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标题: Quantitative Analysis【Reading 13】Sample [打印本页]
作者: kasinkei    时间: 2012-3-27 13:31     标题: [2012 L2] Quantitative Analysis【Session 3 - Reading 13】Sample

David Wellington, CFA, has estimated the following log-linear trend model: LN(xt) = b0 + b1t + εt. Using six years of quarterly observations, 2001:I to 2006:IV, Wellington gets the following estimated equation: LN(xt) = 1.4 + 0.02t. The first out-of-sample forecast of xt for 2007:I is closest to:
A)
1.88.
B)
4.14.
C)
6.69.


Wellington’s out-of-sample forecast of LN(xt) is 1.9 = 1.4 + 0.02 × 25, and e1.9 = 6.69.
作者: kasinkei    时间: 2012-3-27 13:31

Modeling the trend in a time series of a variable that grows at a constant rate with continuous compounding is best done with:
A)
a log-linear transformation of the time series.
B)
a moving average model.
C)
simple linear regression.


The log-linear transformation of a series that grows at a constant rate with continuous compounding (exponential growth) will cause the transformed series to be linear.
作者: kasinkei    时间: 2012-3-27 13:32

In the time series model: yt=b0 + b1 t + εt, t=1,2,…,T, the:
A)
change in the dependent variable per time period is b1.
B)
disturbance terms are autocorrelated.
C)
disturbance term is mean-reverting.

The slope is the change in the dependent variable per unit of time. The intercept is the estimate of the value of the dependent variable before the time series begins. The disturbance term should be independent and identically distributed. There is no reason to expect the disturbance term to be mean-reverting, and if the residuals are autocorrelated, the research should correct for that problem.
作者: kasinkei    时间: 2012-3-27 13:34

Yolanda Seerveld is an analyst studying the growth of sales of a new restaurant chain called Very Vegan. The increase in the public’s awareness of healthful eating habits has had a very positive effect on Very Vegan’s business. Seerveld has gathered quarterly data for the restaurant’s sales for the past three years. Over the twelve periods, sales grew from $17.2 million in the first quarter to $106.3 million in the last quarter. Because Very Vegan has experienced growth of more than 500% over the three years, the Seerveld suspects an exponential growth model may be more appropriate than a simple linear trend model. However, she begins by estimating the simple linear trend model:
(sales)t = α + β × (Trend)t + εt
Where the Trend is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
Regression Statistics
Multiple R0.952640
R20.907523
Adjusted R2

0.898275

Standard Error8.135514
Observations

12

1st order autocorrelation coefficient of the residuals: −0.075

ANOVA
   dfSS
Regression16495.203
Residual10661.8659
Total117157.069

Coefficients Standard Error
Intercept

10.0015

5.0071

Trend

6.7400

0.6803


The analyst then estimates the following model:
(natural logarithm of sales)t = α + β × (Trend)t + εt
Regression Statistics
Multiple R0.952028
R20.906357
Adjusted R20.896992
Standard Error0.166686
Observations12
1st order autocorrelation coefficient of the residuals: −0.348

ANOVA
   dfSS
Regression12.6892
Residual100.2778
Total112.9670

   Coefficients Standard Error
Intercept2.9803 0.1026
Trend0.1371 0.0140

Seerveld compares the results based upon the output statistics and conducts two-tailed tests at a 5% level of significance. One concern is the possible problem of autocorrelation, and Seerveld makes an assessment based upon the first-order autocorrelation coefficient of the residuals that is listed in each set of output. Another concern is the stationarity of the data. Finally, the analyst composes a forecast based on each equation for the quarter following the end of the sample. Are either of the slope coefficients statistically significant?
A)
Yes, both are significant.
B)
The simple trend regression is, but not the log-linear trend regression.
C)
The simple trend regression is not, but the log-linear trend regression is.


The respective t-statistics are 6.7400 / 0.6803 = 9.9074 and 0.1371 / 0.0140 = 9.7929. For 10 degrees of freedom, the critical t-value for a two-tailed test at a 5% level of significance is 2.228, so both slope coefficients are statistically significant. (Study Session 3, LOS 13.a)
Based upon the output, which equation explains the cause for variation of Very Vegan’s sales over the sample period?
A)
Both the simple linear trend and the log-linear trend have equal explanatory power.
B)
The simple linear trend.
C)
The cause cannot be determined using the given information.


To actually determine the explanatory power for sales itself, fitted values for the log-linear trend would have to be determined and then compared to the original data. The given information does not allow for such a comparison. (Study Session 3, LOS 13.b)
With respect to the possible problems of autocorrelation and nonstationarity, using the log-linear transformation appears to have:
A)
improved the results for autocorrelation but not nonstationarity.
B)
improved the results for nonstationarity but not autocorrelation.
C)
not improved the results for either possible problems.


The fact that there is a significant trend for both equations indicates that the data is not stationary in either case. As for autocorrelation, the analyst really cannot test it using the Durbin-Watson test because there are fewer than 15 observations, which is the lower limit of the DW table. Looking at the first-order autocorrelation coefficient, however, we see that it increased (in absolute value terms) for the log-linear equation. If anything, therefore, the problem became more severe. (Study Session 3, LOS 13.b)
The primary limitation of both models is that:
A)
each uses only one explanatory variable.
B)
the results are difficult to interpret.
C)
regression is not appropriate for estimating the relationship.


The main problem with a trend model is that it uses only one variable so the underlying dynamics are really not adequately addressed. A strength of the models is that the results are easy to interpret. The levels of many economic variables such as the sales of a firm, prices, and gross domestic product (GDP) have a significant time trend, and a regression is an appropriate tool for measuring that trend. (Study Session 3, LOS 13.b)
Using the simple linear trend model, the forecast of sales for Very Vegan for the first out-of-sample period is:
A)
$97.6 million.
B)
$113.0 million.
C)
$123.0 million.


The forecast is 10.0015 + (13 × 6.7400) = 97.62. (Study Session 3, LOS 13.a)
Using the log-linear trend model, the forecast of sales for Very Vegan for the first out-of-sample period is:
A)
$117.0 million.
B)
$121.2 million.
C)
$109.4 million.


The forecast is e2.9803 + (13 × 0.1371) = 117.01. (Study Session 3, LOS 13.a)
作者: kasinkei    时间: 2012-3-27 13:35

Clara Holmes, CFA, is attempting to model the importation of an herbal tea into the United States which last year was $ 54 million. She gathers 24 years of annual data, which is in millions of inflation-adjusted dollars.
She computes the following equation:

(Tea Imports)t = 3.8836 + 0.9288 × (Tea Imports)t − 1 + et
t-statistics (0.9328)(9.0025)

R2 = 0.7942
Adj. R2 = 0.7844
SE = 3.0892
N = 23

Holmes and her colleague, John Briars, CFA, discuss the implication of the model and how they might improve it. Holmes is fairly satisfied with the results because, as she says “the model explains 78.44 percent of the variation in the dependent variable.” Briars says the model actually explains more than that.
Briars asks about the Durbin-Watson statistic. Holmes said that she did not compute it, so Briars reruns the model and computes its value to be 2.1073. Briars says “now we know serial correlation is not a problem.” Holmes counters by saying “rerunning the model and computing the Durbin-Watson statistic was unnecessary because serial correlation is never a problem in this type of time-series model.”
Briars and Holmes decide to ask their company’s statistician about the consequences of serial correlation. Based on what Briars and Holmes tell the statistician, the statistician informs them that serial correlation will only affect the standard errors and the coefficients are still unbiased. The statistician suggests that they employ the Hansen method, which corrects the standard errors for both serial correlation and heteroskedasticity.
Given the information from the statistician, Briars and Holmes decide to use the estimated coefficients to make some inferences. Holmes says the results do not look good for the future of tea imports because the coefficient on (Tea Import)t − 1 is less than one. This means the process is mean reverting. Using the coefficients in the output, says Holmes, “we know that whenever tea imports are higher than 41.810, the next year they will tend to fall. Whenever the tea imports are less than 41.810, then they will tend to rise in the following year.” Briars agrees with the general assertion that the results suggest that imports will not grow in the long run and tend to revert to a long-run mean, but he says the actual long-run mean is 54.545. Briars then computes the forecast of imports three years into the future.With respect to the statements made by Holmes and Briars concerning serial correlation and the importance of the Durbin-Watson statistic:
A)
they were both incorrect.
B)
Holmes was correct and Briars was incorrect.
C)
Briars was correct and Holmes was incorrect.


