CFA L2 Apostila 01 Exame 2018 - COMPLETA IMPRESSÃO

Fixed Income Investments Quantitative MethodsQuantitative Methods 11. Time-Series AnalysisLOS 11.a Calculate/Evaluate Time SeriesSchweser B1 pg 194, CFAI V1 pg 415 Time Series Analysis  A time series is a set of values for a particular 66 variable for many time periods.  Examples:  Quarterly sales over the last five years  Monthly CPI over the last 12 years  Daily returns over the last 180 days© Kaplan, Inc. ©2018 FK Pa

SS 03 - Quantitative Methods for Valuation Time Series Topical Overview: Time Series Analysis Financial data can be difficult to model using conventional multiple regression, so what should you do? Trend models  Linear trend model  Log-linear trend model  Lagged models 65  Autoregressive model (AR)© Kaplan, Inc.LOS 11.a Calculate/Evaluate Time SeriesSchweser B1 pg 194, CFAI V1 pg 415 Linear Trend Model Point: Regression with time as the independent variable: y t  b0  b1t   t t  1 to n Observed value Time index: One for(e.g., unemployment) the first observation, two for the second…  The predicted change in y is b1. 67© Kaplan, Inc.artners - Exame CFA 17

LOS 11.a Calculate/Evaluate Time SeriesSchweser B1 pg 194, CFAI V1 pg 415 Linear Trend Model: ExampleThe following inflation trend model was estimatedusing monthly CPI from January 1995 (month 1) toDecember 2008 (month 168): yˆ t = 4 .3 5 4 3 – 0 .0 1 3 2 t + ε t Example: The forecasted value of y for January 68 2009 (month 169) is: y169 = 4.3543 – 0.0132 x 169 = 2.1235© Kaplan, Inc.LOS 11.b Describe/Evaluate Time SeriesSchweser B1 pg 200, CFAI V1 pg 418Log-Linear Trend Models (Visual) LinLienaear rTTrreennddMMoodedlel LLoogg--LLiinneeaar rTrTernednMdoMdeol delyt ytb=0b0++bb11tt ++ t ln(y(Int ))yt =bb00 ++ bb11tt++ ty ln(yt) yt = b0 + b1t ln(yt) = b0 + b1t time time 70© Kaplan, Inc. ©2018 FK Pa

SS 03 - Quantitative Methods for ValuationLOS 11.a Calculate/Evaluate Time SeriesSchweser B1 pg 194, CFAI V1 pg 415 Log-Linear Trend Model Log-linear regression assumes the dependent financial variable exhibits exponential growth: yt = ebo b1t  Taking the natural log of both sides ln yt   b0 + b1t + t yields Slope coefficient, b1, is the constant growth!© Kaplan, Inc. 69LOS 11.b Describe/Evaluate Time SeriesSchweser B1 pg 200, CFAI V1 pg 418Log-Linear Trend Model: Example You have run a log-linear trend analysis of financial data with the following results: ln  yˆ t   5.01 + 0.095t Using the model, your estimates of y1 and y2 are:    ln yˆ1  5.01 + 0.095 1  5.105; yˆ1  e5.105  164.84 71    ln yˆ 2  5.01 + 0.095 2  5.200; yˆ 2  e5.200  181.27© Kaplan, Inc.artners - Exame CFA 18

LOS 11.b Describe/Evaluate Time SeriesSchweser B1 pg 200, CFAI V1 pg 418Serial Correlation in Trend Models Regression errors in current period must be uncorrelated with errors from all other periods. Use Durbin Watson to test for serial correlation. If linear model exhibits autocorrelation, try log- linear (next slide). If log-linear model still exhibits autocorrelation, try autoregression.© Kaplan, Inc. 72LOS 11.d Describe/Calculate Time SeriesSchweser B1 pg 202, CFAI V1 pg 425Forecasting With an AR(1) Model  Assume we estimated the following AR(1): 74 Xt = 0.0748 + 0.8785 xt–1 + et  Question: If X0 is 0.5150, what is the forecast value at time 1 and time 2?  One-step-ahead forecast: X1 = 0.0748 + 0.8785 (0.5150) = 0.5272  Two-step-ahead forecast: X2 = 0.0748 + 0.8785 (0.5272) = 0.5380© Kaplan, Inc. ©2018 FK Pa

SS 03 - Quantitative Methods for ValuationLOS 11.d Describe/Calculate Time SeriesSchweser B1 pg 202, CFAI V1 pg 425 Autoregressive Models Main idea: The dependent variable is regressed on prior value(s) of itself AR(p) model: x t = b0 + b1x t –1 + b2x t – 2 + ... + bp x t –p +  tNote: No longer a distinction between ‘X’ and ‘Y’ variables© Kaplan, Inc. 73LOS 11.c Describe/Explain Time SeriesSchweser B1 pg 201, CFAI V1 pg 425 Covariance Stationarity To use autoregressive models, the time series must be covariance stationary:  Constant and finite expected value  Constant and finite variance  Constant and finite covariance with leading or lagged values A nonstationary time series will produce meaningless regression results.© Kaplan, Inc. 75artners - Exame CFA 19

LOS 11.e Explain Time SeriesSchweser B1 pg 203, CFAI V1 pg 427Serial Correlation in an AR Model x t = b0 + b1x t –1   t 76  Cannot use Durbin Watson to test for serial correlation in AR models.  Use a t-test on residual autocorrelations.  If serial correlation exists, model is incomplete.  Solutions: Increase order of model by adding more lagged variables (e.g., AR(2) from AR(1) and/or adjust for seasonality).© Kaplan, Inc.LOS 11.e Explain Time SeriesSchweser B1 pg 203, CFAI V1 pg 427 AR Model: ExampleLag Autocorrelation SE t-stat 1 0.1884 0.1414 1.33242 0.3363 0.1414 2.37843 –0.1218 0.1414 –0.86144 –0.0285 0.1414 –0.2016 Assuming a critical t-value of 2.0, the model is incorrectly specified. To correct, try an AR(2) model (i.e., add a lag).© Kaplan, Inc. xt = b0 + b1xt –1  b2xt –2   t 78 ©2018 FK Pa

