / 64 (8) Thus P [min i X i ≤ 3 / 4] = 63 / 64. Problem 5.2.2 Solution In this problem, we find the constant c from the requirement that that the integral of the vector PDF over all possible values is 1. That is, integraltext ∞ −∞ ··· integraltext ∞ −∞ f X ( x ) dx 1 ··· dx n = 1 . Since f X ( x ) = c a ′ x = c ∑ n i =1 a i x i , we have that integraldisplay ∞ −∞ ··· integraldisplay ∞ −∞ f X ( x ) dx 1 ··· dx n = c integraldisplay 1 ··· integraldisplay 1 parenleftBigg...
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## This note was uploaded on 02/13/2011 for the course 332 226 taught by Professor Staff during the Spring '08 term at Rutgers.

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Probset10sol - Probability and Stochastic Processes...

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