Answer Set 1 - Fall 2008

Answer Set 1 Fall - Econ 371 Problem Set#1 Answer Sheet 2.1 In this question you are asked to consider the random variable Y which denotes the

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Econ 371 Problem Set #1 Answer Sheet 2.1 In this question, you are asked to consider the random variable Y , which denotes the number of “heads” that occur when two coins are tossed. a. The first part of the question asks you to derive the probability distribution for Y . For a discrete random variable, this is just a listing of all the possible outcomes and the probability that they can occur. Assuming that the two coins are fair, there are only four possible results when we flip the two coins: Heads,Heads ( Y = 2) Heads,Tails ( Y = 1) Tails,Heads ( Y = 1) Tails,Tails ( Y = 0) which happen with equal probability. The corresponding probability distribution is given by the following table. outcome (number of heads) Y = 0 Y = 1 Y = 2 probability 0.25 0.50 0.25 b. The cumulative probability distribution function for Y corresponds to just computing the probabilities of Y being less than or equal to a given value outcome (number of heads) Y < 0 0 Y < 1 1 Y < 2 Y > 2 cumulative probability 0 0.25 0.75 1 c. Finally, you are asked to derive the mean and variance of Y . These are given by: E ( Y ) = k X i =1 Pr ( Y = y i ) y i = k X i =1 p i y i = (0 . 25) · 0 + (0 . 50) · 1 + (0 . 25) · 2 = 1 and V ar ( Y ) = E [( Y - μ Y ) 2 ] = k X i =1 p i ( y i - μ Y ) 2 = (0 . 25) · (0 - 1) 2 + (0 . 50) · (1 - 1) 2 + (0 . 25) · (2 - 1) 2 = (0 . 25) · 1 + (0 . 50) · 0 + (0 . 25) · 1 = 0 . 5 2.3 In this question, you are asked to compute various characteristics of two random variables, W and V that are functions of the random variables in Table 2.2 ( X and Y ). Specifically, we have: W = 3 + 6 X V = 20 - 7 Y a. First, you are asked to compute E ( W ) and E ( V ). From equation (2.12) in the text, we know that: E ( W ) = 3 + 6 E ( X ) (1) 1
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but, using Table 2.2 and our formula for computing means, we have that E ( X ) = k X X i =1 p i x i = (0 . 30) · 0 + (0 . 70) · 1 = 0 . 7 Substituting this into (1) yields. E ( W ) = 3 + 6 E ( X ) = 3 + 6(0 . 7) = 7 . 2 (2) Similarly, E ( Y ) = k Y X i =1 p i x i = (0 . 22) · 0 + (0 . 78) · 1 = 0 . 78 , so that E ( V ) = 20 - 7 E ( Y ) = 20 - 7(0 . 78) = 14 . 54 . (3) b. In part b you are asked to compute the corresponding variances for W and V . Equation (2.13) in the text tells us that: σ 2 W = 6 2 · σ 2 X = 36 σ 2 X σ 2 V = ( - 7) 2 · σ 2 Y = 49 σ 2 Y Turning to our formulas for variances, however, we know that: V ar ( X ) = k X X i =1 p i ( x i - μ X ) 2 = (0 . 30) · (0 - 0 . 7) 2 + (0 . 70) · (1 - 0 . 7) 2 = (0 . 30) · (0 . 49) + (0 . 70) · (0 . 09) = 0 . 21 So that σ 2 W = 6 2 · σ 2 X = 36(0 . 21) = 7
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This note was uploaded on 08/09/2009 for the course ECON 120B taught by Professor Jeon during the Spring '08 term at UCSD.

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Answer Set 1 Fall - Econ 371 Problem Set#1 Answer Sheet 2.1 In this question you are asked to consider the random variable Y which denotes the

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