SS 03 - Quantitative Methods for ValuationLOS 11.e Explain Time SeriesSchweser B1 pg 203, CFAI V1 pg 427 Testing for Serial Correlation in an AR Model SE= 1 nComputed autocorrelations (AR1 model):Lag Autocorrelation SE t-stat 1 0.1884 0.1414 1.3324 2.3784 2 0.3363 0.1414 –0.8614 –0.2016 3 –0.1218 0.1414 Continued → 4 –0.0285 0.1414 77 Since 1 n  0.1414  n  50© Kaplan, Inc.LOS 11.e Explain Time SeriesSchweser B1 pg 203, CFAI V1 pg 427 AR Model: Solution The AR(2) model better fits Since all t-stats are the data. insignificant, there is not serial correlation. No significant correlations Lag Autocorrelation SE t-stat 1 –0.0174 0.1429 –0.1217 2 0.0957 0.1429 3 –0.1472 0.1429 0.6701 4 –0.1520 0.1429 –1.0301 –1.0641© Kaplan, Inc. 79artners - Exame CFA 20

LOS 11.l Explain/Calculate/Interpret Time SeriesSchweser B1 pg 212, CFAI V1 pg 452 Seasonality What if changing the order of the model failed to correct the serial correlation? Check for seasonality A time series shows consistent seasonal patterns  For instance, the monthly sales for a retailer In the following example, an AR(1) model is applied to monthly sales data for a retailer. Objective: Bring the seasonal component into themodel to improve forecasting accuracy.© Kaplan, Inc. 80LOS 11.l Explain/Calculate/Interpret Time SeriesSchweser B1 pg 212, CFAI V1 pg 452 Does This Suggest Seasonality?Lag Autocorrelation t-stat Lag Autocorrelations t-stat s 1 –2.3907 7 0.2137 1.7883 2 –0.2857 –2.4460 8 –0.1471 –1.2310 3 –0.2923 9 4 0.2619 10 0.0297 0.2435 5 0.0313 –1.2192 11 –0.2904 –2.4301 6 –0.1457 12 –0.03061 –2.5615 1.8736 0.2239 –0.2117 0.9948 8.3247 –0.0253 SE = 0.1195© Kaplan, Inc. 82 ©2018 FK Pa

SS 03 - Quantitative Methods for ValuationLOS 11.l Explain/Calculate/Interpret Time SeriesSchweser B1 pg 212, CFAI V1 pg 452 Seasonality: ExampleANOVA F 45.3393 An AR(1) moddefl provides thSisS output: MSRegression 1 25.3814 25.3814Residual 68 38.0670 0.5598Total 69 63.4484 Intercept Coefficients Standard Error t-stat P-value lag 1 –0.0029 0.0894 –0.0324 –0.6334 0.0940 –6.7334 0.97© Kaplan, Inc. <0.01 81LOS 11.l Explain/Calculate/Interpret Time SeriesSchweser B1 pg 212, CFAI V1 pg 452 Correcting Seasonality The effects of seasonality in the data can be removed by incorporating lags in the AR model. After looking at the autocorrelations, we first try including lag terms for months 1 and 12: xt = b0 + b1xt –1 + b2xt –12 + et Note that the choice of including lags 1 and 12 involves some “art”. We could have tried other lags as well.© Kaplan, Inc. 83artners - Exame CFA 21

LOS 11.l Explain/Calculate/Interpret Time SeriesSchweser B1 pg 212, CFAI V1 pg 452After Including the Lagged VariablesANOVA df SS MS F 2 53.77100003 26.8855 3126.58228Regression 56 0.00859901Residual 58 0.48154434Total 54.25254437Intercept Coefficients Standard Error t-stat P-valuelag 1 0.000487577 0.012073415 0.04038434 0.97lag 12 0.016319792 –7.7756100 –0.1268963 0.016736381 51.93251157 <0.01 © Kaplan, Inc. 0.8691623 <0.01 84LOS 11.l Explain/Calculate/Interpret Time SeriesSchweser B1 pg 212, CFAI V1 pg 452AR(1) Model With a Seasonal Lagxt = b0 + b1xt–1 + b2xt–12 + t 2008 Jan 0.9239 2008 Feb –0.2363 Coefficients 2008 Mar 2008 Apr 0.1097Intercept 0.000487577 2008 May –0.0174 2008 Jun –0.0570lag 1 –0.1268963 2008 Jul –0.1167 2008 Aug –1.4074lag 12 0.8691623 2008 Sep 2008 Oct 2.6702Predicted value for Jan 2009: 2008 Nov –1.2110 2008 Dec0.0004876 – 0.1268963 (–0.5420) 0.0070+ 0.8691623 (0.9239)= 0.8723 –0.0290 –0.548620© Kaplan, Inc. ©2018 FK Pa

SS 03 - Quantitative Methods for ValuationLOS 11.l Explain/Calculate/Interpret Time SeriesSchweser B1 pg 212, CFAI V1 pg 452 After Including the Lagged VariablesLag Autocorrelations t-stat Lag Autocorrelations t-stat 1 –0.1491 –1.1452 7 –0.1284 –0.9862 2 –0.0654 –0.5023 8 0.1409 3 0.2248 9 –0.1928 1.0822 4 –0.1354 1.7266 –0.0730 –1.4808 5 0.0591 –1.0399 10 0.1643 –0.5607 6 0.0550 11 –0.1306 0.4539 12 1.2619 0.4224 –1.0031 SE = 0.1302 None significant 85© Kaplan, Inc.LOS 11.f Explain/Calculate/Interpret Time SeriesSchweser B1 pg 204, CFAI V1 pg 430 Mean Reversion If a time series is mean-reverting:  The value of the dependent variable tends to fall when above its mean and rise when below its mean. For an AR(1) model: b0 MRL = 1 – b1 Note that we would have problems if b1 = 1 (see later).© Kaplan, Inc. 87artners - Exame CFA 22

