Contingency Table 1926-2015 (90 vears Stocks Below average Above average g Below average 24 20 Ab...
Contingency Table 1926-2015 (90 vears Stocks Below average Above average g Below average 24 20 Above average 28 1. Calculate the marginal frequency distributions and complete the five blanks in the above table. Demonstrate that the probability stocks are above average equals 53.3%? Hint: the answer uses all the observations; hence the denominator is 90. 2. What is the probability that stocks are above average given that (conditional on) bonds being above average? Hint: the answer uses only the observations where bonds are above average. 3. What is the probability that the stock market returns above average for three successive years? Hint: use the 53.3% the probability from first question and the fact that annual stock returns are I.I.D 4. I've been investing for 20 years. How many years should I expect to have above average stock returns? Note this is a question about the expected value of a binomial random variable where the probability in a single year equals the 53.3% value from question 1
Contingency Table 1926-2015 (90 vears Stocks Below average Above average g Below average 24 20 Above average 28 1. Calculate the marginal frequency distributions and complete the five blanks in the above table. Demonstrate that the probability stocks are above average equals 53.3%? Hint: the answer uses all the observations; hence the denominator is 90. 2. What is the probability that stocks are above average given that (conditional on) bonds being above average? Hint: the answer uses only the observations where bonds are above average. 3. What is the probability that the stock market returns above average for three successive years? Hint: use the 53.3% the probability from first question and the fact that annual stock returns are I.I.D 4. I've been investing for 20 years. How many years should I expect to have above average stock returns? Note this is a question about the expected value of a binomial random variable where the probability in a single year equals the 53.3% value from question 1