Briars was incorrect because the DW statistic is not appropriate for testing serial correlation in an autoregressive model of this sort. Holmes was incorrect because serial correlation can certainly be a problem in such a model. They need to analyze the residuals and compute autocorrelation coefficients of the residuals to better determine if serial correlation is a problem. (Study Session 3, LOS 12.i)
With respect to the statement that the company’s statistician made concerning the consequences of serial correlation, assuming the company’s statistician is competent, we would most likely deduce that Holmes and Briars did not tell the statistician:
A)
the sample size.
B)
the value of the Durbin-Watson statistic.
C)
the model’s specification.


Serial correlation will bias the standard errors. It can also bias the coefficient estimates in an autoregressive model of this type. Thus, Briars and Holmes probably did not tell the statistician the model is an AR(1) specification. (Study Session 3, LOS 12.k)
The statistician’s statement concerning the benefits of the Hansen method is:
A)
not correct, because the Hansen method only adjusts for problems associated with serial correlation but not heteroskedasticity.
B)
not correct, because the Hansen method only adjusts for problems associated with heteroskedasticity but not serial correlation.
C)
correct, because the Hansen method adjusts for problems associated with both serial correlation and heteroskedasticity.


The statistician is correct because the Hansen method adjusts for problems associated with both serial correlation and heteroskedasticity. (Study Session 3, LOS 12.i)
Using the model’s results, Briar’s forecast for three years into the future is:
A)
$54.543 million.
B)
$47.151 million.
C)
$54.108 million.

Briars’ forecasts for the next three years would be:
year one: 3.8836 + 0.9288 × 54 = 54.0388
year two: 3.8836 + 0.9288 × (54.0388) = 54.0748
year three: 3.8836 + 0.9288 × (54.0748) = 54.1083
(Study Session 3, LOS 13.a)


With respect to the comments of Holmes and Briars concerning the mean reversion of the import data, the long-run mean value that:
A)
Briars computes is correct.
B)
Briars computes is not correct, and his conclusion is probably not accurate.
C)
Briars computes is not correct, but his conclusion is probably accurate.


Briars has computed a value that would be correct if the results of the model were reliable. The long-run mean would be 3.8836 / (1 − 0.9288)= 54.5450. (Study Session 3, LOS 13.a)
Given the nature of their analysis, the most likely potential problem that Briars and Holmes need to investigate is:
A)
autocorrelation.
B)
unit root.
C)
multicollinearity.


Multicollinearity cannot be a problem because there is only one independent variable. For a time series AR model, autocorrelation is a bigger worry. The model may have been misspecified leading to statistically significant autocorrelations. Unit root does not seem to be a problem given the value of b1<1. (Study Session 3, LOS 13.e)
作者: kasinkei    时间: 2012-3-27 13:36

Dianne Hart, CFA, is considering the purchase of an equity position in Book World, Inc, a leading seller of books in the United States. Hart has obtained monthly sales data for the past seven years, and has plotted the data points on a graph. Which of the following statements regarding Hart’s analysis of the data time series of Book World’s sales is most accurate? Hart should utilize a:
A)
log-linear model to analyze the data because it is likely to exhibit a compound growth trend.
B)
mean-reverting model to analyze the data because the time series pattern is covariance stationary.
C)
linear model to analyze the data because the mean appears to be constant.


A log-linear model is more appropriate when analyzing data that is growing at a compound rate. Sales are a classic example of a type of data series that normally exhibits compound growth.
作者: kasinkei    时间: 2012-3-27 13:37

Trend models can be useful tools in the evaluation of a time series of data. However, there are limitations to their usage. Trend models are not appropriate when which of the following violations of the linear regression assumptions is present?

A) Heteroskedasticity.

B) Model misspecification.

C) Serial correlation.





--------------------------------------------------------------------------------

One of the primary assumptions of linear regression is that the residual terms are not correlated with each other. If serial correlation, also called autocorrelation, is present, then trend models are not an appropriate analysis tool.
作者: kasinkei    时间: 2012-3-27 14:03

Rhonda Wilson, CFA, is analyzing sales data for the TUV Corp, a current equity holding in her portfolio. She observes that sales for TUV Corp. have grown at a steadily increasing rate over the past ten years due to the successful introduction of some new products. Wilson anticipates that TUV will continue this pattern of success. Which of the following models is most appropriate in her analysis of sales for TUV Corp.?
A)
A linear tend model, because the data series is equally distributed above and below the line and the mean is constant.
B)
A log-linear trend model, because the data series can be graphed using a straight, upward-sloping line.
C)
A log-linear trend model, because the data series exhibits a predictable, exponential growth trend.


The log-linear trend model is the preferred method for a data series that exhibits a trend or for which the residuals are predictable. In this example, sales grew at an exponential, or increasing rate, rather than a steady rate.
作者: kasinkei    时间: 2012-3-27 14:04

To qualify as a covariance stationary process, which of the following does not have to be true?
A)
Covariance(xt, xt-1) = Covariance(xt, xt-2).
B)
E[xt] = E[xt+1].
C)
Covariance(xt, xt-2) = Covariance(xt, xt+2).


If a series is covariance stationary then the unconditional mean is constant across periods. The unconditional mean or expected value is the same from period to period: E[xt] = E[xt+1]. The covariance between any two observations equal distance apart will be equal, e.g., the t and t-2 observations with the t and t+2 observations. The one relationship that does not have to be true is the covariance between the t and t-1 observations equaling that of the t and t-2 observations
作者: kasinkei    时间: 2012-3-27 14:05

To qualify as a covariance stationary process, which of the following does not have to be true?
A)
Covariance(xt, xt-1) = Covariance(xt, xt-2).
B)
E[xt] = E[xt+1].
C)
Covariance(xt, xt-2) = Covariance(xt, xt+2).


If a series is covariance stationary then the unconditional mean is constant across periods. The unconditional mean or expected value is the same from period to period: E[xt] = E[xt+1]. The covariance between any two observations equal distance apart will be equal, e.g., the t and t-2 observations with the t and t+2 observations. The one relationship that does not have to be true is the covariance between the t and t-1 observations equaling that of the t and t-2 observations
作者: kasinkei    时间: 2012-3-27 14:05

Which of the following statements regarding covariance stationarity is CORRECT?
A)
A time series that is covariance stationary may have residuals whose mean changes over time.
B)
The estimation results of a time series that is not covariance stationary are meaningless.
C)
A time series may be both covariance stationary and have heteroskedastic residuals.


Covariance stationarity requires that the expected value and the variance of the time series be constant over time.
作者: kasinkei    时间: 2012-3-27 14:06

Which of the following statements regarding covariance stationarity is CORRECT?
A)
A time series that is covariance stationary may have residuals whose mean changes over time.
B)
The estimation results of a time series that is not covariance stationary are meaningless.
C)
A time series may be both covariance stationary and have heteroskedastic residuals.


Covariance stationarity requires that the expected value and the variance of the time series be constant over time.
作者: kasinkei    时间: 2012-3-27 14:06

The model xt = b0 + b1 xt-1 + b2 xt-2 + b3 xt-3 + b4 xt-4 + εt is:
A)
an autoregressive model, AR(4).
B)
an autoregressive conditional heteroskedastic model, ARCH.
C)
a moving average model, MA(4).


