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Source: http://www.doksinet School of Business and Economics RESIT EXAM Course Code Date Location : : : : Quantitative Business EBC2025 9 / 10 / 11 / 12 July 2013 MECC Communication devices are not allowed – put them in your bag (under your table) or on the floor, not within arm’s reach! Otherwise this will be reported as possible fraud to the Board of Examiners. Any changes to the exam after the official end will be reported as possible fraud to the Board of Examiners. Sanctions by the Board of Examiners on fraud or irregularities include such measures as: - complete or partial voidance or annulment of the relevant exam; - exclusion from participation or further participation for one or more exams at the SBE for a period of time to be determined by the Board of Examiners, with a maximum period of one year; - termination of the student’s registration for the study programme concerned. This exam consists of: 20 pages (front page, formulas and statistical tables included) 4

open questions, i.e 2 for mathematics and 2 for statistics You are allowed to make use of: Scrap paper and a calculator (if non-programmable and non-graphical). Books, manuals, lecture slides etc. are not allowed How to answer the questions: Be sure to write legibly, and to answer the mathematics and statistics parts on separate sheets of paper. Grading: This exam consists of two parts: mathematics and statistics. Both parts contain 2 open questions, each composed of distinct items numbered as a), b) etc. Both parts of the exam can be rewarded with 50 points at the most, for a maximum total of 100 points. Your unrounded exam grade will be determined by dividing the number of points scored by 10. Concerning your final grade for the course, 30% of it will be determined by the unrounded grades for your three discussion team performances (10% each), and 70% by your unrounded grade for this exam. The resulting weighted average will be rounded into half points, subject to two qualifications:

i) You need to score at least 25 points for both parts of this exam separately. If not, then your final grade can never be higher than 5.0 ii) To pass, you need a weighted average of at least 5.5 Scores like 54 will be rounded down Procedure for objections: The consultation session will be on Wednesday July 17, 9.00-1200h, in room A309: the provisional marks will be announced on Monday July 15 at the latest. All issues raised by this resit exam can be put forward during this consultation hour. Protests or other forms of objection, not brought in during this consultation hour, will be rejected. SUCCESS !!! 1 Source: http://www.doksinet MATHEMATICS PART Question 1 (Case W4): Portfolio Optimization (25pt) Consider an investor who can invest in two assets: A1 and A2. The payoff of either asset depends on which of the two states S1 or S2 is realized. Each state Si occurs with probability Pi The payoffs of each asset in each state and the probabilities Pi are reported in the table below.

The symbol a in the table is a parameter with a positive value. S1 S2 Pi 1 3 2 3 A1 a 6 A2 9 6 a) (5pt) i) Show that the expected payoffs of assets A1 and A2 are: 1 3 µ1 = a + 4 µ2 = 7 (1) ii) Show that the variances of the returns on assets A1 and A2 are: σ12 = 92 a 2 − 83 a + 8 σ 22 = 2 (2) Now imagine the following portfolio P, composed of assets A1 and A2: P = λA1 + (1 − λ) A2 with 0 ≤ λ ≤1 (3) while the investor has the following utility function: U ( P ) = µ( P ) − 12 σ 2 ( P ) (4) We will consider two different cases, based on two different values for the parameter a. (Questions b-e) Case i): take a = 6 b) (1pt) Motivate that the correlation between the returns on both assets, ρ( A1 , A2 ) , is equal to 0. c) (4pt) i) Starting from eq. (1), show that µ( P ) = 7 − λ (5) ii) Starting from eq. (2), show that σ 2 ( P ) = 2(1 − λ) 2 (6) Hint: Use the fact that ρ( A1 , A2 ) = 0 in Case i). 1 2 d) (3pt) Show that the

investor’s utility function attains its maximum value at λ = . e) (4pt) In the familiar (σ,µ)-plane, sketch question d)’s optimization problem and its solution as explicitly as you can. 2 Source: http://www.doksinet (Questions f-g) Case ii): take a = 3 f) (3pt) i) Motivate that, in this case, asset A1 is inferior compared to asset A2. ii) Motivate that the correlation between the returns on both assets, ρ( A1 , A2 ) , is equal to –1. 1 4 g) (5pt) i) Show that the investor’s utility function attains its maximum value at λ = . Hint: Don’t forget to take your result at item f) into account when determining σ 2 ( P ) . ii) Motivate intuitively, why it makes sense for the investor to mix in a positive amount of asset A1, even though it is inferior to asset A2. (please turn to the next page) 3 Source: http://www.doksinet Question 2 (Case W6): Game Theory and Economics (25pt) Consider the following classroom game with 10 students. Each student can vote “aye” or