LOS 11.f Explain/Calculate/Interpret Time SeriesSchweser B1 pg 204, CFAI V1 pg 430Mean Reverting Level: Example Calculate the mean-reverting level (MRL) for our previous example: xt = 0.0748 +0.8785xt–1 + t MRL = 0.0748 = 0.6156 1– 0.8785 What does this mean? If X0 is below 0.6156, the forecast value for X1 will increase towards 0.6156. The forecast values will always revert to 0.6156!© Kaplan, Inc. 88LOS 11.i Describe/Contrast Time SeriesSchweser B1 pg 207, CFAI V1 pg 439 Random Walks Two types: No drift and drift  Defining characteristic: b1 = 1 note: b0 = 0 Random walk without a drift: note: b1 = 1 Xt = Xt – 1 + εt Or, equivalently: Xt = 0 + (1)Xt – 1 + εt Random walk with a drift: note: b0 ≠ 0 Xt = b0 + Xt – 1 + εt© Kaplan, Inc. 90 ©2018 FK Pa

SS 03 - Quantitative Methods for ValuationLOS 11.j/l Describe/Explain/Demonstrate Time SeriesSchweser B1 pg 208, CFAI V1 pg 443 Unit Roots In an AR(1) model, coefficient must be < 1.0. If coefficient = 1 unit root. Unit root is defining feature of random walks (next slide). Unit roots are common in series that consistently increase or decrease over time (e.g., 1990s stock market).© Kaplan, Inc. 89LOS 11.i Describe/Contrast Time SeriesSchweser B1 pg 207, CFAI V1 pg 439Why are Unit Roots Problematic? If b1 = 1, then mean reverting level is undefined: b0  b0  b0 1 b1 1 1 0 Without a mean reverting level, the time series isnonstationary. Question: How do we test for a unit root?© Kaplan, Inc. 91artners - Exame CFA 23

LOS 11.i Describe/Contrast Time SeriesSchweser B1 pg 207, CFAI V1 pg 439Dickey-Fuller Test for a Unit Root Test is based on a transformed version of the AR(1) model. Subtracting xt–1 from both sides:xt – xt–1 = b0 + (b1 – 1)xt–1 + εt orxt – xt–1 = b0 + g1xt–1 + εt where g1 = (b1 – 1) If there is a unit root in the AR(1) model, then g1 will be 0.© Kaplan, Inc. 92LOS 11.i Describe/Contrast Time SeriesSchweser B1 pg 207, CFAI V1 pg 439 First Differencing Use First Differencing to remove unit roots/nonstationarity Create a new dependent variable, y, defined as the change in x: yt = xt – xt –1 and  yt = b0 + b1 yt–1 + t© Kaplan, Inc. Note: b0 and b1 = 0 94 ©2018 FK Pa

SS 03 - Quantitative Methods for ValuationLOS 11.i Describe/Contrast Time SeriesSchweser B1 pg 207, CFAI V1 pg 439 Dickey Fuller Test (continued) DF test: H0: g1 = 0, time series has a unit root and is nonstationary Ha: g1 < 0, time series does not have a unit root Test is conducted by calculating a t-statistic for g1 and using a revised set of critical values computed by DF. Using a revised set of t-stats, if the Null is rejected, there is no evidence of a unit root.© Kaplan, Inc. 93LOS 11.i Describe/Contrast Time SeriesSchweser B1 pg 207, CFAI V1 pg 439 First DifferencingNote how first differencing removes the upward trend: linear trend yt=xt-xt-1 40 15 35 10 30 25 5 20 0 15 -5 10 -10 -15 5 0 10 20 30 40 -5-10 0© Kaplan, Inc. 95artners - Exame CFA 24

LOS 11.g Contrast/Compare Time SeriesSchweser B1 pg 206, CFAI V1 pg 434 Comparing Model Accuracy In-sample: Data used to develop model Out-of-sample: Any data outside above range Forecasting accuracy is measured by square root of the mean squared error (RMSE). Use the model with the lowest RMSE based on out-of-sample forecasting errors.© Kaplan, Inc. 96LOS 11.o Determine/Justify Time SeriesSchweser B1 pg 219, CFAI V1 pg 466 Time Series Analysis Is series covariance Yes Estimate model Serial correlation stationary? (start with AR(1)) Examine autocorrelations Yes (look at graph) Keep adding lags to Yes (not DW) No model until resids. no Yes No Take first diffs. longer significant Seasonality present? (and Ln if exponential) (examine autocorrelations) Add lags for© Kaplan, Inc. seasonality No Correct using generalized Heteroskedasticity? least squares, or (use ARCH model) similar method No Test model on out-of-sample data 98 ©2018 FK Pa

SS 03 - Quantitative Methods for ValuationLOS 11.h Explain Time SeriesSchweser B1 pg 207, CFAI V1 pg 436Regression Coefficient Instability Point: Estimated regression coefficients change from one time period to another.  Creates a trade-off between statistical reliability of long time series and stability of short time series  Has economic process or environment changed?© Kaplan, Inc. 97 Fixed Income Investments Study Session 3 Discussion Questions CFA Institute Program curriculum, Level II, Volume 1, page 387 Questions 17–22artners - Exame CFA 25











SS 03 - Quantitative Methods for Valuation Discussion Question Solution 17IR = 0.0477 + (0.0150 × 6) + (0.435 × 0.04) + (– 0.0009 × 40) + (0.05 × 0.70)= 0.1541© Kaplan, Inc. 109 Discussion Question QQuueesstitoionn118818. The 95 percent confidence interval for the regression coefficient for the pre-offer price adjustment is closest to: A. 0.156 to 0.714. B. 0.395 to 0.475 C. 0.402 to 0.468 © Kaplan, Inc. 111artners - Exame CFA 28

SSoolulutitoionn1188 Discussion Question Confidence interval 112 = bj ± (std error × critical stat) = 0.435 ± (0.0202 × 1.96) = (0.395 to 0.475)© Kaplan, Inc. Discussion Question QQuueesstitoionn1199 19. The most appropriate null hypothesis and the most appropriate conclusion regarding A. Hansen’s belief about the magnitude of the B. initial return relative to that of the pre-offer price C. adjustment (reflected by the coefficient bj) are:© Kaplan, Inc. Null Hypothesis Conclusion about bj (0.05 Level of Significance) Ho: bj = 0.5 Reject Ho Ho: bj ≥ 0.5 Fail to Reject Ho Ho: bj ≥ 0.5 Reject Ho 114 ©2018 FK Pa