This is an autoregressive model (i.e., lagged dependent variable as independent variables) of order p=4 (that is, 4 lags).
作者: kasinkei    时间: 2012-3-27 14:06

The model xt = b0 + b1 xt − 1 + b2 xt − 2  + εt is:
A)
an autoregressive conditional heteroskedastic model, ARCH.
B)
a moving average model, MA(2).
C)
an autoregressive model, AR(2).




This is an autoregressive model (i.e., lagged dependent variable as independent variables) of order p = 2 (that is, 2 lags).
作者: kasinkei    时间: 2012-3-27 14:09

Diem Le is analyzing the financial statements of McDowell Manufacturing. He has modeled the time series of McDowell’s gross margin over the last 15 years. The output is shown below. Assume 5% significance level for all statistical tests.

Autoregressive Model
Gross Margin – McDowell Manufacturing
Quarterly Data: 1st Quarter 1985 to 4th Quarter 2000

Regression Statistics

R-squared

0.767

Standard Error

0.049

Observations

64

Durbin-Watson

1.923 (not statistically significant)



Coefficient

Standard Error

t-statistic


Constant

0.155

0.052

?????


Lag 1

0.240

0.031

?????


Lag 4

0.168

0.038

?????

Autocorrelation of Residuals

Lag

Autocorrelation

Standard Error

t-statistic

1

0.015

0.129

?????

2

-0.101

0.129

?????

3

-0.007

0.129

?????

4

0.095

0.129

?????

Partial List of Recent Observations

Quarter

Observation

4th Quarter 2002

0.250

1st Quarter 2003

0.260

2nd Quarter 2003

0.220

3rd Quarter 2003

0.200

4th Quarter 2003

0.240

Abbreviated Table of the Student’s t-distribution (One-Tailed Probabilities)

df

p = 0.10

p = 0.05

p = 0.025

p = 0.01

p = 0.005

50

1.299

1.676

2.009

2.403

2.678

60

1.296

1.671

2.000

2.390

2.660

70

1.294

1.667

1.994

2.381

2.648

This model is best described as:
A)
an AR(1) model with a seasonal lag.
B)
an ARMA(2) model.
C)
an MA(2) model.


This is an autoregressive AR(1) model with a seasonal lag. Remember that an AR model regresses a dependent variable against one or more lagged values of itself. (Study Session 3, LOS 13.o)
Which of the following can Le conclude from the regression? The time series process:
A)
includes a seasonality factor and a unit root.
B)
includes a seasonality factor, has significant explanatory power, and is mean reverting.
C)
includes a seasonality factor and has significant explanatory power.


The gross margin in the current quarter is related to the gross margin four quarters (one year) earlier. To determine whether there is a seasonality factor, we need to test the coefficient on lag 4. The t-statistic for the coefficients is calculated as the coefficient divided by the standard error with 61 degrees of freedom (64 observations less three coefficient estimates). The critical t-value for a significance level of 5% is about 2.000 (from the table). The computed t-statistic for lag 4 is 0.168/0.038 = 4.421. This is greater than the critical value at even alpha = 0.005, so it is statistically significant. This suggests an annual seasonal factor.
Both slope coefficients are significantly different from one:
first lag coefficient: t = (1-0.24)/0.031 = 24.52

second lag coefficient: t = (1-0.168)/0.038 =21.89


Thus, the process does not contain a unit root, is stationary, and is mean reverting. The process has significant explanatory power since both slope coefficients are significant and the coefficient of determination is 0.767. (Study Session 3, LOS 13.l)

Le can conclude that the model is:
A)
properly specified because there is no evidence of autocorrelation in the residuals.
B)
not properly specified because there is evidence of autocorrelation in the residuals and the Durbin-Watson statistic is not significant.
C)
properly specified because the Durbin-Watson statistic is not significant.


The Durbin-Watson test is not an appropriate test statistic in an AR model, so we cannot use it to test for autocorrelation in the residuals. However, we can test whether each of the four lagged residuals autocorrelations is statistically significant. The t-test to accomplish this is equal to the autocorrelation divided by the standard error with 61 degrees of freedom (64 observations less 3 coefficient estimates). The critical t-value for a significance level of 5% is about 2.000 from the table. The appropriate t-statistics are:
None of these are statically significant, so we can conclude that there is no evidence of autocorrelation in the residuals, and therefore the AR model is properly specified. (Study Session 3, LOS 13.d)

What is the 95% confidence interval for the sales in the first quarter of 2004?
A)
0.197 to 0.305.
B)
0.158 to 0.354.
C)
0.168 to 0.240.


The forecast for the following quarter is 0.155 + 0.240(0.240) + 0.168(0.260) = 0.256. Since the standard error is 0.049 and the corresponding t-statistic is 2, we can be 95% confident that sales will be within 0.256 – 2 × (0.049) and 0.256 + 2 × (0.049) or 0.158 to 0.354. (Study Session 3, LOS 11.h)
With respect to heteroskedasticity, we can say:
A)
heteroskedasticity is not a problem because the DW statistic is not significant.
B)
nothing.
C)
an ARCH process exists because the autocorrelation coefficients of the residuals have different signs.


None of the information in the problem provides information concerning heteroskedasticity. Note that heteroskedasticity occurs when the variance of the error terms is not constant. When heteroskedasticity is present in a time series, the residuals appear to come from different distributions (model seems to fit better in some time periods than others). (Study Session 3, LOS 12.i)
Using the provided information, the forecast for the 2nd quarter of 2004 is:
A)
0.192.
B)
0.253.
C)
0.250.


To get the 2nd quarter forecast, we use the one period forecast for the 1st quarter of 2004, which is 0.155 + 0.240(0.240) + 0.168(0.260) = 0.256. The 4th lag for the 2nd quarter is 0.22. Thus the forecast for the 2nd quarter is 0.155 + 0.240(0.256) + 0.168(0.220) = 0.253. (Study Session 3, LOS 12.c)
作者: MarginofSafety    时间: 2012-3-27 14:15

Troy Dillard, CFA, has estimated the following equation using semiannual data: xt = 44 + 0.1×xt–1 – 0.25×xt–2 - 0.15×xt–3 + et. Given the data in the table below, what is Dillard’s best forecast of the second half of 2007?
Time

Value

2003: I31
2003: II31
2004: I33
2004: II33
2005: I36
2005: II35
2006: I32
2006: II33

A)
34.36.
B)
34.05.
C)
60.55.


To get the answer, Dillard must first make the forecast for 2007:I
E[x2007:I]= 44 + 0.1 × xt–1 - 0.25 × xt–2 - 0.15 × xt–3
E[x2007:I] = 44 + 0.1×33 - 0.25×32 - 0.15×35
E[x2007:I] = 34.05
Then, use this forecast in the equation for the first lag:
E[x2007:II] = 44 + 0.1×34.05 - 0.25×33 - 0.15×32
E[x2007:II] = 34.36
作者: MarginofSafety    时间: 2012-3-27 14:16

Troy Dillard, CFA, has estimated the following equation using quarterly data: xt = 93 - 0.5×xt–1 + 0.1×xt–4 + et. Given the data in the table below, what is Dillard's best estimate of the first quarter of 2007?
Time

Value

2005: I62
2005: II62
2005: III66
2005: IV66
2006: I72
2006: II70
2006: III64
2006: IV66

A)
66.40.
B)
66.60.
C)
67.20.


