# We are given that the proportion of defectives

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We are given that the proportion of defectives follows a uniform distribution on (0, 1), sothe unconditional probability thatY= 2 can be found by=======1032101010132)(3)1(3)()|2(),2()2(dpppdpppdppfpYPdppYPYP= 1/4.5.42(Similar to Ex. 5.41)LetY= # of defects per yard.Then,( )1210!00)()|(),()(+λλ=λ=λλλ==λλ==λyyededfyYPdyYPypy,y= 0, 1, 2, … .Note that this is essentially a geometric distribution (see Ex. 3.88).5.43Assume).()|(1121yfyyf=Then,)()()()|(),(2211222121yfyfyfyyfyyf==so thatY1andY2are independent.Now assume thatY1andY2are independent.Then, thereexists functionsgandhsuch that)()(),(2121yhygyyf=so that∫ ∫×==22112121)()(),(1dyyhdyygdydyyyf.Then, the marginals forY1andY2can be defined by=×=111222112111)()()()()()()(dyygygdydyyhdyygyhygyf, so=22222)()()(dyyhyhyf.Thus,)()(),(221121yfyfyyf=.Now it is clear that)()(/)()()(/),()|(11222211222121yfyfyfyfyfyyfyyf===,provided that)(22yf> 0 as was to be shown.5.44The argument follows exactly as Ex. 5.43 with integrals replaced by sums and densitiesreplaced by probability mass functions.5.45No.Counterexample:P(Y1= 2,Y2= 2) = 0P(Y1= 2)P(Y2= 2) = (1/9)(1/9).5.46No.Counterexample:P(Y1= 3,Y2= 1) = 1/8P(Y1= 3)P(Y2= 1) = (1/8)(4/8).
102Chapter 5: Multivariate Probability DistributionsInstructor’s Solutions Manual5.47Dependent.For example:P(Y1= 1,Y2= 2)P(Y1= 1)P(Y2= 2).5.48Dependent.For example:P(Y1= 0,Y2= 0)P(Y1= 0)P(Y2= 0).5.49Note that10,33)(121021111==yydyyyfy,10],1[3)(22223111221==yydyyyfy.Thus,)()(),(221121yfyfyyfso thatY1andY2are dependent.5.50a.Note that10,11)(110211==ydyyfand10,11)(210122==ydyyf.Thus,)()(),(221121yfyfyyf=so thatY1andY2are independent.b.Yes, the conditional probabilities are the same as the marginal probabilities.5.51a.Note that0,)(102)(11121>==+yedyeyfyyyand0,)(201)(22221>==+yedyeyfyyy.Thus,)()(),(221121yfyfyyf=so thatY1andY2are independent.b.Yes, the conditional probabilities are the same as the marginal probabilities.5.52Note that),(21yyfcan be factored and the ranges ofy1andy2do not depend on eachother so by Theorem 5.5Y1andY2are independent.5.53The ranges ofy1andy2depend on each other soY1andY2cannot be independent.5.54The ranges ofy1andy2depend on each other soY1andY2cannot be independent.5.55The ranges ofy1andy2depend on each other soY1andY2cannot be independent.5.56The ranges ofy1andy2depend on each other soY1andY2cannot be independent.5.57The ranges ofy1andy2depend on each other soY1andY2cannot be independent.5.58Following Ex. 5.32, it is seen that)()(),(221121yfyfyyfso thatY1andY2aredependent.5.59The ranges ofy1andy2depend on each other soY1andY2cannot be independent.5.60From Ex. 5.36,21111)(+=yyf, 0y11, and21222)(+=yyf, 0y21.But,)()(),(221121yfyfyyfsoY1andY2are dependent.5.61Note that),(21yyfcan be factored and the ranges ofy1andy2do not depend on eachother so by Theorem 5.5,Y1andY2are independent.

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Term
Spring
Professor
Xu,Lily
Tags
Probability theory, INSTRUCTOR S SOLUTIONS