t3818m1ans - University of Colorado Department of Economics...

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- 1 - University of Colorado Department of Economics Economics 3818 Midterm Examination Prof. Jeffrey S. Zax 27 September 2005 Solutions (14) 1. Consider two outcomes C and D, where Pr(C)=.3, Pr(D)=.5, Pr(C 1 D)=.3 (3) a. What is Pr(C c D)? Why? Pr(C c D) = Pr(C) + Pr(D) ! Pr(C 1 D) = .3 + .5 ! .3 = .5 (3) b. What is Pr(C * D)? Pr(C * D) = Pr(C 1 D) ' Pr(D) = .3 ' .5 = 3 ' 5 (2) c. What is Pr(D * C)? Pr(D * C) = Pr(C 1 D) ' Pr(C) = .3 ' .3 = 1 (2) d. Would you expect Pr(C * D) and Pr(D * C) to be the same? Why or why not? Ordinarily they would not be the same. They share the same numerator, Pr(C 1 D) , but their denominators would usually be different. They would be equal only if the numerator were zero or if Pr(C)=Pr(D). (2) e, Are C and D mutually exclusive? Why or why not? C and D are not mutually exclusive because the probability of their intersection is not zero. (2) f. Are C and D independent? Why or why not? C and D are not independent because .3 = P(C) P(C|D) = 3 ' 5.
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- 2 - (5) 2. X is a continuous random variable whose outcomes range between one and four. It has a uniform probability density function, f(x)=.33. (2) a. What is Pr(1<x<2.5)? Pr(1<x<2.5) = (2.5 ! 1)(.33) = (1.5)(.33) = .5 (2) b. What is Pr(3>x>1)? Pr(3>x>1) = (3 ! 1)(.33) = (2)(.33) = .66 (1) c. What is Pr(x=2)? This is the probability of a single point. It is effectively zero. (6) 3. Prove that () xx x i i n −= = 1 0. Because is a constant, it can be factored out of the summation: x i i n i i n == ∑∑ 11 . The summation can be split into two separate summations: x x i i n i i n i n = 1 . We can multiply the first summation by one, in the form of n ' n. Because is a x constant, it can be factored out of the second summation: n n i i n i n i i n i n = = =− 1 1 1. Rearrange the first summation. Recognize that one summed to itself n times is n in the second summation:
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- 3 - x n n xx x n x n nx i i n i ni i n == = ∑∑ =− 11 1 1.
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t3818m1ans - University of Colorado Department of Economics...

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