To get the answer, Dillard will use the data for 2006: IV and 2006: I, xt–1 = 66 and xt–4 = 72 respectively:
E[x2007:I] = 93– 0.5×xt–2 + 0.1×xt–4
E[x2007:I] = 93– 0.5×66 + 0.1×72
E[x2007:I] = 67.20
作者: MarginofSafety    时间: 2012-3-27 14:18

Albert Morris, CFA, is evaluating the results of an estimation of the number of wireless phone minutes used on a quarterly basis within the territory of Car-tel International, Inc. Some of the information is presented below (in billions of minutes):

Wireless Phone Minutes (WPM)t = bo + b1 WPMt-1 + ε t

ANOVA

Degrees of Freedom

Sum of Squares

Mean Square


Regression

1

7,212.641

7,212.641


Error

26

3,102.410

119.324


Total

27

10,315.051




Coefficients

Coefficient

Standard Error of the Coefficient


Intercept

-8.0237

2.9023


WPM t-1

1.0926

0.0673


The variance of the residuals from one time period within the time series is not dependent on the variance of the residuals in another.The value for WPM this period is 544 billion. Using the results of the model, the forecast for three periods in the future is:
A)
683.18.
B)
691.30.
C)
586.35.


The one-period forecast is −8.023 + (1.0926 × 544) = 586.35.
The two-period forecast is then −8.023 + (1.0926 × 586.35) = 632.62.
Finally, the three-period forecast is then −8.023 + (1.0926 × 632.62) = 683.18.

Is the time series of WPM covariance stationary?
A)
Yes, because the computed t-statistic for a slope of 1 is significant.
B)
Yes, because the computed t-statistic for a slope of 1 is not significant.
C)
No, because the Coefficient of WPMt-1 is not less than 1.


For an AR(1) model − the type specified in this problem, when b1 is not less than 1, the time series is said to be covariance nonstationary.
The above model was specified as a(n):
A)
Autoregressive (AR) Model.
B)
Moving Average (MA) Model.
C)
Autoregressive (AR) Model with a seasonal lag.


The model is specified as an AR Model, but there is no seasonal lag. No moving averages are employed in the estimation of the model.
Based upon the information provided, Morris would get more meaningful statistical results by:
A)
adding more lags to the model.
B)
first differencing the data.
C)
doing nothing. No information provided suggests that any of these will improve the specification.


Since the slope coefficient is greater than one, the process is not covariance stationary. A common technique to correct for this is to first difference the variable to perform the following regression: Δ(WPM)t = bo + b1 Δ(WPM)t-1 + ε t.
作者: MarginofSafety    时间: 2012-3-27 14:18

Consider the estimated model xt = −6.0 + 1.1 xt − 1 + 0.3 xt − 2 + εt that is estimated over 50 periods. The value of the time series for the 49th observation is 20 and the value of the time series for the 50th observation is 22. What is the forecast for the 52nd observation?
A)
24.2.
B)
42.
C)
27.22.


Using the chain-rule of forecasting,
Forecasted x51 = −6.0 + 1.1(22) + 0.3(20) = 24.2.
Forecasted x52 = −6.0 + 1.1(24.2) + 0.3(22) = 27.22.
作者: MarginofSafety    时间: 2012-3-27 14:19

Consider the estimated model xt = -6.0 + 1.1 xt-1 + 0.3 xt-2 + εt that is estimated over 50 periods. The value of the time series for the 49th observation is 20 and the value of the time series for the 50th observation is 22. What is the forecast for the 51st observation?
A)
30.2.
B)
23.
C)
24.2.


Forecasted x51 = -6.0 + 1.1 (22) + 0.3 (20) = 24.2.
作者: MarginofSafety    时间: 2012-3-27 14:19

The regression results from fitting an AR(1) model to the first-differences in enrollment growth rates at a large university includes a Durbin-Watson statistic of 1.58. The number of quarterly observations in the time series is 60. At 5% significance, the critical values for the Durbin-Watson statistic are dl = 1.55 and du = 1.62. Which of the following is the most accurate interpretation of the DW statistic for the model?
A)
Since dl < DW < du, the results of the DW test are inconclusive.
B)
The Durbin-Watson statistic cannot be used with AR(1) models.
C)
Since DW > dl, the null hypothesis of no serial correlation is rejected.


The Durbin-Watson statistic is not useful when testing for serial correlation in an autoregressive model where one of the independent variables is a lagged value of the dependent variable. The existence of serial correlation in an AR model is determined by examining the autocorrelations of the residuals.
作者: MarginofSafety    时间: 2012-3-27 14:20

The table below includes the first eight residual autocorrelations from fitting the first differenced time series of the absenteeism rates (ABS) at a manufacturing firm with the model ΔABSt = b0 + b1ΔABSt-1 + εt. Based on the results in the table, which of the following statements most accurately describes the appropriateness of the specification of the model, ΔABSt = b0 + b1ΔABSt-1 + εt?

Lagged Autocorrelations of the Residuals of the First Differences in Absenteeism Rates

Lag

Autocorrelation

Standard Error

t-Statistic

1

−0.0738

0.1667

−0.44271

2

−0.1047

0.1667

−0.62807

3

−0.0252

0.1667

−0.15117

4

−0.0157

0.1667

−0.09418

5

−0.1262

0.1667

−0.75705

6

0.0768

0.1667

0.46071

7

0.0038

0.1667

0.02280

8

−0.0188

0.1667

−0.11278

A)
The negative values for the autocorrelations indicate that the model does not fit the time series.
B)
The Durbin-Watson statistic is needed to determine the presence of significant correlation of the residuals.
C)
The low values for the t-statistics indicate that the model fits the time series.


The t-statistics are all very small, indicating that none of the autocorrelations are significantly different than zero. Based on these results, the model appears to be appropriately specified. The error terms, however, should still be checked for heteroskedasticity.
作者: MarginofSafety    时间: 2012-3-27 14:20

The procedure for determining the structure of an autoregressive model is:
A)
estimate an autoregressive model (for example, an AR(1) model), calculate the autocorrelations for the model's residuals, test whether the autocorrelations are different from zero, and add an AR lag for each significant autocorrelation.
B)
test autocorrelations of the residuals for a simple trend model, and specify the number of significant lags.
C)
estimate an autoregressive model (e.g., an AR(1) model), calculate the autocorrelations for the model's residuals, test whether the autocorrelations are different from zero, and revise the model if there are significant autocorrelations.


The procedure is iterative: continually test for autocorrelations in the residuals and stop adding lags when the autocorrelations of the residuals are eliminated. Even if several of the residuals exhibit autocorrelation, the lags should be added one at a time.
作者: MarginofSafety    时间: 2012-3-27 14:20

An analyst modeled the time series of annual earnings per share in the specialty department store industry as an AR(3) process. Upon examination of the residuals from this model, she found that there is a significant autocorrelation for the residuals of this model. This indicates that she needs to:
A)
switch models to a moving average model.
B)
revise the model to include at least another lag of the dependent variable.
C)
alter the model to an ARCH model.


She should estimate an AR(4) model, and then re-examine the autocorrelations of the residuals
作者: MarginofSafety    时间: 2012-3-27 14:22

A monthly time series of changes in maintenance expenses (ΔExp) for an equipment rental company was fit to an AR(1) model over 100 months. The results of the regression and the first twelve lagged residual autocorrelations are shown in the tables below. Based on the information in these tables, does the model appear to be appropriately specified? (Assume a 5% level of significance.)

Regression Results for Maintenance Expense Changes

Model: DExpt = b0 + b1DExpt–1 + et


Coefficients

Standard Error

t-Statistic

p-value


Intercept
1.3304
0.0089
112.2849
< 0.0001

Lag-1
0.1817
0.0061
30.0125
< 0.0001

Lagged Residual Autocorrelations for Maintenance Expense Changes

Lag

Autocorrelation

t-Statistic

Lag

Autocorrelation

t-Statistic

1


−0.239
−2.39

7


−0.018
−0.18

2


−0.278
−2.78

8


−0.033
−0.33

3


−0.045
−0.45

9


0.261
2.61

4


−0.033
−0.33

10


−0.060
−0.60

5


−0.180
−1.80

11


0.212
2.12

6


−0.110
−1.10

12


0.022
0.22
A)
No, because several of the residual autocorrelations are significant.
B)
Yes, because the intercept and the lag coefficient are significant.
C)
Yes, because most of the residual autocorrelations are negative.