“nay”; they all cast their votes simultaneously, without knowing the votes of their classmates. After the votes have been counted, each student receives € 1 for every student that votes “aye”, provided that at least 6 students vote “aye”. If fewer than 6 students vote “aye”, nobody gets any payment. Apart from the payment that you (may) receive, voting “aye” costs you € 1.50 So, to illustrate: - If 8 students vote “aye” and 2 students vote “nay”, then each “nay”-voter has a net payoff of € 8, and each “aye”-voter has a net payoff of € 6.50 (ie € 8 minus € 150) - If 5 students vote “aye” and 5 students vote “nay”, then each “nay”-voter has a net payoff of € 0, while each “aye”-voter has a net payoff of minus € 1.50 (ie they each have to pay one and a half euro) a) (6pt) Are there dominant strategies in this game? Motivate your answer! Hint: Imagine you are one of the students. Let a denote how many of the other nine

students vote “aye”, so that a = 0, 1, , 9. Then identify your best response for all possible values of a, and draw your conclusion. b) (6pt) Describe all the Nash equilibria of this game. Hint: All strategy profiles are fully characterized by the total number of students voting “aye”, say k, with the remaining 10 – k students voting “nay”. So we really have eleven distinct strategy profiles, for k = 0, 1, , 10. We suggest that you systematically evaluate these profiles, to determine whether they constitute a Nash equilibrium or not. c) (5pt) What is the Pareto optimal strategy profile? That is, which strategy profile maximizes the sum of all net payoffs? Motivate your answer. (Questions d-e) Now consider the following sequential version of this game. First student 1 publicly, in the presence of all other students, announces whether he votes “aye” or “nay”. Second, knowing the vote of student 1, student 2 publicly casts his vote. This procedure continues until all

students have voted In the following two questions, we will solve this sequential game by using backwards induction, a concept you should remember from previous courses (or from the Stackelberg model discussed in Case W7). d) (3pt) Suppose you are student number 10, i.e the last one to cast a vote Let a denote the number of “aye” votes among the nine students that voted before you, so that a = 0, 1, , 9. Motivate carefully, how your best response depends on the value of a. e) (5pt) What is the (unique) backwards induction outcome of this game? Motivate your answer. Hint: To get an idea of the sequential logic involved here, you may start by thinking about a simplified version of this game: only 3 students, who need a majority of 2 “ayes” to get a payment of € 1 per “aye”, while each “aye” has to pay a € 1.50 fee (please turn to the next page) 4 Source: http://www.doksinet STATISTICS PART PLEASE START ON A NEW SHEET OF PAPER! Question 1 (Case W5): The CAPM

(25pt) The Capital Asset Pricing Model or CAPM is one of the cornerstones of Finance. For any stock i, the CAPM relationship can be estimated by running a regression of the following form: ( Rit − R0t ) = α i + β i ( R Mt − R0t ) + ε it t = 1,,T (1) where Rit is the actual return on stock i in period t, R0t the actual return on the risk-free bond in period t, and RMt the actual return on the market portfolio in period t. According to the CAPM, the intercept alpha of this regression equation should be zero, while its slope beta can basically be anything (in particular, it is theoretically possible for beta to be larger than 1, and even smaller than 0). In Case W5, we tested the validity of the CAPM for two stocks that are represented in the AEX, the leading Dutch stock market index: Aegon and Heineken. In this exam question, we focus on the behavior of Akzo-Nobel stock during the period 1992-1 till 2008-4 (i.e for T = 196 consecutive months). In the rest of this question,

whenever we use terms like “the return on Akzo” or “the market return”, we actually mean: the return in excess of the risk-free rate, as addressed in eq. (1) The scatterplot alongside provides a first impression of the relationship between the return on Akzo and the market return. The regression results for model (1), obtained with SPSS, are reported below. Note that we have asked for the Durbin-Watson statistic to be reported; moreover, the model’s residuals have been saved under the name “ResAkzo”. 5 Source: http://www.doksinet (Question a) As it turns out, model (1)’s most pronounced positive outlier (with a massive studentized residual of 3.97) occurs in period 88, i.e in April 1999 In this period, indicated by the arrow in the scatterplot on the previous page, the regular residual equals e88 = 21.06 a) (4pt) i) Describe in words, as explicitly as you can, what it means that “e88 = 21.06” ii) On the basis of the scatterplot on the previous page, motivate as

explicitly as you can, whether the observation for period 88 is likely to be influential or not. (Question b) In the rest of this exam, we will take model (1) as our starting point. But since the CAPM involves time series data, model (1) is potentially subject to autocorrelation. If so, this would invalidate our standard inferential procedures. In this question, like in the original Case W5, we focus on first-order autocorrelation We have already noted that our SPSS-output for the basic CAPM-model (1) includes the Durbin-Watson statistic. Below, in addition, we report a regression (without intercept) of model (1)’s residuals against their first lags. Now consider the following verbally stated null hypothesis: “Model (1) does not suffer from first-order autocorrelation.” b) (6pt) i) Provide a strictly verbal but adequate definition of first-order autocorrelation. ii) Motivate as explicitly as you can, that the Durbin-Watson statistic of D = 1.978, reported for the residual

regression above, is of no help to assess the veracity of our verbally stated null hypothesis above. iii) Motivate how the SPSS-output for the residual regression above can be used to test our verbally stated null hypothesis, against an appropriate alternative, at the 10% significance level. (Questions c-f) In the rest of this question, we assume that the standard inferential procedures are valid in model (1) and its sequels. (Question c) According to the CAPM, the intercept of our regression equation, alpha, should be zero. This implies the following null hypothesis in terms of the coefficients of model (1): 6 Source: http://www.doksinet H0: αi = 0 You can easily check that, for Akzo-Nobel, this null cannot be rejected at any reasonable significance level. When we repeat our analysis for all the 16 stocks that were part of the AEX during our entire observation period, using a significance level of 10%, it turns out that the null hypothesis H0: αi = 0 is rejected for two of those

16 stocks (in particular: for Heineken, as we already saw in class, and for SBM Offshore). c) (3pt) To what extent does this result cast doubt on the validity of the CAPM? Try to motivate your answer as explicitly as possible. (Question d) Consider the null hypothesis that the Akzo-Nobel stock follows the market portfolio one-for-one (apart from diversifiable risk, captured by the error term), against a two-sided alternative: H0: βi = 1 vs. HA: βi ≠ 1 d) (3pt) Use the SPSS-output for model (1) to determine a range for the p-value of this null hypothesis, against the stated alternative (e.g: 1% < p < 5%) (Questions e-f) An important consideration in empirical finance is the issue of parameter stability. As firm policies or market conditions change over time (sometimes quite abruptly), so may the betas or alphas of those firms. In April 2000 the “new economy” bubble burst: the technology-heavy NASDAQ index started to collapse, pulling the rest of the stock market (and

the world economy!) down with it. It may therefore be interesting to distinguish the following two subperiods: i) 1992-1 until 2000-3 (the first 99 observations); ii) 2000-4 until 2008-4 (the remaining 97 observations). In the actual Case W5, we discussed two different ways to examine whether there were any parameter breaks between these two subperiods: I) Split the dataset into two subsets, corresponding to the two subperiods above; then estimate the basic CAPM-model (1) for both periods separately. II) Fit the following dummy-interaction model: ~ ( Rit − R0t ) = α i + α~i break t + β i ( R Mt − R0t ) + βi ( RMt − R0t ) ⋅ break t + ε it with breakt = 0 breakt = 1 for t = 1,,99 for t = 100,,196 (i.e 1992-1 until 2000-3) (i.e 2000-4 until 2008-4) The SPSS-results for model (2) are reported below and on the next page. 7 t = 1,,T (2) Source: http://www.doksinet e) (3pt) i) On the basis of the output for model (2), determine for both subperiods in turn: - the

estimated value of the CAPM-intercept alpha; - the estimated value of the CAPM-slope beta. ii) Sketch the implied CAPM-lines for both subperiods into a system of axes, like the one alongside. Akzo Market (Question f) Consider the following joint null hypothesis in terms of the coefficients of model (2): ~ H0: α~i = βi = 0 f) (6pt) i) Describe the meaning of this joint null hypothesis in words, as explicitly as you can. ii) Use the available information to perform a formal test of this joint null hypothesis at the 10% significance level. iii) Describe an approach which would allow us to test the joint null hypothesis above, which is about the coefficients of the dummy-interaction model (2), without actually estimating that model. (please turn to the next page) 8 Source: http://www.doksinet Question 2 (Case W7i): A Survey of Student Opinions in Maastricht, 2011-2012, part II (25pt) At the end of course period 4 in academic year 2011-2012, all first-year IB students at the SBE

