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29#
发表于 2012-3-26 17:12
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Kathy Williams, CFA, and Nigel Faber, CFA, have been managing a hedge fund over the past 18 months. The fund’s objective is to eliminate all systematic risk while earning a portfolio return greater than the return on Treasury Bills. Williams and Faber want to test whether they have achieved this objective. Using monthly data, they find that the average monthly return for the fund was 0.417%, and the average return on Treasury Bills was 0.384%. They perform the following regression (Equation I):
(fund return)t = b0 + b1 (T-bill return) t + b2 (S&P 500 return) t + b3 (global index return) t + et
The correlation matrix for the independent variables appears below:
| S&P 500 | Global Index |
T-bill | 0.163 | 0.141 |
S&P 500 |
| 0.484 |
In performing the regression, they obtain the following results for Equation I: Variable | Coefficient | Standard Error |
Intercept | 0.232 | 0.098 |
T-bill return | 0.508 | 0.256 |
S&P 500 Return | −0.0161 | 0.032 |
Global index return | 0.0037 | 0.034 |
R2 = 22.44%
adj. R2 = 5.81%
standard error of forecast = 0.0734 (percent)
In addition to the regular summary statistics, Williams computes the correlation coefficient for the residuals, i.e., correlation of the last 17 residuals on the lag of those residuals. The value of the correlation coefficient is 0.605.
Williams argues that the equation may suffer from multicollinearity and reruns the regression omitting the return on the global index. This time, the regression (Equation II) is:
(fund return) t = b0 + b1 (T-bill return) t + b2 (S&P 500 return) t +et
The results for Equation II are: Variable | Coefficient | Standard Error |
Intercept | 0.232 | 0.095 |
T-bill return | 0.510 | 0.246 |
S&P 500 return | −0.015 | 0.028 |
R2 = 22.37%
adj. R2 = 12.02%
standard error of forecast = 0.0710 (percent)
The correlation of the residuals on their lagged values for this regression 0.558.
Finally, Williams reruns the regression omitting the return on the S&P 500 as well. This time, the regression (Equation III) is:
(fund return) t = b0 + b1 (T-bill return) t +et
The results for Equation III are: Variable |
Coefficient |
Standard Error |
Intercept | 0.229 | 0.093 |
T-bill return | 0.4887 | 0.2374 |
R2 = 20.94%
adj. R2 = 16.00%
standard error of forecast = 0.0693 (percent)
The correlation of the residuals on their lagged values for this regression 0.604. In the regression using Equation I, which of the following hypotheses can be rejected at a 5% level of significance in a two-tailed test? (The corresponding independent variable is indicated after each null hypothesis.) | | C)
| H0: b0 = 0 (intercept) |
|
The critical t-value for 18 − 3 − 1 = 14 degrees of freedom in a two-tailed test at a 5% significance level is 2.145. Although the t-statistic for T-bill is close at 0.508 / 0.256 = 1.98, it does not exceed the critical value. Only the intercept’s coefficient has a significant t-statistic for the indicated test: t = 0.232 / 0.098 = 2.37. (Study Session 3, LOS 12.b)
In the regression using Equation II, which of the following hypothesis or hypotheses can be rejected at a 5% level of significance in a two-tailed test? (The corresponding independent variable is indicated after each null hypothesis.) A)
| H0: b0 = 0 (intercept) and b1 = 0 (T-bill) only. |
| B)
| H0: b0 = 0 (intercept) only. |
| C)
| H0: b1 = 0 (T-bill) and H0: b2 = 0 (S&P 500) only. |
|
The critical t-value for 18 − 2 − 1 = 15 degrees of freedom in a two-tailed test at a 5% significance level is 2.131. The t-statistics on the intercept, T-bill and S&P 500 coefficients are 2.442, 2.073, −0.536, respectively. Therefore, only the coefficient on the intercept is significant. (Study Session 3, LOS 12.b)
With respect to multicollinearity and Williams’ removal of the global index variable when running regression Equation II, Williams had: A)
| reason to be suspicious, but she took the wrong step to cure the problem. |
| B)
| no reason to be suspicious, but took a correct step to improve the analysis. |
| C)
| reason to be suspicious and took the correct step to cure the problem. |
|
Investigating multicollinearity is justified for two reasons. First, the S&P 500 and the global index have a significant degree of correlation. Second, neither of the market index variables are significant in the first specification. The correct step is to remove one of the variables, as Williams did, to see if the remaining variable becomes significant. (Study Session 3, LOS 12.j)
At a 5% level of significance, which of the equations suffers from serial correlation? | B)
| Equations I, II, and III. |
| C)
| Equations I and III only. |
|
Using the correlations of the residuals, the DW statistics are 2 × (1 − 0.605) = 0.79, 0.88, and 0.79 for Equations I, II, III, respectively. The critical values for the DW test are 0.93 for Equation I, 1.05 for Equation II, and 1.16 for Equation III. Note that in the calculation of the DW statistics with the correlation coefficient of the residuals, we have made the simplifying assumption that the sample size is large enough to use the DW = 2(1 − r) method. (Study Session 3, LOS 12.i)
Which of the following problems, multicollinearity and/or serial correlation, can bias the estimates of the slope coefficients? A)
| Serial correlation, but not multicollinearity. |
| B)
| Both multicollinearity and serial correlation. |
| C)
| Multicollinearity, but not serial correlation. |
|
Multicollinearity can bias the coefficients because the shared movement of the independent variables. Serial correlation biases the standard errors of the slope coefficients. (Study Session 3, LOS 12.j)
If we expect that next month the T-bill rate will equal its average over the last 18 months, using Equation III, calculate the 95% confidence interval for the expected fund return.
The forecast is 0.417 = 0.229 + 0.4887 × (0.384). The 95% confidence interval is Y ± (tc × sf) and tc for 16 degrees of freedom for a 2 tailed test = 2.120. The 95% confidence interval = 0.417 ± (2.120)(.0693) = 0.270 to 0.564. (Study Session 3, LOS 12.c) |
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