At a 5% level of significance, the critical t-value is 1.98. Since the absolute values of several of the residual autocorrelation’s t-statistics exceed 1.98, it can be concluded that significant serial correlation exists and the model should be respecified. The next logical step is to estimate an AR(2) model, then test the associated residuals for autocorrelation. If no serial correlation is detected, seasonality and ARCH behavior should be tested.
作者: MarginofSafety    时间: 2012-3-27 14:22

David Brice, CFA, has used an AR(1) model to forecast the next period’s interest rate to be 0.08. The AR(1) has a positive slope coefficient. If the interest rate is a mean reverting process with an unconditional mean, a.k.a., mean reverting level, equal to 0.09, then which of the following could be his forecast for two periods ahead?
A)
0.113.
B)
0.072.
C)
0.081.


As Brice makes more distant forecasts, each forecast will be closer to the unconditional mean. So, the two period forecast would be between 0.08 and 0.09, and 0.081 is the only possible answer.
作者: MarginofSafety    时间: 2012-3-27 14:23

Which of the following statements regarding a mean reverting time series is least accurate?
A)
If the current value of the time series is above the mean reverting level, the prediction is that the time series will decrease.
B)
If the current value of the time series is above the mean reverting level, the prediction is that the time series will increase.
C)
If the time-series variable is x, then xt = b0 + b1xt-1.


If the current value of the time series is above the mean reverting level, the prediction is that the time series will decrease; if the current value of the time series is below the mean reverting level, the prediction is that the time series will increase.
作者: MarginofSafety    时间: 2012-3-27 14:23

Suppose that the time series designated as Y is mean reverting. If Yt+1 = 0.2 + 0.6 Yt, the best prediction of Yt+1 is:
A)
0.5.
B)
0.3.
C)
0.8.


The prediction is Yt+1 = b0 / (1-b1) = 0.2 / (1-0.6) = 0.5
作者: MarginofSafety    时间: 2012-3-27 14:24

The regression results from fitting an AR(1) to a monthly time series are presented below. What is the mean-reverting level for the model?

Model: ΔExpt = b0 + b1ΔExpt–1 + εt


Coefficients

Standard Error

t-Statistic

p-value


Intercept
1.3304
0.0089
112.2849
< 0.0001

Lag-1
0.1817
0.0061
30.0125
< 0.0001
A)
0.6151.
B)
1.6258.
C)
7.3220.


The mean-reverting level is b0 / (1 − b1) = 1.3304 / (1 − 0.1817) = 1.6258.
作者: MarginofSafety    时间: 2012-3-27 14:25

Which of the following statements regarding an out-of-sample forecast is least accurate?
A)
There is more error associated with out-of-sample forecasts, as compared to in-sample forecasts.
B)
Out-of-sample forecasts are of more importance than in-sample forecasts to the analyst using an estimated time-series model.
C)
Forecasting is not possible for autoregressive models with more than two lags.


Forecasts in autoregressive models are made using the chain-rule, such that the earlier forecasts are made first. Each later forecast depends on these earlier forecasts.
作者: MarginofSafety    时间: 2012-3-27 14:29

Frank Batchelder and Miriam Yenkin are analysts for Bishop Econometrics. Batchelder and Yenkin are discussing the models they use to forecast changes in China’s GDP and how they can compare the forecasting accuracy of each model. Batchelder states, “The root mean squared error (RMSE) criterion is typically used to evaluate the in-sample forecast accuracy of autoregressive models.” Yenkin replies, “If we use the RMSE criterion, the model with the largest RMSE is the one we should judge as the most accurate.”
With regard to their statements about using the RMSE criterion:
A)
Batchelder is incorrect; Yenkin is incorrect.
B)
Batchelder is correct; Yenkin is incorrect.
C)
Batchelder is incorrect; Yenkin is correct.


The root mean squared error (RMSE) criterion is used to compare the accuracy of autoregressive models in forecasting out-of-sample values (not in-sample values). Batchelder is incorrect. Out-of-sample forecast accuracy is important because the future is always out of sample, and therefore out-of-sample performance of a model is critical for evaluating real world performance.Yenkin is also incorrect. The RMSE criterion takes the square root of the average squared errors from each model. The model with the smallest RMSE is judged the most accurate.
作者: MarginofSafety    时间: 2012-3-27 14:29

William Zox, an analyst for Opal Mountain Capital Management, uses two different models to forecast changes in the inflation rate in the United Kingdom. Both models were constructed using U.K. inflation data from 1988-2002. In order to compare the forecasting accuracy of the models, Zox collected actual U.K. inflation data from 2004-2005, and compared the actual data to what each model predicted. The first model is an AR(1) model that was found to have an average squared error of 10.429 over the 12 month period. The second model is an AR(2) model that was found to have an average squared error of 11.642 over the 12 month period. Zox then computed the root mean squared error for each model to use as a basis of comparison. Based on the results of his analysis, which model should Zox conclude is the most accurate?
A)
Model 1 because it has an RMSE of 3.23.
B)
Model 2 because it has an RMSE of 3.41.
C)
Model 1 because it has an RMSE of 5.21.


The root mean squared error (RMSE) criterion is used to compare the accuracy of autoregressive models in forecasting out-of-sample values. To determine which model will more accurately forecast future values, we calculate the square root of the mean squared error. The model with the smallest RMSE is the preferred model. The RMSE for Model 1 is √10.429 = 3.23, while the RMSE for Model 2 is √11.642 = 3.41. Since Model 1 has the lowest RMSE, that is the one Zox should conclude is the most accurate.
作者: MarginofSafety    时间: 2012-3-27 14:30

Consider the estimated AR(2) model, xt = 2.5 + 3.0 xt-1 + 1.5 xt-2 + εt t=1,2,…50. Making a prediction for values of x for 1 ≤ t ≤ 50 is referred to as:
A)
an out-of-sample forecast.
B)
requires more information to answer the question.
C)
an in-sample forecast.


An in-sample (a.k.a. within-sample) forecast is made within the bounds of the data used to estimate the model. An out-of-sample forecast is for values of the independent variable that are outside of those used to estimate the model.
作者: MarginofSafety    时间: 2012-3-27 14:30

The primary concern when deciding upon a time series sample period is which of the following factors?
A)
Current underlying economic and market conditions.
B)
The length of the sample time period.
C)
The total number of observations.


There will always be a tradeoff between the increase statistical reliability of a longer time period and the increased stability of estimated regression coefficients with shorter time periods. Therefore, the underlying economic environment should be the deciding factor when selecting a time series sample period.
作者: MarginofSafety    时间: 2012-3-27 14:32

Which of the following statements regarding the instability of time-series models is most accurate? Models estimated with:
A)
a greater number of independent variables are usually more stable than those with a smaller number.
B)
shorter time series are usually more stable than those with longer time series.
C)
longer time series are usually more stable than those with shorter time series.


Those models with a shorter time series are usually more stable because there is less opportunity for variance in the estimated regression coefficients between the different time periods.
作者: JoeyDVivre    时间: 2012-3-27 14:35

The main reason why financial and time series intrinsically exhibit some form of nonstationarity is that:

A) most financial and economic relationships are dynamic and the estimated regression coefficients can vary greatly between periods.

B) serial correlation, a contributing factor to nonstationarity, is always present to a certain degree in most financial and time series.

C) most financial and time series have a natural tendency to revert toward their means.





--------------------------------------------------------------------------------

Because all financial and time series relationships are dynamic, regression coefficients can vary widely from period to period. Therefore, financial and time series will always exhibit some amount of instability or nonstationarity.
作者: JoeyDVivre    时间: 2012-3-27 14:36

Given an AR(1) process represented by xt+1 = b0 + b1×xt + et, the process would not be a random walk if:
A)
b1 = 1.
B)
E(et)=0.
C)
the long run mean is b0 + b1.


For a random walk, the long-run mean is undefined. The slope coefficient is one, b1=1, and that is what makes the long-run mean undefined: mean = b0/(1-b1).
作者: JoeyDVivre    时间: 2012-3-27 14:36

David Brice, CFA, has tried to use an AR(1) model to predict a given exchange rate. Brice has concluded the exchange rate follows a random walk without a drift. The current value of the exchange rate is 2.2. Under these conditions, which of the following would be least likely?
A)
The residuals of the forecasting model are autocorrelated.
B)
The forecast for next period is 2.2.
C)
The process is not covariance stationary.


The one-period forecast of a random walk model without drift is E(xt+1) = E(xt + et ) = xt + 0, so the forecast is simply xt = 2.2. For a random walk process, the variance changes with the value of the observation. However, the error term et = xt - xt-1 is not autocorrelated.
作者: JoeyDVivre    时间: 2012-3-27 14:36

Which of the following statements regarding time series analysis is least accurate?
A)
We cannot use an AR(1) model on a time series that consists of a random walk.
B)
If a time series is a random walk, first differencing will result in covariance stationarity.
C)
An autoregressive model with two lags is equivalent to a moving-average model with two lags.


An autoregression model regresses a dependent variable against one or more lagged values of itself whereas a moving average is an average of successive observations in a time series. A moving average model can have lagged terms but these are lagged values of the residual.
作者: JoeyDVivre    时间: 2012-3-27 14:37

A time series x that is a random walk with a drift is best described as:
A)
xt = b0 + b1xt − 1 + εt.
B)
xt = b0 + b1 xt − 1.
C)
xt = xt − 1 + εt.


The best estimate of random walk for period t is the value of the series at (t − 1). If the random walk has a drift component, this drift is added to the previous period’s value of the time series to produce the forecast.
作者: JoeyDVivre    时间: 2012-3-27 14:37

Barry Phillips, CFA, has the following time series observations from earliest to latest: (5, 6, 5, 7, 6, 6, 8, 8, 9, 11). Phillips transforms the series so that he will estimate an autoregressive process on the following data (1, -1, 2, -1, 0, 2, 0, 1, 2). The transformation Phillips employed is called:
A)
beta drift.
B)
first differencing.
C)
moving average.


Phillips obviously first differenced the data because the 1=6-5, -1=5-6, .... 1 = 9 - 9, 2 = 11 - 9.
作者: JoeyDVivre    时间: 2012-3-27 14:38

Barry Phillips, CFA, has estimated an AR(1) relationship (xt = b0 + b1 × xt-1 + et) and got the following result: xt+1 = 0.5 + 1.0xt + et. Phillips should:
A)
first difference the data because b1 = 1.
B)
not first difference the data because b0 = 0.5 < 1.
C)
not first difference the data because b1 − b0 = 1.0 − 0.5 = 0.5 < 1.


The condition b1 = 1 means that the series has a unit root and is not stationary. The correct way to transform the data in such an instance is to first difference the data.
作者: JoeyDVivre    时间: 2012-3-27 14:38

A time series that has a unit root can be transformed into a time series without a unit root through:
A)
calculating moving average of the residuals.
B)
first differencing.
C)
mean reversion.


First differencing a series that has a unit root creates a time series that does not have a unit root
作者: JoeyDVivre    时间: 2012-3-27 14:39

Suppose that the following time-series model is found to have a unit root:

Salest = b0 + b1 Sales t-1+ εt

What is the specification of the model if first differences are used?
A)
Salest = b0 + b1 Sales t-1 + b2 Sales t-2 + εt.  
B)
(Salest - Salest-1)= b0 + b1 (Sales t-1 - Sales t-2) + εt.
C)
Salest = b1 Sales t-1+ εt.  


Estimation with first differences requires calculating the change in the variable from period to period.
作者: JoeyDVivre    时间: 2012-3-27 14:39

Which of the following statements regarding unit roots in a time series is least accurate?
A)
A time series that is a random walk has a unit root.
B)
Even if a time series has a unit root, the predictions from the estimated model are valid.
C)
A time series with a unit root is not covariance stationary.


The presence of a unit root means that the least squares regression procedure that we have been using to estimate an AR(1) model cannot be used without transforming the data first.
A time series with a unit root will follow a random walk process. Since a time series that follows a random walk is not covariance stationary, modeling such a time series in an AR model can lead to incorrect statistical conclusions, and decisions made on the basis of these conclusions may be wrong. Unit roots are most likely to occur in time series that trend over time or have a seasonal element.
作者: JoeyDVivre    时间: 2012-3-27 14:40

Marvin Greene is interested in modeling the sales of the retail industry. He collected data on aggregate sales and found the following:

Salest = 0.345 + 1.0 Salest-1

The standard error of the slope coefficient is 0.15, and the number of observations is 60. Given a level of significance of 5%, which of the following can we NOT conclude about this model?
A)
The model has a unit root.
B)
The slope on lagged sales is not significantly different from one.
C)
The model is covariance stationary.


The test of whether the slope is different from one indicates failure to reject the null H0: b1=1 (t-critical with df = 58 is approximately 2.000, t-calculated = (1.0 - 1.0)/0.15 = 0.0).  This is a 2-tailed test and we cannot reject the null since 0.0 is not greater than 2.000. This model is nonstationary because the 1.0 coefficient on Salest-1 is a unit root. Any time series that has a unit root is not covariance stationary which can be corrected through the first-differencing process.
作者: JoeyDVivre    时间: 2012-3-27 14:40

An AR(1) autoregressive time series model:
A)
can be used to test for a unit root, which exists if the slope coefficient equals one.
B)
cannot be used to test for a unit root.
C)
can be used to test for a unit root, which exists if the slope coefficient is less than one.


If you estimate the following model xt = b0 + b1 × xt-1 + et and get b1 = 1, then the process has a unit root and is nonstationary.
作者: JoeyDVivre    时间: 2012-3-27 14:42

Winston Collier, CFA, has been asked by his supervisor to develop a model for predicting the warranty expense incurred by Premier Snowplow Manufacturing Company in servicing its plows. Three years ago, major design changes were made on newly manufactured plows in an effort to reduce warranty expense. Premier warrants its snowplows for 4 years or 18,000 miles, whichever comes first. Warranty expense is higher in winter months, but some of Premier’s customers defer maintenance issues that are not essential to keeping the machines functioning to spring or summer seasons. The data that Collier will analyze is in the following table (in $ millions):

Quarter

Warranty
Expense

Change in
Warranty
Expense
yt

Lagged Change in
Warranty Expense
yt-1

Seasonal Lagged
Change in
Warranty
Expense
yt-4


2002.1

103





2002.2

52

-51




2002.3

32

-20

-51



2002.4

68

+36

-20



2003.1

91

+23

+36



2003.2

44

-47

+23

-51


2003.3

30

-14

-47

-20


2003.4

60

+30

-14

+36


2004.1

77

+17

+30

+23


2004.2

38

-39

+17

-47


2004.3

29

-9

-39

-14


2004.4

53

+24

-9

+30


Winston submits the following results to his supervisor. The first is the estimation of a trend model for the period 2002:1 to 2004:4. The model is below. The standard errors are in parentheses.
(Warranty expense)t = 74.1 - 2.7* t + et
R-squared = 16.2%
(14.37) (1.97)

Winston also submits the following results for an autoregressive model on the differences in the expense over the period 2004:2 to 2004:4. The model is below where “y” represents the change in expense as defined in the table above. The standard errors are in parentheses.

yt = -0.7 - 0.07* yt-1 + 0.83* yt-4 + et
R-squared = 99.98%
(0.643) (0.0222) (0.0186)

After receiving the output, Collier’s supervisor asks him to compute moving averages of the sales data. Collier’s supervisors would probably not want to use the results from the trend model for all of the following reasons EXCEPT:
A)
it does not give insights into the underlying dynamics of the movement of the dependent variable.
B)
the model is a linear trend model and log-linear models are always superior.
C)
the slope coefficient is not significant.


Linear trend models are not always inferior to log-linear models. To determine which specification is better would require more analysis such as a graph of the data over time. As for the other possible answers, Collier can see that the slope coefficient is not significant because the t-statistic is 1.37=2.7/1.97. Also, regressing a variable on a simple time trend only describes the movement over time, and does not address the underlying dynamics of the dependent variable. (Study Session 3, LOS 13.a)
The mean reverting level for the first equation is closest to:
A)
-0.8.
B)
43.6.
C)
20.0.


The mean reverting level is X1 = bo/(1-b1)
X1 = 74.1/[1-(-2.7)] = 20.03
(Study Session 3, LOS 13.f)

Based upon the output provided by Collier to his supervisor and without any further calculations, in a comparison of the two equations’ explanatory power of warranty expense it can be concluded that:
A)
the autoregressive model on the first differenced data has more explanatory power for warranty expense.
B)
the provided results are not sufficient to reach a conclusion.
C)
the two equations are equally useful in explaining warranty expense.


Although the R-squared values would suggest that the autoregressive model has more explanatory power, there are a few problems. First, the models have different sample periods and different numbers of explanatory variables. Second, the actual input data is different. To assess the explanatory power of warranty expense, as opposed to the first differenced values, we must transform the fitted values of the first-differenced data back to the original level data to assess the explanatory power for the warranty expense. (Study Session 3, LOS 12.f)
Based on the autoregressive model, expected warranty expense in the first quarter of 2005 will be closest to:
A)
$65 million.
B)
$78 million.
C)
$60 million.

Substituting the 1-period lagged data from 2004.4 and the 4-period lagged data from 2004.1 into the model formula, change in warranty expense is predicted to be higher than 2004.4.
11.73 =-0.7 - 0.07*24+ 0.83*17.
The expected warranty expense is (53 + 11.73) = $64.73 million. (Study Session 3, LOS 13.d)

Based upon the results, is there a seasonality component in the data?
A)
No, because the slope coefficients in the autoregressive model have opposite signs.
B)
Yes, because the coefficient on yt-4 is large compared to its standard error.
C)
Yes, because the coefficient on yt is small compared to its standard error.


The coefficient on the 4th lag tests the seasonality component. The t-ratio is 44.6. Even using Chebychev’s inequality, this would be significant. Neither of the other answers are correct or relate to the seasonality of the data. (Study Session 3, LOS 13.l)
Collier most likely chose to use first-differenced data in the autoregressive model:
A)
to increase the explanatory power.
B)
because the time trend was significant.
C)
in order to avoid problems associated with unit roots.


Time series with unit roots are very common in economic and financial models, and unit roots cause problems in assessing the model. Fortunately, a time series with a unit root may be transformed to achieve covariance stationarity using the first-differencing process. Although the explanatory power of the model was high (but note the small sample size), a model using first-differenced data often has less explanatory power. The time trend was not significant, so that was not a possible answer. (Study Session 3, LOS 13.k)
作者: JoeyDVivre    时间: 2012-3-27 14:42

Barry Phillips, CFA, is analyzing quarterly data. He has estimated an AR(1) relationship (xt = b0 + b1 × xt-1 + et) and wants to test for seasonality. To do this he would want to see if which of the following statistics is significantly different from zero?
A)
Correlation(et, et-1).
B)
Correlation(et, et-5).
C)
Correlation(et, et-4).


Although seasonality can make the other correlations significant, the focus should be on correlation(et, et-4) because the 4th lag is the value that corresponds to the same season as the predicted variable in the analysis of quarterly data.
作者: JoeyDVivre    时间: 2012-3-27 14:43

The table below shows the autocorrelations of the lagged residuals for quarterly theater ticket sales that were estimated using the AR(1) model: ln(salest) = b0 + b1(ln salest − 1) + et. Assuming the critical t-statistic at 5% significance is 2.0, which of the following is the most likely conclusion about the appropriateness of the model? The time series:

Lagged Autocorrelations of the Log of Quarterly Theater Ticket Sales

Lag

Autocorrelation

Standard Error

t-Statistic

1


−0.0738
0.1667
−0.44271

2


−0.1047
0.1667
−0.62807

3


−0.0252
0.1667
−0.15117

4


0.5528
0.1667
3.31614
A)
contains ARCH (1) errors.
B)
contains seasonality.
C)
would be more appropriately described with an MA(4) model.


The time series contains seasonality as indicated by the strong and significant autocorrelation of the lag-4 residual.
作者: JoeyDVivre    时间: 2012-3-27 14:44

Which of the following statements regarding seasonality is least accurate?
A)
Not correcting for seasonality when, in fact, seasonality exists in the time series results in a violation of an assumption of linear regression.
B)
The presence of seasonality makes it impossible to forecast using a time-series model.
C)
A time series that is first differenced can be adjusted for seasonality by incorporating the first-differenced value for the previous year's corresponding period.


Forecasting is no different in the case of seasonal component in the time-series model than any other forecasting.
作者: JoeyDVivre    时间: 2012-3-27 14:44

Which of the following is a seasonally adjusted model?
A)
Salest = b0 + b1 Sales t-1 + b2 Sales t-2 + εt.
B)
Salest = b1 Sales t-1+ εt.
C)
(Salest - Sales t-1)= b0 + b1 (Sales t-1 - Sales t-2) + b2 (Sales t-4 - Sales t-5) + εt.


This model is a seasonal AR with first differencing.
作者: JoeyDVivre    时间: 2012-3-27 14:45

The table below shows the autocorrelations of the lagged residuals for the first differences of the natural logarithm of quarterly motorcycle sales that were fit to the AR(1) model: (ln salest − ln salest − 1) = b0 + b1(ln salest − 1 − ln salest − 2) + εt. The critical t-statistic at 5% significance is 2.0, which means that there is significant autocorrelation for the lag-4 residual, indicating the presence of seasonality. Assuming the time series is covariance stationary, which of the following models is most likely to CORRECT for this apparent seasonality?

Lagged Autocorrelations of First Differences in the Log of Motorcycle Sales

Lag

Autocorrelation

Standard Error

t-Statistic

1


−0.0738
0.1667
−0.44271

2


−0.1047
0.1667
−0.62807

3


−0.0252
0.1667
−0.15117

4


0.5528
0.1667
3.31614
A)
ln salest = b0 + b1(ln salest − 1) − b2(ln salest − 4) + εt.
B)
(ln salest − ln salest − 4) = b0 + b1(ln salest − 1 − ln salest − 2) + εt.
C)
(ln salest − ln salest − 1) = b0 + b1(ln salest − 1 − ln salest − 2) + b2(ln salest − 4 − ln salest − 5) + εt.


Seasonality is taken into account in an autoregressive model by adding a seasonal lag variable that corresponds to the seasonality. In the case of a first-differenced quarterly time series, the seasonal lag variable is the first difference for the fourth time period. Recognizing that the model is fit to the first differences of the natural logarithm of the time series, the seasonal adjustment variable is (ln salest − 4 − ln salest − 5).
作者: JoeyDVivre    时间: 2012-3-27 14:45

Which of the following is least likely a consequence of a model containing ARCH(1) errors? The:
A)
model's specification can be corrected by adding an additional lag variable.
B)
variance of the errors can be predicted.
C)
regression parameters will be incorrect.


