Problem Set 4 with Solutions.docx - Problem Set 4 With Solutions ECO 207 3 and 4 8 October 2019 What is the expected value for a game in which you roll

Problem Set 4 with Solutions.docx - Problem Set 4 With...

This preview shows page 1 - 3 out of 7 pages.

Problem Set 4 With Solutions : ECO 207: 3 and 4 8 October 2019 1. What is the expected value for a game in which you roll two fair dice and get paid the amount in the lower number if the two numbers, or the number if they are equal. Ans: If X is the random variable which is the number displayed when one fair dice is thrown, then the expected value of X is E(X) = (1 + 2+ 3+ 4+ 5+ 6)/6 = 21/6 = 7/2 or 3.5 If 2 fair dice are thrown, then the possible outcomes are: (1,1) .... (1,6) . . (6,1) ... (6,6) If Y is the random variable which is equal to the lower number, then E(Y) = [(11*1) + (9*2) + (7*3) + (5*4) + (3*5) + (1*6)]/36 = (11 + 18+ 21+ 20 + 15 + 6)/36 = 91/36 = approx 2.5278 2. The joint distribution of X and Y is as follows. X takes 3 values: 0,1,2 and Y (1,2, …6) Y X 0 1 2 1 0.025 0.05 0.025 2 0.025 0.05 0.025 3 0.025 0.05 0.025 4 0.025 0.05 0.025 5 0.05 0.1 0.05 6 0.1 0.2 0.1 1
Image of page 1
Find the marginal PMFs of X and Y. Also find E(X), E (Y) and E[min(X, Y, 1)]. What is E (X|Y =6) Answer: The reason they are called marginal distributions is that they are written into the margins of the table. So just make a column for the total of Y and a row for the total of X and add across rows and down columns. Also add the marginal totals to make sure they are both 1. I’ll do Y, you do X. I’ll write the Y totals as a row, but they should be in a column. For Y = 1, 2, …, 6 you should get 0.1, 0.1, 0.1, 0.1, 0.2 and 0.4. And the sum is 1.
Image of page 2
Image of page 3

You've reached the end of your free preview.

Want to read all 7 pages?

  • Summer '18
  • Sagar Arora

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture