Quantitative Analysis【Reading 13】Sample 2012, following

kasinkei

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.

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Wellington’s out-of-sample forecast of LN(xt) is 1.9 = 1.4 + 0.02 × 25, and e1.9 = 6.69.

chunty

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.

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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.

chunty

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.

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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

chunty

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.

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If the residuals have an ARCH process, then the correct remedy is generalized least squares which will allow Popov to better interpret the results.

chunty

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.

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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.

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The value that Johnson should enter in the table in place of "w" is:

A) −48.

B) 164.

C) −115.

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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.

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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.

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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.

chunty

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:

t	Period	Sales
T	2000.2	$1,000
T-1	2000.1	$900
T-2	1999.4	$1,200
T-3	1999.3	$1,400
T-4	1999.2	$1,000
T-5	1999.1	$800

The forecasted Sales amount for 2000.3 is closest to:

A) $1,730.

B) $1,430.

C) $730.

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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

chunty

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:

t	Period	Sales
T	2000.2	$2,000
T-1	2000.1	$1,800
T-2	1999.4	$1,500
T-3	1999.3	$1,400
T-4	1999.2	$1,900
T-5	1999.1	$1,700

The forecasted Sales amount for 2000.3 is closest to:

A) $2,625.

B) $2,270.

C) $1,730.

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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

chunty

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.

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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.

chunty

The data below yields the following AR(1) specification: xt = 0.9 – 0.55xt-1 + Et , and the indicated fitted values and residuals.

Time	xt	fitted values	residuals
1	1	-	-
2	-1	0.35	-1.35
3	2	1.45	0.55
4	-1	-0.2	-0.8
5	0	1.45	-1.45
6	2	0.9	1.1
7	0	-0.2	0.2
8	1	0.9	0.1
9	2	0.35	1.65

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)

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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).

JoeyDVivre

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.

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The variance at t = t + 1 is 0.25 + [0.60 (0.9)2] = 0.25 + 0.486 = 0.736. See also, ARCH models.