ecm06-hw1-solutions

# ecm06-hw1-solutions - Winter 06 Econometrics ASSIGNMENT 1...

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Winter 06 1 Econometrics ASSIGNMENT 1 - SOLUTIONS 100 POINTS TOTAL DUE: Thursday, January 26, 3:20 p.m. 1. [5] Let X be the random variable distributed as Normal (5,4). Find the probabilities of the following events: (i) P( X 6) (ii) P( X > 4) (iii) P(| X – 5| > 1) (i) P( X 6) = P[( X – 5)/2 (6 – 5)/2] = P( Z .5) .692, where Z denotes a Normal (0,1) random variable. [We obtain P( Z .5) from Table 1.] (ii) P( X > 4) = P[( X – 5)/2 > (4 – 5)/2] = P( Z > .5) = P( Z .5) .692. (iii) P(| X – 5| > 1) = P( X – 5 > 1) + P( X – 5 < –1) = P( X > 6) + P( X < 4) (1 – .692) + (1 – .692) = .616, where we have used answers from parts (i) and (ii). 2. [5] Let X denote the prison sentence, in years, for people convicted of auto theft in California. Suppose that the pdf of X is given by 2 1 9 () () ,0 3 . = << fx x x Use integration to find the expected prison sentence. [ hint : requires calculus.] E( X ) = 3 0 () x fxd x = 3 0 x x = (1/9) 3 3 0 x dx . But 3 3 0 x dx = (1/4) x 4 3 0 | = 81/4. Therefore, E( X ) = (1/9)(81/4) = 9/4, or 2.25 years. 3.[5] Let X denote the annual salary on university professors in the United States, measured in thousands of dollars . Suppose that the average salary is 52.3, with a standard deviation of 14.6. Find the mean and standard deviation when salary is measured in dollars . If Y is salary in dollars then Y = 1000 X , and so the expected value of Y is 1,000 times the expected value of X , and the standard deviation of Y is 1,000 times the standard deviation of X . Therefore, the expected value and standard deviation of salary, measured in dollars, are \$52,300 and \$14,600, respectively. 4. [20] Suppose a fair coin is tossed 3 times and let a random variable X denote the number of heads. Define Y, a Bernoulli random variable, as Y=1 if the first tossed coin turns up head, and Y=0 otherwise. (i) Determine and sketch the pdf of X and cdf of X. (ii) Determine the joint probability density function of (X, Y) (iii) Are random variables X and Y independent? Explain. (iv) Compute P(X=2|Y=1), Cov(X,Y) and Corr(X, Y)

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Winter 06 2 (i) The possible outcomes of X = 0, 1, 2, 3 - X=0 if (T,T,T) - X=1 if (H,T,T), (T,H,T), (T,T,H) - X=2 if (H,H,T), (H,T,H), (T,H,H) - X=3 if (H,H,H) The pdf of X is ( ) ( ) , 1,2,. .. jj j f xP X x p i f j k == = = Thus, the pdf of X : x f (0) = P(X=0) = 1/8 x f (1) = P(X=1) = 3/8 x f (2) = P(X=2) = 3/8 x f (3) = P(X=3) = 1/8 The cdf of X is ( ) ( ) Fx PX x =≤ Thus, the pdf of X : F(0) = P(X 0) = P(X=0) = 1/8 F(1) = P(X 1) = P(X=0)+P(X=1) = 4/8 F(2) = P(X
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## This note was uploaded on 08/27/2011 for the course ECON 7043 taught by Professor Projim during the Three '11 term at University of Adelaide.

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ecm06-hw1-solutions - Winter 06 Econometrics ASSIGNMENT 1...

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