SS 03 - Quantitative Methods for Valuation18. Discussion Question QSuoelsuttiioonn 1188 The 95 percent confidence interval for the regression coefficient for the pre-offer price adjustment is closest to: A. 0.156 to 0.714. B. 0.395 to 0.475 C. 0.402 to 0.468© Kaplan, Inc. 113 Discussion Question SSoolulutitoionn1199Null Hypothesis“He also believes that for each 1 percent increasein pre-offer price adjustment, the initial return willincrease by less than 0.5 percent, holding othervariables constant.”If the return increases by less than 0.5% as a resultof a 1% increase in the dependent variable, thecoefficient must be strictly less than 0.5. The null hypothesis must contain equality, so: 115 Ho: bj ≥ 0.5 (and Ha: bj < 0.5)© Kaplan, Inc.artners - Exame CFA 29

Discussion Question SSoolulutitoionn1199DecisionTest statistic: t = (bj – H0) / (std error) = (0.435 − 0.50) / 0.0202 = −3.22Critical t-stat: one-tailed, 0.05 sig: Area in t-stat The test statistic is significant, and right tail 0.050 1.645 the null hypothesis can be rejected 1.960 0.025 2.326 at the 0.05 level of significance as 0.010 2.576 the test stat –3.22 is less than the 0.005 critical stat of –1.645 116© Kaplan, Inc. Discussion Question QQuueesstitoionn220020. The most appropriate interpretation of the multiple R-squared for Hansen’s model isA. that:B.C. unexplained variation in the dependent variable is 36 percent of total variation. correlation between predicted and actual values of the dependent variable is 0.36. correlation between predicted and actual values of the dependent variable is 0.60.© Kaplan, Inc. 118 ©2018 FK Pa

SS 03 - Quantitative Methods for Valuation Discussion Question QSuoelsuttiioonn 1199 19. The most appropriate null hypothesis and the most appropriate conclusion regarding A. Hansen’s belief about the magnitude of the B. initial return relative to that of the pre-offer price C. adjustment (reflected by the coefficient bj) are:© Kaplan, Inc. Null Hypothesis Conclusion about bj (0.05 Level of Significance) Ho: bj = 0.5 Reject Ho Ho: bj ≥ 0.5 Fail to Reject Ho Ho: bj ≥ 0.5 Reject Ho 117 Discussion Question SSoolulutitoionn2200A multiple R-squared of 0.36 means that 36% of thevariation in the dependent variable is explained bythe independent variables.The correlation between the dependent andindependent variables is √0.36 = 0.60 © Kaplan, Inc. 119artners - Exame CFA 30

Discussion Question QSuoelsuttiioonn 220020. The most appropriate interpretation of the multiple R-squared for Hansen’s model isA. that:B.C. unexplained variation in the dependent variable is 36 percent of total variation. correlation between predicted and actual values of the dependent variable is 0.36. correlation between predicted and actual values of the dependent variable is 0.60.© Kaplan, Inc. 120 Discussion Question SSoolulutitoionn2211Conditional heteroskedasticity results in: consistent parameter estimates biased (up or down) standard errors biased t-statistics biased F-statistics© Kaplan, Inc. 122 ©2018 FK Pa

SS 03 - Quantitative Methods for Valuation Discussion Question QQuueesstitoionn221121. Is Chang’s statement 1 correct?A.B. Yes. No, because the model’s F-statistic will not beC. biased. No, because the model’s t-statistics will not be biased.© Kaplan, Inc. 121 Discussion Question QSuoelsuttiioonn 221121. Is Chang’s statement 1 correct?A.B. Yes.C. No, because the model’s F-statistic will not be biased. No, because the model’s t-statistics will not be biased. © Kaplan, Inc. 123artners - Exame CFA 31

Discussion Question QQuueesstitoionn222222. Is Chang’s statement 2 correct?A.B. Yes.C. No, because the model’s coefficient estimates will be unbiased. No, because the model’s coefficient estimates will be consistent.© Kaplan, Inc. 124 Discussion Question QSuoelsuttiioonn 222222. Is Chang’s statement 2 correct?A.B. Yes.C. No, because the model’s coefficient estimates will be unbiased. No, because the model’s coefficient estimates will be consistent.© Kaplan, Inc. 126 ©2018 FK Pa

SS 03 - Quantitative Methods for Valuation Discussion Question SSoolulutitoionn2222Correlated omitted variables result in: biased and inconsistent parameter estimates inconsistent standard errors© Kaplan, Inc. 125artners - Exame CFA 32

Questions – SS 3 - Quantitative Methods for ValuationQuestions 1 – 6Albert Morris, CFA, is evaluating the results of an estimation of the number of wireless ph oneminutes used on a quarterly basis within the territory of Car-tel International, Inc. Some of theinformation is presented below (in billions of minutes):Wireless Phone Minutes (WPM)t = bo + b1 WPMt-1 + εtANOVA Degrees of Freedom Sum of Squares Mean SquareRegression 1 7,212.641 7,212.641Error 26 3,102.410 119.324Total 27 10,315.051Coefficients Coefficient Standard Error of the CoefficientIntercept -8.0237 2.9023WPM t-1 1.0926 0.0673The variance of the residuals from one time period within the time series is not dependent on thevariance of the residuals in another time period.Morris also models the monthly revenue of Car-tel using data over 96 monthly observations. Themodel is shown below:Sales (CAD$ millions) = b0 + b1 Salest−1 + εtCoefficients Coefficient Standard Error of the CoefficientIntercept 43.2 12.32Salest−1 0.8867 0.4122Question 1The value for WPM this period is 544 billion. Using the results of the model, the forecast WirelessPhone Minutes three periods in the future is:A) 683.18.B) 586.35.C) 691.30.Question 2The R-squared for the WPM model is closest to:A) 70%.B) 97%.C) 33%.Question 3The WPM model was specified as a(n):A) Autoregressive (AR) Model with a seasonal lag.B) Autoregressive (AR) ModelC) Moving Average (MA) ModelQuestion 4Based upon the information provided, Morris would most likely get more meaningful statisticalresults by: ______________________ 1 ©2018 FK PartnersTodos os direitos reservados – É proibida a reprodução total ou parcial, de qualquer forma ou por qualquer meio.

A) Doing nothing. No information provided suggests that any of these will improve the specification B) First differencing the data. C) Adding more lags to the model.Question 5The mean reverting level of monthly sales is closest to: A) 381.29 million. B) 8.83 million. C) 43.2 million.Question 6Morris concludes that the current price of Car-tel stock is consistent with single stage constantgrowth model (with g=3%). Based on this information, the sales model is most likely: A) Correctly specified. B) Incorrectly specified and taking the natural log of the data would be an appropriate remedy. C) Incorrectly specified and first differencing the data would be an appropriate remedy.Questions 7 – 12Bill Johnson, CFA, has prepared data concerning revenues from sales of winter clothing made byPolar Corporation. This data is presented (in $ millions) in the following table:Question 7The 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.Question 8The value that Johnson should enter in the table in place of \"w\" is: A) 164. B) −48. C) −115.Question 9Imagine that Johnson prepares a change-in-sales regression analysis model with seasonality, whichincludes the Following: ______________________ 2 ©2018 FK PartnersTodos os direitos reservados – É proibida a reprodução total ou parcial, de qualquer forma ou por qualquer meio.

Intercept Coefficients Lag 1 −6.032 Lag 4 0.017 0.983Based on the model, expected sales in the first quarter of 2015 will be closest to: A) 155. B) 210. C) 190.Question 10Johnson's model was most likely designed to incorporates correction for: A) Heteroskedasticity of model residuals. B) Nonstationarity in time series data. C) Cointegration in the time series.Question 11To test for covariance-stationarity in the data, Johnson would most likely use a: A) Dickey-Fuller test. B) t-test. C) Durbin-Watson test.Question 12The presence of conditional heteroskedasticity of residuals in Johnson's model is would most likelyto lead to: A) Invalid standard errors of regression coefficients and invalid statistical tests. B) Invalid estimates of regression coefficients, but the standard errors will still be valid. C) Invalid standard errors of regression coefficients, but statistical tests will still be valid.Question 13 - 18Miles Mason, CFA, works for ABC Capital, a large money management company based in New York.Mason has several years of experience as a financial analyst, but is currently working in themarketing department developing materials to be used by ABC's sales team for both existing andprospective clients. ABC Capital's client base consists primarily of large net worth individuals andFortune 500 companies. ABC invests its clients' money in both publicly traded mutual funds as wellas its own investment funds that are managed in-house. Five years ago, roughly half of its assetsunder management were invested in the publicly traded mutual funds, with the remaining half inthe funds managed by ABC's investment team. Currently, approximately 75% of AB C's assetsunder management are invested in publicly traded funds, with the remaining 25% being distributedamong ABC's private funds. The managing partners at ABC would like to shift more of its client'sassets away from publicly-traded funds into ABC's proprietary funds, ultimately returning to a 50/50split of assets between publicly traded funds and ABC funds. There are three key reasons for thisshift in the firm's asset base. First, ABC's in -house funds have outperformed other fundsconsistently for the past five years. Second, ABC can offer its clients a reduced fee structure onfunds managed in-house relative to other publicly traded funds. Lastly, ABC has recently hired atop fund manager away from a competing investment company and would like to increase his assetsunder management.ABC Capital's upper management requested that current clients be surveyed in order to determinethe cause of the shift of assets away from ABC funds. Results of the survey indicated that clientsfeel there is a lack of information regarding ABC's funds. Clients would like to see extensiveinformation about ABC's past performance, as well as a sensitivity analysis showing how the funds ______________________ 3 ©2018 FK Partners Todos os direitos reservados – É proibida a reprodução total ou parcial, de qualquer forma ou por qualquer meio.

will perform in varying market scenar ios. Mason is part of a team that has been charged by uppermanagement to create a marketing program to present to both current and potential clients of ABC.He needs to be able to demonstrate a history of strong performance for the ABC funds, and, whilen ot promising any measure of future performance, project possible return scenarios. He decides toconduct a regression analysis on all of ABC's in-house funds. He is going to use 12 independenteconomic variables in order to predict each particular fund's return. Mason is very aware of themany factors that could minimize theeffectiveness of his regression model, and if any are present, he knows he must determine if anycorrective actions are necessary.Mason is using a sample size of 121 monthly returns.Question 13In order to conduct an F-test, what would be the degrees of freedom used (dfnumerator; dfdenominator)? A) 11; 120. B) 108; 12. C) 12; 108.Question 14In regard to multiple regression analysis, which of the following statements is most accurate? A) Adjusted R2 always decreases as independent variables increase. B) Adjusted R2 is less than R2. C) R2 is less than adjusted R2.Question 15Which of the following tests is most likely to be used to detect autocorrelation? A) Durbin-Watson. B) Dickey-Fuller. C) Breusch-Pagan.Question 16One of the most popular ways to correct heteroskedasticity is to: A) Adjust the standard errors. B) Use robust standard errors. C) Improve the specification of the model.Question 17Which of the following statements regarding the Durbin-Watson statistic is most accurate? TheDurbin-Watson statistic: A) Is approximately equal to 1 if the error terms are not serially correlated. B) Can only be used to detect positive serial correlation. C) Only uses error terms in its computations.Question 18If a regression equation shows that no individual t-tests are significant, but the F-statistic issignificant, the regression probably exhibits: A) Multicollinearity. B) Serial correlation. C) Heteroskedasticity.Question 19 - 24Damon Washburn, CFA, is currently enrolled as a part-time graduate student at State University. ______________________ 4 ©2018 FK PartnersTodos os direitos reservados – É proibida a reprodução total ou parcial, de qualquer forma ou por qualquer meio.

One of his recent assignments for his course on Quantitative Analysis is to perform a regressionanalysis utilizing the concepts covered during the semester. He must interpret the results of theregression as well as the test statistics. Washburn is confident in his ability to calculate the statisticsbecause the class is allowed to use statistical software. However, he realizes that the interpretationof the statistics will be the true test of his knowledge of regression analysis. His professor has givento the students a list of questions that must be answered by the results of the analysis.Washburn has estimated a regression equation in which 160 quarterly returns on the S&P 500 areexplained by three macroeconomic variables: employment growth (EMP) as measured by nonfarmpayrolls, gross domestic product (GDP) growth, and private investment (INV). The results of theregression analysis are as follows:Other Data:  Regression sum of squares (RSS) = 126.00  Sum of squared errors (SSE) = 267.00  Durbin-Watson statistic (DW) = 1.34Question 19How many of the three independent variables (not including the intercept term) are statisticallysignificant in explaining quarterly stock returns at the 5.0% level?A) All three are statistically significant.B) One of the three is statistically significant.C) Two of the three are statistically significant.Question 20Can the null hypothesis that the GDP growth coefficient is equal to 3.50 be rejected at the 1.0% ______________________ 5 ©2018 FK PartnersTodos os direitos reservados – É proibida a reprodução total ou parcial, de qualquer forma ou por qualquer meio.

confidence level versus the alternative that it is not equal to 3.50? The null hypothesis is: A) Rejected because the t-statistic is less than 2.617. B) Not rejected because the t-statistic is equal to 0.92. C) Accepted because the t-statistic is less than 2.617.Question 21The percentage of the total variation in quarterly stock returns explained by the independentvariables is closest to: A) 32%. B) 42%. C) 47%.Question 22According to the Durbin-Watson statistic, there is: A) Significant heteroskedasticity in the residuals. B) Significant positive serial correlation in the residuals. C) No significant positive serial correlation in the residuals.Question 23What is the predicted quarterly stock return, given the following forecasts? Employment growth = 2.0% GDP growth = 1.0% Private investment growth = -1.0% A) 4.7%. B) 5.0%. C) 4.4%.Question 24What is the standard error of the estimate? A) 1.31. B) 0.81. C) 1.71.Question 25 - 30Craig Standish, CFA, is investigating the validity of claims associated with a fund that his companyoffers. The company advertises the fund as having low turnover and, hence, low management fees.The fund was created two years ago with only a few uncorrelated assets. Standish randomly drawstwo stocks from the fund, Grey Corporation and Jars Inc., and measures the variances andcovariance of their monthly returns over the past two years. The resulting variance covariancematrix is shown below. Standish will test whether it is reasonable to believe that the returns of Greyand Jars are uncorrelated. In doing the analysis, he plans to address the issue of spurious correlationand outliers. Grey JarsGrey 42.2 20.8Jars 20.8 36.5Standish wants to learn more about the performance of the fund. He performs a linear regressionof the fund's monthly returns over the past two years on a large capitalization index. The resultsare below: ______________________ 6 ©2018 FK PartnersTodos os direitos reservados – É proibida a reprodução total ou parcial, de qualquer forma ou por qualquer meio.

ANOVA df SS MS FRegression 1 92.53009 92.53009 28.09117Residual 22 72.46625 3.293921Total 23 164.9963Intercept Coefficients Standard Error t-statistic P-valueLarge Cap 0.148923 0.391669 0.380225 0.707424Index 1.205602 0.227467 5.30011 2.56E-05Standish forecasts the fund's return, based upon the prediction that the return to the largecapitalization index used in the regression will be 10%. He also wants to quantify the degree of theprediction error, as well as the minimum and maximum sensitivity that the fund actually has withrespect to the index.He plans to summarize his results in a report. In the report, he will also include caveats concerningthe limitations of regression analysis. He lists four limitations of regression analysis that he feelsare important: relationships between variables can change over time, multicollinearity leads toinconsistent estimates of regression coefficie nts, if the error terms are heteroskedastic the standarderrors for the regression coefficient may not be reliable, and if the error terms are correlated witheach other over time the test statistics may not be reliable.Question 25Given the variance/covariance matrix for Grey and Jars, in a one-sided hypothesis test that thereturns are positively correlated H0: ρ ≤ 0 vs. H1: ρ > 0, Standish would: A) Reject the null at the 5% but not the 1% level of significance. B) Reject the null at the 1% level of significance. C) Need to gather more information before being able to reach a conclusion concerning significance.Question 26In using the correlation coefficient between returns on Grey and Jars, Standish would mostappropriately question the issue of: A) Issue of outliers but not the issue of spurious correlation. B) Both spurious correlation and outliers. C) Spurious correlation but not the issue of outliers.Question 27If the large capitalization index has a 10% return, then the forecast of the fund's return will be: A) 13.5. B) 12.2. C) 16.1.Question 28The standard deviation of monthly fund returns is closest to: A) 7.17. B) 12.84. C) 2.68.Question 29A 95% confidence interval for the slope coefficient is: A) 0.760 to 1.650. 7 B) 0.734 to 1.677. ______________________ ©2018 FK Partners Todos os direitos reservados – É proibida a reprodução total ou parcial, de qualquer forma ou por qualquer meio.

C) 0.905 to 1.506.Question 30Of the four caveats of regression analysis listed by Standish, the least accurate is: A) The relationships of variables change over time. B) Multicollinearity leads to inconsistent estimates of the regression coefficients. C) If the error terms are heteroskedastic the standard errors for the regression coefficients may not be reliable. ______________________ 8 ©2018 FK PartnersTodos os direitos reservados – É proibida a reprodução total ou parcial, de qualquer forma ou por qualquer meio.

Answers – SS 03 - Quantitative Methods for ValuationQuestions 1 -6Questions 1The correct answer was A) 683.18.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 −8.023 + (1.0926 × 632.62) = 683.18.Question 2The correct answer was A) 70%.R-squared = SSR/SST = 7,212.641/10,315.051 = 70%.Question 3The correct answer was B) Autoregressive (AR) Model.The model is specified as an AR Model, but there is no seasonal lag. No moving averages areemployed in the estimation of the model.Question 4The correct answer was B) first differencing the data.Since the slope coefficient is greater than one, the process may not be covariance stationary (wewould have to test this to be definitive). A common technique to correct for this is to first differencethe variable to perform the following regression: ∆(WPM)t = bo + b1 ∆(WPM)t-1 + ε t.Question 5The correct answer was A) 381.29 million.Question 6The correct answer was B) Incorrectly specified and taking the natural log of the data would be anappropriate remedy.If constant growth rate is an appropriate model for Car-tel, its dividends (as well as earnings andrevenues) will grow at a constant rate. In such a case, the time series needs to be adjusted bytaking the natural log of the time series. First differencing would remove the trending componentof a covariance non -stationary time series but would not be appropriate for transforming anexponentially growing time series.Questions 7 - 12Question 7The correct answer was C) an autoregressive model with a seasonal lag.Johnson will use the table to forecast values using an autoregressive model for periods in successionsince each successive forecast relies on the forecast for the preceding period. The seasonal lag isintroduced to account for seasonal variations in the observed data.Question 8The correct answer was A) 164.The seasonal lagged change in sales shows the change in sales from the period 4 quarters beforethe current period. Sales in the year 2013 quarter 4 increased $164 million over the prior period.Question 9The correct answer was B) 210.Substituting the 1-period lagged data from 2014.4 and the 4-period lagged data from 2014.1 intothe 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. _________________________ 1 ©2018 FK PartnersTodos os direitos reservados – É proibida a reprodução total ou parcial, de qualquer forma ou por qualquer meio

Question 10The correct answer was B) nonstationarity in time series data.Johnson's model transforms raw sales data by first differencing it and then modeling change insales. This is most likely an adjustment to make the data stationary for use in an AR model.Question 11The correct answer was A) Dickey-Fuller test.The Dickey-Fuller test for unit roots could be used to test whether the data is covariance non -stationarity. The Durbin-Watson test is used for detecting serial correlation in the residuals of trendmodels but cannot be used in AR models. A t-test is used to test for residual autocorrelation in ARmodelsQuestion 12The correct answer was A) invalid standard errors of regression coefficients and invalid statisticaltests.The presence of conditional heteroskedasticity may leads to incorrect estimates of standard errorsof regression coefficients and hence invalid tests of significance of the coefficients.Questions 13 - 18Question 13The correct answer was C) 12; 108.Degrees of freedom for the F-statistic is k for the numerator and n − k − 1 for the denominator.k = 12n − k − 1 = 121 − 12 − 1 = 108Question 14The correct answer was B) Adjusted R2 is less than R2.Whenever there is more than one independent variable, adjusted R2 is less than R2. Adding a newindependent variable will increase R2, but may either increase or decrease adjusted R2.R2 adjusted = 1 − [((n − 1) / (n − k − 1)) × (1 − R2)]Where:n = number of observationsK = number of independent variables R2 = unadjusted R2Question 15The correct answer was A) Durbin-Watson.Durbin-Watson is used to detect autocorrelation. The Breusch-Pagan test is used to detectheteroskedasticity. The Dickey Fuller test is a test for unit root.Question 16The correct answer was B) use robust standard errors.Using generalized least squares and calculating robust standard errors are possible remedies forheteroskedasticity. Improving specifications remedies serial correlation. The standard error cannotb e adjusted, only the coefficient of the standard errors.Question 17The correct answer was C) only uses error terms in its computations.The formula for the Durbin-Watson statistic uses error terms in its calculation. The Durbin -Watsonstatistic is approximately equal to 2 if there is no serial correlation. A Durbin-Watson statisticsignificantly less than 2 may indicate positive serial correlation, while a Durbin-Watson statisticsignificantly greater then 2 may indicate negative serial correlation.Question 18 2The correct answer was A) multicollinearity. _________________________ ©2018 FK PartnersTodos os direitos reservados – É proibida a reprodução total ou parcial, de qualquer forma ou por qualquer meio

Common indicators of multicollinearity include: high correlation (>0.7) between independentvariables, no individual t-tests are significant but the F-statistic is, and signs on the coefficients thatare opposite of what is expected.Questions 19 - 24Question 19The correct answer was C) Two of the three are statistically significant.To determine whether the independent variables are statistically significant, we use the student'st-statistic, where t equals the coefficient estimate divided by the standard error of the coefficient.This is a two -tailed test. The critical value for a 5.0% significance level and 156 degrees of freedom(160 -3-1) is about 1.980, according to the table.The t-statistic for employment growth = -4.50/1.25 = -3.60. The t-statistic for GDP growth =4.20/0.76 = 5.53.The t-statistic for investment growth = -0.30/0.16 = -1.88.Therefore, employment growth and GDP growth are statistically significant because the absolutevalues of their t-statistics are larger than the critical value, which means two of the threeindependent variables are statistically significantly different from zero.Question 20The correct answer was B) not rejected because the t-statistic is equal to 0.92.The hypothesis is:H0: bGDP = 3.50 Ha: bGDP ≠ 3.50This is a two-tailed test. The critical value for the 1.0% significance level and 156 degrees of freedom(160 − 3 − 1) is about 2.617. The t-statistic is (4.20 − 3.50)/0.76 = 0.92. Because the t-statisticis less than the critical value, we cannot reject the null hypothesis. Notice we cannot say that thenull hypothesis is accepted; only th at it is not rejected.Question 21The correct answer was A) 32%.The R2 is the percentage of variation in the dependent variable explained by the independentvariables. The R 2 is equal to the SSRegession/SSTotal, where the SSTotal is equal to SSRegression + SSError.R2 = 126.00/(126.00+267.00) = 32%.Question 22The correct answer was B) significant positive serial correlation in the residuals.The Durbin-Watson statistic tests for serial correlation in the residuals. According to the table, dl =1.61 and du = 1.74 for three independent variables and 160 degrees of freedom. Because the DW(1.34) is less than the lower value (1.61), the null hypothesis of no significant positive serialcorrelation can be rejected. This means there is a problem with serial correlation in the regression,which affects the interpretation of the results.Question 23The correct answer was B) 5.0%.Predicted quarterly stock return is 9.50% + (−4.50)(2.0%) + (4.20)(1.0%) + (−0.30)(−1.0%) =5.0%.Question 24The correct answer was A) 1.31.The standard error of the estimate is equal to [SSE/(n − k − 1)]1/2 = [267.00/156]1/2 =approximately 1.31. _________________________ 3 ©2018 FK PartnersTodos os direitos reservados – É proibida a reprodução total ou parcial, de qualquer forma ou por qualquer meio