The presence of autoregressive conditional heteroskedasticity (ARCH) indicates that the variance of the error terms is not constant. This is a violation of the regression assumptions upon which time series models are based. The addition of another lag variable to a model is not a means for correcting for ARCH (1) errors.
作者: JoeyDVivre    时间: 2012-3-27 14:46

Suppose you estimate the following model of residuals from an autoregressive model:
εt2 = 0.25 + 0.6ε2t-1 + µt, where ε = ε^

If the residual at time t is 0.9, the forecasted variance for time t+1 is:
A)
0.790.
B)
0.736.
C)
0.850.


The variance at t = t + 1 is 0.25 + [0.60 (0.9)2] = 0.25 + 0.486 = 0.736. See also, ARCH models.
作者: chunty    时间: 2012-3-27 14:49

The data below yields the following AR(1) specification: xt = 0.9 – 0.55xt-1 + Et , and the indicated fitted values and residuals.
Timextfitted valuesresiduals
11 --
2-10.35-1.35
321.450.55
4-1-0.2-0.8
501.45-1.45
620.91.1
70-0.20.2
810.90.1
920.351.65

The following sets of data are ordered from earliest to latest. To test for ARCH, the researcher should regress:
A)
(1.8225, 0.3025, 0.64, 2.1025, 1.21, 0.04, 0.01) on (0.3025, 0.64, 2.1025, 1.21, 0.04, 0.01, 2.7225).
B)
(-1.35, 0.55, -0.8, -1.45, 1.1, 0.2, 0.1, 1.65) on (0.35, 1.45, -0.2, 1.45, 0.9, -0.2, 0.9, 0.35)
C)
(1, 4, 1, 0, 4, 0, 1, 4) on (1, 1, 4, 1, 0, 4, 0, 1)


The test for ARCH is based on a regression of the squared residuals on their lagged values. The squared residuals are (1.8225, 0.3025, 0.64, 2.1025, 1.21, 0.04, 0.01, 2.7225). So, (1.8225, 0.3025, 0.64, 2.1025, 1.21, 0.04, 0.01) is regressed on (0.3025, 0.64, 2.1025, 1.21, 0.04, 0.01, 2.7225). If coefficient a1 in: is statistically different from zero, the time series exhibits ARCH(1).
作者: chunty    时间: 2012-3-27 14:49

One choice a researcher can use to test for nonstationarity is to use a:
A)
Breusch-Pagan test, which uses a modified t-statistic.
B)
Dickey-Fuller test, which uses a modified χ2 statistic.
C)
Dickey-Fuller test, which uses a modified t-statistic.


The Dickey-Fuller test estimates the equation (xt – xt-1) = b0 + (b1 - 1) * xt-1 + et and tests if H0: (b1 – 1) = 0. Using a modified t-test, if it is found that (b1–1) is not significantly different from zero, then it is concluded that b1 must be equal to 1.0 and the series has a unit root.
作者: chunty    时间: 2012-3-27 14:50

Consider the following estimated model:
(Salest - Sales t-1) = 30 + 1.25 (Sales t-1 - Sales t-2) + 1.1 (Sales t-4 - Sales t-5) t=1,2,.. T

and Sales for the periods 1999.1 through 2000.2:
tPeriodSales
T2000.2$2,000
T-12000.1$1,800
T-21999.4$1,500
T-31999.3$1,400
T-41999.2$1,900
T-51999.1$1,700

The forecasted Sales amount for 2000.3 is closest to:
A)
$2,625.
B)
$2,270.
C)
$1,730.


Note that since we are forecasting 2000.3, the numbering of the "t" column has changed.
Change in sales = $30 + 1.25 ($2,000-1,800) + 1.1 ($1,400-1,900)
Change in sales = $30 + 250 - 550 = -$270
Sales = $2,000 – 270 = $1,730
作者: chunty    时间: 2012-3-27 14:50

Consider the following estimated model:
(Salest - Sales t-1)= 100 - 1.5 (Sales t-1 - Sales t-2) + 1.2 (Sales t-4 - Sales t-5) t=1,2,.. T
and Sales for the periods 1999.1 through 2000.2:

tPeriodSales
T2000.2$1,000
T-12000.1$900
T-21999.4$1,200
T-31999.3$1,400
T-41999.2$1,000
T-51999.1$800

The forecasted Sales amount for 2000.3 is closest to:
A)
$1,730.
B)
$1,430.
C)
$730.


Change in sales = $100 - 1.5 ($1,000-900) + 1.2 ($1,400-1,000)
Change in sales = $100 - 150 + 480 =$430
Sales = $1,000 + 430 = $1,430
作者: chunty    时间: 2012-3-27 14:51

Bill Johnson, CFA, has prepared data concerning revenues from sales of winter clothing made by Polar Corporation for presentation in the following table (in $ millions):



Change In Sales
Lagged Change
In Sales
Seasonal Lagged
Change In Sales

Quarter
Sales
Y
Y + (−1)
Y + (−4)

2006.1
182



2006.2
74
−108


2006.3
78
4
−108

2006.4
242
164
4

2007.1
194
−48
164

2007.2
79
−115
−48
−108

2007.3
90
11
−115
4

2007.4
260
170
11
w


The preceding table will be used by Johnson to forecast values using:
A)
a log-linear trend model with a seasonal lag.
B)
a serially correlated model with a seasonal lag.
C)
an autoregressive model with a seasonal lag.


Johnson will use the table to forecast values using an autoregressive model for periods in succession since each following forecast relies on the forecast for the preceding period. The seasonal lag is introduced to account for seasonal variations in the observed data.
The value that Johnson should enter in the table in place of "w" is:
A)
−48.
B)
164.
C)
−115.


The seasonal lagged change in sales shows the change in sales from the period 4 quarters before the current period. Sales in the year 2006 quarter 4 increased $164 million over the prior period.
Assume that Johnson prepares a change in sales regression analysis model with seasonality which includes the following:

Coefficients

Intercept

−6.032

Lag 1

0.017

Lag 4

0.983

Based on the model, expected sales in the first quarter of 2008 will be closest to:

A)
155.
B)
190.
C)
210.


Substituting the 1-period lagged data from 2007.4 and the 4-period lagged data from 2007.1 into the model formula, change in sales is predicted to be −6.032 + (0.017 × 170) + (0.983 × −48) = −50.326. Expected sales are 260 + (−50.326) = 209.674.
作者: chunty    时间: 2012-3-27 14:52

Alexis Popov, CFA, is analyzing monthly data. Popov has estimated the model xt = b0 + b1 × xt-1 + b2 × xt-2 + et. The researcher finds that the residuals have a significant ARCH process. The best solution to this is to:
A)
re-estimate the model with generalized least squares.
B)
re-estimate the model using only an AR(1) specification.
C)
re-estimate the model using a seasonal lag.


If the residuals have an ARCH process, then the correct remedy is generalized least squares which will allow Popov to better interpret the results.
作者: chunty    时间: 2012-3-27 14:52

Alexis Popov, CFA, has estimated the following specification: xt = b0 + b1 × xt-1 + et. Which of the following would most likely lead Popov to want to change the model’s specification?
A)
Correlation(et, et-2) is significantly different from zero.
B)
Correlation(et, et-1) is not significantly different from zero.
C)
b0 < 0.


If correlation(et, et-2) is not zero, then the model suffers from 2nd order serial correlation. Popov may wish to try an AR(2) model. Both of the other conditions are acceptable in an AR(1) mod
作者: chunty    时间: 2012-3-27 14:53

Alexis Popov, CFA, wants to estimate how sales have grown from one quarter to the next on average. The most direct way for Popov to estimate this would be:
A)
an AR(1) model.
B)
a linear trend model.
C)
an AR(1) model with a seasonal lag.


If the goal is to simply estimate the dollar change from one period to the next, the most direct way is to estimate xt = b0 + b1 × (Trend) + et, where Trend is simply 1, 2, 3, ....T. The model predicts a change by the value b1 from one period to the next.



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