1. Christopher Dougherty EC220 - Introduction to econometrics (review chapter) Slideshow: exercise r.4 Original citation: Dougherty, C. (2012) EC220 - Introduction to econometrics (review chapter). [Teaching Resource] © 2012 The Author This version available at: http://learningresources.lse.ac.uk/141/http://learningresources.lse.ac.uk/141/ Available in LSE Learning Resources Online: May 2012 This work is licensed under a Creative Commons Attribution-ShareAlike 3.0 License. This license allows the user to remix, tweak, and build upon the work even for commercial purposes, as long as the user credits the author and licenses their new creations under the identical terms. http://creativecommons.org/licenses/by-sa/3.0/ http://creativecommons.org/licenses/by-sa/3.0/ http://learningresources.lse.ac.uk/

2. EXERCISE R.4 1 R.4*Find the expected value of X in Exercise R.2. [R.2*A random variable X is defined to be the larger of the numbers when two dice are thrown, or the number if they are the same. Find the probability distribution for X.]

3. Definition of E(X), the expected value of X: 2 The expected value of a random variable, also known as its population mean, is the weighted average of its possible values, the weights being the probabilities attached to the values. EXERCISE R.4

4. The left side of the table above shows in abstract how an expected value should be calculated. x i p i x 1 p 1 x 2 p 2 x 3 p 3......... x n p n  x i p i = E(X) 3

5. EXERCISE R.4 The random variable X defined in Exercise R.2 could be any of the integers from 1 to 6 with probabilities as shown. x i p i x i p i x i p i x 1 p 1 x 1 p 1 11/36 x 2 p 2 x 2 p 2 23/36 x 3 p 3 x 3 p 3 35/36......... 47/36......... 59/36 x n p n x n p n 611/36  x i p i = E(X) 4

6. EXERCISE R.4 Before calculating the expected value, it is a good idea to make an initial guess. The higher values of X have the greatest probabilities, so the expected value is likely to be between 4 and 5. x i p i x i p i x i p i x 1 p 1 x 1 p 1 11/36 x 2 p 2 x 2 p 2 23/36 x 3 p 3 x 3 p 3 35/36......... 47/36......... 59/36 x n p n x n p n 611/36  x i p i = E(X) 5

7. EXERCISE R.4 X could be equal to 1 with probability 1/36, so the first entry in the calculation of the expected value is 1/36. x i p i x i p i x 1 p 1 x 1 p 1 11/361/36 x 2 p 2 x 2 p 2 23/36 x 3 p 3 x 3 p 3 35/36......... 47/36......... 59/36 x n p n x n p n 611/36  x i p i = E(X) 6

8. EXERCISE R.4 Similarly for the other 5 possible values. x i p i x i p i x 1 p 1 x 1 p 1 11/361/36 x 2 p 2 x 2 p 2 23/366/36 x 3 p 3 x 3 p 3 35/3615/36......... 47/3628/36......... 59/3645/36 x n p n x n p n 611/3666/36  x i p i = E(X) 7

9. EXERCISE R.4 To obtain the expected value, we sum the entries in this column. x i p i x i p i x 1 p 1 x 1 p 1 11/361/36 x 2 p 2 x 2 p 2 23/366/36 x 3 p 3 x 3 p 3 35/3615/36......... 47/3628/36......... 59/3645/36 x n p n x n p n 611/3666/36  x i p i = E(X) 161/36 8

10. x i p i x i p i x 1 p 1 x 1 p 1 11/361/36 x 2 p 2 x 2 p 2 23/366/36 x 3 p 3 x 3 p 3 35/3615/36......... 47/3628/36......... 59/3645/36 x n p n x n p n 611/3666/36  x i p i = E(X) 161/36 = 4.47 The expected value turns out to be 4.47, which is in line with our initial guess. EXERCISE R.4 9

11. Copyright Christopher Dougherty 1999–2006. This slideshow may be freely copied for personal use. 26.08.06