Questions 25 – 30Question 25The correct answer was B) reject the null at the 1% level of significance.First, we must compute the correlation coefficient, which is 0.53 = 20.8 / (42.2 × 36.5) 0.5.The t-statistic is: 2.93 = 0.53 × [(24 - 2) / (1 − 0.53 × 0.53)]0.5, and for df = 22 = 24 − 2, the t-statistics for the 5% and 1% level are 1.717 and 2.508 respectively.Question 26The correct answer was B) Both spurious correlation and outliers.Both these issues are important in performing correlation analysis. A single outlier observation canchange the correlation coefficient from significant to not significant and even from negative (positive)to positive (negative). Even if the correlation coefficient is significant, the researcher would want tomake sure there is a reason for a relationship and that the correlation is not spurious (i.e., causedby chance).Question 27The correct answer was B) 12.2.The forecast is 12.209 = 0.149 + 1.206 × 10, so the answer is 12.2.Question 28The correct answer was C) 2.68.Variance of fund returns = SST/(n-1) = 164.9963/23 = 7.17. Standard deviation = (7.17)0.5 = 2.68Question 29The correct answer was B) 0.734 to 1.677.The 95% confidence interval is 1.2056 ± (2.074 × 0.2275). Remember, to use 2-tailed t-statisticfor confidence intervals.Question 30The correct answer was B) multicollinearity leads to inconsistent estimates of the regressioncoefficients.In the presence of multicollinearlity, the regression coefficients would still be consistent butunreliable. The other possible shortfalls listed are valid. _________________________ 4 ©2018 FK PartnersTodos os direitos reservados – É proibida a reprodução total ou parcial, de qualquer forma ou por qualquer meio

Fixed Income Investments Study Session 4Economics for Valuation Topic Weight: 5–10% Fixed Income InvestmEecnotnsomicsEconomics for Valuation 13. Currency Exchange Rates: Understanding Equilibrium Value ©2018 FK Partne

SS 04 - Economics for Valuation Fixed Income Investments Study Session 4 Economics for Valuation 13. Currency Exchange Rates: Understanding Equilibrium Value 14. Economic Growth and the Investment Decision 15. Economics of RegulationLOS 13.a Calculate/Interpret/Describe Currency Exchange RatesSchweser B1 pg 246, CFAI V1 pg 516 Foreign Exchange QuotationsQuote on Swiss francs (CHF):USD/CHF 0.9521 – 0.9536 (= 0.9521/36 = 0.9521-36) Bid: Dealer sells Ask (offer): Dealer sells quoted and buys base base and buys quoted CHF is the base currency, and USD is the priced 3 currency (quoted) Spread (in dollars): 0.9536 – 0.9521 = USD 0.0015 (or 15 ‘pips’ = 0.0015 × 10,000)© Kaplan, Inc.ers - Exame CFA 1

LOS 13.b Identify/Calculate Currency Exchange RatesSchweser B1 pg 248, CFAI V1 pg 520 Cross RatesDefinition: The exchange rate between twocurrencies implied by their exchange rates with acommon third currency.Example: Suppose you have the following quotes: USD/GBP CHF/USDThe common currency is the USDThe cross rate defines the value of the GBP vs. theCHF© Kaplan, Inc. 4LOS 13.a Calculate/Interpret/Describe Currency Exchange RatesSchweser B1 pg 246, CFAI V1 pg 516 The Triangle: A Visual RepresentationTo convert CHF into GBP, there are two paths:  Path #1: Exchange CHF for GBP at CHF/GBP 2.3182  Path #2: Exchange CHF for USD at CHF/USD 1.4860 and then convert the USD into GBP at USD/GBP 1.5600No arbitrage requirement: The two paths to getfrom CHF to GBP must yield the same results© Kaplan, Inc. Visualize the currency triangle 6 ©2018 FK Partne

SS 04 - Economics for ValuationLOS 13.b Identify/Calculate Currency Exchange RatesSchweser B1 pg 248, CFAI V1 pg 520 Example: Cross Rates Given the following FX rates: USD/GBP 1.5600 and CHF/USD 1.4860 Calculate the GBP/CHF cross exchange rate USD CHF = CHF Just algebra: Set up the GBP  USD GBP quotes so the common currency cancels 1.5600 USD  1.4860 CHF = CHF 2.3182 GBP USD GBPIf you want GBP/CHF, take the reciprocal of 2.3182:© Kaplan, Inc. 1 / (2.3182) ≈ GBP/CHF 0.4314 5LOS 13.b Identify/Calculate Currency Exchange RatesSchweser B1 pg 248, CFAI V1 pg 520 The Triangle USD CHF CHF/GBP 2.3182 GBP© Kaplan, Inc. 7ers - Exame CFA 2

LOS 13.b Identify/Calculate Currency Exchange RatesSchweser B1 pg 248, CFAI V1 pg 520 Example: Cross Rates With Bid-Ask QuotesCHF / USD 1.5000  1.5010 USD / GBP 1.3500  1.3510 Compute CHF/GBP cross bid and ask rates Process is identical, multiply bid prices to get bid cross rate and multiply ask prices to get ask cross rate. Remember: The equation needs to be set up tocancel the common currency.  CHF  = 1  USD  offer  USD   CHF  bid© Kaplan, Inc. 8LOS 13.b Identify/Calculate Currency Exchange RatesSchweser B1 pg 248, CFAI V1 pg 520Triangular Currency Arbitrage: A Trick But how do you know whether the bid or the ask is the appropriate rate? Up-the-bid, multiply, and down-the-ask, divide BID $ ASK means € means turning turning € into $ $ into €© Kaplan, Inc. 10 ©2018 FK Partne

SS 04 - Economics for ValuationLOS 13.b Identify/Calculate Currency Exchange RatesSchweser B1 pg 248, CFAI V1 pg 520Calculating Cross Bid and Ask RatesCHF / USD 1.5000  1.5010 USD / GBP 1.3500  1.3510 USD  CHF = CHF GBP USD GBP 1.3500 USD  1.5000 CHF = 2.0250 CHF (bid) GBP USD GBP 1.5010 CHF  1.3510 USD = 2.0279 CHF (ask) USD GBP GBP© Kaplan, Inc. 9LOS 13.b Identify/Calculate Currency Exchange RatesSchweser B1 pg 248, CFAI V1 pg 520 Dealing With Arbitrage QuestionsExam questions dealing with arbitrage revolvearound three issues: 1. Verify the arbitrage (does an opportunity exist, meaning quoted rate ≠ calculated rate?) 2. Structure the trades to exploit the opportunity (sell overvalued at quoted rate, buy undervalued). The direction is the key! 3. Calculate the profit given an initial investment (may start with the third currency such as USD)© Kaplan, Inc. 11ers - Exame CFA 3