AK6_230 - ECO 230 Economic Statistics Homework #6 Due date:...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
ECO 230 Marianna Kudlyak Economic Statistics Spring 2007 Homework #6 Due date: Tuesday, March 6 th Total points: 10. Graded exercises: Ex.4 (2 points), Ex.6 (2 points), Ex. 8 (3 points), and 0.7 points were subtracted for the absence of each of the remaining 3 exercises. Bonus points: Ex. 1 (5 points) Updated: March 1 st , 2007 The HW contains 8 exercises; however, you are required to submit only 7 exercises: from Ex.2 to Ex.8. Exercise 1 is optional and is not to be submitted. Exercise 1 (Bonus). Solution will be posted later. Exercise 2. Show the following results, justifying each step in details: a) For any random variables X and Y and constants a, b, c, d, show that COV(aX + b, cY + d) = acCOV(X,Y) COV(aX + b, cY + d) = E[(aX + b – E(aX + b))( cY + d – E(cY + d))] = E[(aX + b – aE(X) - b)( cY + d – cE(Y) - d)] = E[(aX – aE(X))( cY – cE(Y)] = E[ac(X – E(X))( Y – E(Y)] = ac E[(X – E(X))( Y – E(Y)] = acCOV(X,Y) b) var(X - Y) = var(X) + var(Y) – 2cov(X,Y). Hint: Start from the definition of variance. var(X - Y) = E[(X – Y - E(X – Y)) 2 ] = E[(X – Y - E(X) + E(Y)) 2 ] = E[{(X – E(X)) – (Y - E(Y))} 2 ] = E[(X – E(X)) 2 + (Y - E(Y)) 2 - 2(X – E(X)) (Y - E(Y))] = E[(X – E(X)) 2 ]+ E[(Y - E(Y)) 2 ]- 2E[(X – E(X)) (Y - E(Y))] = var(X) + var(Y) – 2cov(X,Y). Last step follows from the definition of var and cov. Exercise 3. Let X and Y be a pair of jointly distributed discrete random variables. Develop a numerical example (different from the one in the textbook) to demonstrate that if covariance between X and Y is 0, it does not imply that X and Y are statistically independent. Answers vary. Please see ask instructor or TA if you have question. Exercise 4. Newbold, Carlson and Thorne, Problem 5.84 p.173. a. Here you need to find marginal distribution function of Y Py(0) = .08 + .03 + .01 = .12 Py(1) = .13 + .08 + .03 = .24 Py(2) = .09 + .08 + .06 = .23 Py(3) = .06 + .09 + .08 = .23 Py(4) = .03 + .07 + .08 = .18
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
b. Conditional probability function: P Y|X (y|3) = 1/26; 3/26; 6/26; 8/26; 8/26 for y = 0,1,2,3,4 respectively. c. No, because Px,y(3,4) = .08 .0468 = Px(3)Py(4) In addition answer the following questions: d) Find the probability that a randomly chosen person from this group does not make any purchases in a given week.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/09/2008 for the course ECO 230 taught by Professor Kudlyak during the Spring '08 term at Rochester.

Page1 / 6

AK6_230 - ECO 230 Economic Statistics Homework #6 Due date:...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online