23.When you use @RISK’s correlation feature to generate correlated random numbers, how can you verify that they are correlated? Try the following. Use the RiskCorrmat function to generate two normally distributed random numbers, each with mean 100 and standard deviation 10, and with correlation 0.7. To run a simulation, you need an output variable, so sum these two numbers and designate the sum as an output variable. Now run @RISK. Click @RISK’s Excel Reports button and check the Simulation Data option to see the actual simulated data. a. Use Excel’s CORREL function to calculate the correlation between the two input variables. It should be close to 0.7. Then create a scatterplot of these two input variables. The plot should indicate a definite positive relationship. b.Are the two input variables correlated with the output? Use Excel’s CORREL function to find out. Interpret your results intuitively.
26.. Suppose you are going to invest equal amounts in three stocks. The annual return from each stock is normally distributed with mean 0.01 11%2 and standard deviation 0.06. The annual return on your portfolio, the output variable of interest, is the average of the three stock returns. Run @RISK, using 1000 iterations, on each of the following scenarios. a.The three stock returns are highly correlated. The correlation between each pair is 0.9. b.The three stock returns are practically independent. The correlation between each pair is 0.1. c. The first two stocks are moderately correlated. The correlation between their returns is 0.4. The third stock’s return is negatively correlated with the other two. The correlation between its return and each of the first two is 20.8. d.Compare the portfolio distributions from @RISK for these three scenarios. What do you conclude? e. You might think of a fourth scenario, where the correlation between each pair of returns is a large negative number such as 20.8. But explain intuitively why this makes no sense. Try to run the simulation with these negative correlations and see what happens
An investor is considering the purchase of zero-coupon U.S. Treasury bonds. A 30-year zero-coupon bond yielding 8% can be purchased today for $9.94. At the end of 30 years, the owner of the bond will receive $100. The yield of the bond is related to its price by the following equation: Here, P is the price of the bond, y is the yield of the bond, and t is the maturity of the bond measured in years. Evaluating this equation for t = 30 and y = 0.08 gives P = 9.94. The investor is planning to purchase a bond today and sell it one year from now. The investor is interested in evaluating the return on the investment in the bond. Suppose, for example, that the yield of the bond one year from now is 8.5%. Then the price of the bond one year later will be $9.39 [=100/(1+0.085)^29]. The time remaining to maturity is t 5 29 because one year has passed. The return for the year is -5.54% [= (9.39 - 9.94)/9.94]. In addition to the 30-year-maturity zero-coupon bond, the investor is considering the purchase of zero-coupon bonds with maturities of 2, 5, 10, or 20 years. All of the bonds are currently yielding 8.0%. (Bond investors describe this as a flat yield curve.) The investor cannot predict the future yields of the bonds with certainty. However, the investor believes that the yield of each bond one year from now can be modeled by a normal distribution with mean 8% and standard deviation 1%.
Questions 1. Suppose that the yields of the five zero-coupon bonds are all 8.5% one year from today. What are the returns of each bond over the period?
- Using a simulation with 1000 iterations, estimate the expected return of each bond over the year. Estimate the standard deviations of the returns.
- Comment on the following statement: “The expected yield of the 30-year bond one year from today is 8%. At that yield, its price would be $10.73. The return for the year would be 8% [= (10.73 - 9.94)/9.94]. Therefore, the average return for the bond should be 8% as well. A simulation isn’t really necessary. Any difference between 8% and the answer in Question 2 must be due to simulation error.”