Problem Set 6 Suggested Solutions - Computational Finance | ECON 424, Assignments of Economics

Download Problem Set 6 Suggested Solutions - Computational Finance | ECON 424 and more Assignments Economics in PDF only on Docsity! Eric Zivot Econ 424 Fall 2008 Problem Set #6 Suggested Solutions Hypothesis Testing in the Constant Expected Return Model Instructions In this part of the lab you will use R to • Estimate parameters of the constant expected return (CER) model and test various hypotheses about the parameters and assumptions of the model. Exercises 1. Consider the constant expected return model (CER) 2 , 1, , ~ (0, ) cov( , ) it i it it i it jt ij R t T iid N μ ε ε σ ε ε σ = + = = K where Rit denotes the continuously compounded return on asset i, i =Vanguard long term bond index fund (VBLTX), Fidelity Magellan stock mutual fund (FMAGX), General Motors stock (GM). The commands in the R script file lab6.r will be helpful in completing this assignment. a) Using sample descriptive statistics, give estimates for the model parameters , , and i i ijμ σ ρ . Arrange these estimates nicely in a table. > cbind(muhat.vals,sigmahat.vals) muhat.vals sigmahat.vals vbltx 0.006469 0.02313 fmagx 0.005820 0.04680 gm 0.002323 0.10415 > rhohat.vals vbltx,fmagx vbltx,gm fmagx,gm -0.08004 -0.07185 0.4757 All of the mean values are close to zero. As expected, the bond fund has the lowest SD and gm has the highest SD. Notice that vbltx is negatively correlated (weakly) with fmagx and gm, but fmagx and gm are moderately positively correlated. Typically, bond returns are nearly uncorrelated with stock returns. b) For each iμ (i = VBLTX, FMAGX and GM) test the hypotheses 0 1: 0 vs. : 0i iH Hμ μ= ≠ using a 5% significance level. Perform the test using the t-statistic as well as the 95% confidence. What do you conclude? (note: you can use the R function t.test for this problem) > t.test(lab4.z[,"vbltx"]) One-sample t-Test data: lab4.z[, "vbltx"] = 3.064, df = 119, p-value = 0.0027 alternative hypothesis: mean is not equal to 0 95 percent confidence interval: 0.002288 0.010649 sample estimates: mean of x 0.006469 Here, |t-stat| = 3.064 > 2 so we reject H0: μ=0 at the 5 % level. Also, the p-value is less than 0.05 and 0 is not in the 95% confidence interval. > t.test(lab4.z[,"fmagx"]) One-sample t-Test data: lab4.z[, "fmagx"] = 1.362, df = 119, p-value = 0.1757 alternative hypothesis: mean is not equal to 0 95 percent confidence interval: -0.002639 0.014279 sample estimates: mean of x 0.00582 Here, |t-stat| = 1.362 < 2 so we do not reject H0: μ=0 at the 5 % level. Also, the p-value is greater than 0.05 and 0 is in the 95% confidence interval. > t.test(lab4.z[,"gm"]) One-sample t-Test data: lab4.z[, "gm"] = 0.2443, df = 119, p-value = 0.8074 alternative hypothesis: mean is not equal to 0 95 percent confidence interval: Eric Zivot Econ 424 Fall 2008 Problem Set #6 Suggested Solutions Hypothesis Testing in the Constant Expected Return Model Instructions In this part of the lab you will use R to • Estimate parameters of the constant expected return (CER) model and test various hypotheses about the parameters and assumptions of the model. Exercises 1. Consider the constant expected return model (CER) 2 , 1, , ~ (0, ) cov( , ) it i it it i it jt ij R t T iid N μ ε ε σ ε ε σ = + = = K where Rit denotes the continuously compounded return on asset i, i =Vanguard long term bond index fund (VBLTX), Fidelity Magellan stock mutual fund (FMAGX), General Motors stock (GM). The commands in the R script file lab6.r will be helpful in completing this assignment. a) Using sample descriptive statistics, give estimates for the model parameters , , and i i ijμ σ ρ . Arrange these estimates nicely in a table. > cbind(muhat.vals,sigmahat.vals) muhat.vals sigmahat.vals vbltx 0.006469 0.02313 fmagx 0.005820 0.04680 gm 0.002323 0.10415 > rhohat.vals vbltx,fmagx vbltx,gm fmagx,gm -0.08004 -0.07185 0.4757 All of the mean values are close to zero. As expected, the bond fund has the lowest SD and gm has the highest SD. Notice that vbltx is negatively correlated (weakly) with fmagx and gm, but fmagx and gm are moderately positively correlated. Typically, bond returns are nearly uncorrelated with stock returns. b) For each iμ (i = VBLTX, FMAGX and GM) test the hypotheses 0 1: 0 vs. : 0i iH Hμ μ= ≠ using a 5% significance level. Perform the test using the t-statistic as well as the 95% confidence. What do you conclude? (note: you can use the R function t.test for this problem) > t.test(lab4.z[,"vbltx"]) One-sample t-Test data: lab4.z[, "vbltx"] = 3.064, df = 119, p-value = 0.0027 alternative hypothesis: mean is not equal to 0 95 percent confidence interval: 0.002288 0.010649 sample estimates: mean of x 0.006469 Here, |t-stat| = 3.064 > 2 so we reject H0: μ=0 at the 5 % level. Also, the p-value is less than 0.05 and 0 is not in the 95% confidence interval. > t.test(lab4.z[,"fmagx"]) One-sample t-Test data: lab4.z[, "fmagx"] = 1.362, df = 119, p-value = 0.1757 alternative hypothesis: mean is not equal to 0 95 percent confidence interval: -0.002639 0.014279 sample estimates: mean of x 0.00582 Here, |t-stat| = 1.362 < 2 so we do not reject H0: μ=0 at the 5 % level. Also, the p-value is greater than 0.05 and 0 is in the 95% confidence interval. > t.test(lab4.z[,"gm"]) One-sample t-Test data: lab4.z[, "gm"] = 0.2443, df = 119, p-value = 0.8074 alternative hypothesis: mean is not equal to 0 95 percent confidence interval: -0.006478 0.010301 sample estimates: mean of x mean of y 0.007424 0.005513 Here, |t-stat| = 0.4512 < 2 so we do not reject H0: μ1=μ2 at the 5 % level. Also, the p- value is greater than 0.05 and 0 is in the 95% confidence interval. # fmagx > t.test(x=lab4.z[smpl.1,"fmagx"], + y=lab4.z[smpl.2,"fmagx"], + var.equal=T) Pooled-Variance Two-Sample t-Test data: lab4.z[smpl.1, "fmagx"] and lab4.z[smpl.2 , "fmagx"] t = 0.5231, df = 118, p-value = 0.6019 alternative hypothesis: difference in means is no t equal to 0 95 percent confidence interval: -0.01249 0.02145 sample estimates: mean of x mean of y 0.008062 0.003578 Here, |t-stat| = 0.5231 < 2 so we do not reject H0: μ1=μ2 at the 5 % level. Also, the p- value is greater than 0.05 and 0 is in the 95% confidence interval. # gm > t.test(x=lab4.z[smpl.1,"gm"], + y=lab4.z[smpl.2,"gm"], + var.equal=T) Pooled-Variance Two-Sample t-Test data: lab4.z[smpl.1, "gm"] and lab4.z[smpl.2, " gm"] t = 0.2574, df = 118, p-value = 0.7973 alternative hypothesis: difference in means is no t equal to 0 95 percent confidence interval: -0.03289 0.04272 sample estimates: mean of x mean of y 0.00478 -0.0001348 Here, |t-stat| = 0.2574 < 2 so we do not reject H0: μ1=μ2 at the 5 % level. Also, the p- value is greater than 0.05 and 0 is in the 95% confidence interval. e) For each asset VBLTX, FMAGX and GM compute and plot 24-month rolling estimates of and i iμ σ . What do you conclude about the stability of the expected returns and standard deviations over time? (note: use the R function rollapply to compute the rolling estimates) 24-month rolling means and sds for VBLTX 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 -0 .0 8 -0 .0 6 -0 .0 4 -0 .0 2 0. 00 0. 02 0. 04 Rolling means Rolling sds vbltx returns The rolling means for the bond returns are always positive. They dip close to zero for a short period around 2000. Otherwise, they appear fairly constant. The rolling stdevs show some variation. In particular, the stdevs increase toward the end of 2003. In general, there is not too much variation in the rolling estimates, so the CER model assumptions are not too bad. 24-month rolling means and sds for FMAGX 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 -0 .1 5 -0 .1 0 -0 .0 5 0. 00 0. 05 Rolling means Rolling sds fmagx returns The rolling means for fmagx show quite a lot of variability. They start positive, then dip negative during the tech bust, and become positive again at the end of 2004. Clearly, the mean is not constant over time. The rolling sds also show time variation. They decrease rather dramatically at the end of 2004. The CER model is not very good for fmagx over the 10 year period. 24-month rolling means and sds for GM 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 -0 .2 5 -0 .2 0 -0 .1 5 -0 .1 0 -0 .0 5 0. 00 0. 05 0. 10 0. 15 0. 20 Rolling means Rolling sds gm returns The rolling means for gm fluctuate between small positive and negative numbers. Overall, the mean appears to be roughly equal to zero. GM was not a good stock to invest in over this ten year period. The rolling stdev values are fairly constant as well. The CER model appears to be a reasonable model for GM.