were asked to fill in an extensive questionnaire. Below, you find a radically shortened version, reporting only question Q9, which addresses eight different motives which students may have for studying in Maastricht. Abstracting from missing values, our dataset contains 495 respondents. The first questions are about your reasons for studying in Maastricht and different goals you might have in life. Please rate your answers on a 1 to 5 scale, where 1 = very unimportant; 2 unimportant; 3 = somewhat important; 4 = important; 5 = very important I chose this university 1 Q9 1 Because of the study programs offered. Q9 2 Because it is close to my hometown. Q9 3 Because of the PBL educational system. Q9 4 Because of its good reputation. Q9 5 Because I know people here. Q9 6 Because of the city of Maastricht. Q9 7 Because of student life here. Q9 8 Because I like to study in an international environment. 2 3 4 5 We will use factor analysis to have a closer look at these

eight motives. Although we will skip the preliminary data analysis here, we do report the full correlation table below, for later use. On the next page, we provide the SPSS-output for the basic factor solution. Note: - We have included the “Scree Plot” and the “Component Plot”. - We have deviated from Kaiser’s criterion: using the “Extraction” button, we retained only two (rather than three) factors. Apart from this, the output reflects the SPSS default settings: in particular, our eight original variables have been standardized before running the factor analysis, and the factor solution has not been rotated. 9 Source: http://www.doksinet (Question a) As you know, each factor (or “component”) is a linear combination of the 8 original, standardized, variables. The “Total Variance Explained” table above shows that SPSS has determined 8 such factors in succession. a) (6pt) i) Use the “Initial Eigenvalues” part of this table to describe in words, how SPSS

determines these linear combinations, in turn. ii) Relate the eigenvalue for the first factor, 1.981, as explicitly as you can to the factor loadings that are reported in the “Component Matrix” table. iii) Motivate intuitively why SPSS distinguishes precisely 8 factors. 10 Source: http://www.doksinet (Question b) In the case at hand, the standard Kaiser criterion would suggest to retain three factors. As mentioned, we deviated from this and decided to retain only two factors. b) (3pt) On the basis of the available information, defend this decision as carefully as you can. (Question c) Let’s have a closer look at the “Communalities” table. You can easily verify that two motives are not captured to an acceptable degree by our two retained factors: Q9 2 “Close to hometown”, and Q9 3 “Education system”. c) (6pt) i) Relate the communalities of 0.079 and 0146 for motives Q9 2 and Q9 3 as explicitly as you can to the factor loadings as reported in the “Component

Matrix” table. ii) Use the correlation table on p. 9 of this exam to elaborate on the question, why precisely these two motives have such a low communality, compared to the other motives. iii) Imagine that we would insist that all eight motives have an adequate communality. How many factors do you expect we’d have to retain to achieve this purpose: three or four? Motivate your answer as carefully as you can. Hint: The correlation table may again be useful. (Questions d-e) Let’s subject our factor solution to a Direct Oblimin rotation, i.e one of the oblique rotations that are available under SPSS’s “Rotation” button. Of course, most of the resulting output is the same as before Below, we therefore only report the “Pattern Matrix” and the “Component Plot in Rotated Space”. (Question d) As you can see, both the previous “Component Plot” and the new “Component Plot in Rotated Space” contain two axes, which correspond with the two retained factors, before and

after the rotation. d) (6pt) i) Describe how the eight dots in the “Component Plot” and the “Component Plot in Rotated Space” are related to the numbers in the “Component Matrix” and the “Pattern Matrix”. ii) Use these plots to motivate as clearly as you can, what the purpose of rotation is. iii) On the basis of these plots, do you think the rotation has to be oblique here, or would an orthogonal rotation be adequate? 11 Source: http://www.doksinet (Question e) Using the “Scores” button, we have determined the “official” rotated factor scores: in the SPSS data window, they are referred to as FAC1 1 and FAC2 1. Moreover, using the menu Transform>Compute, we have defined the following two variables: F1 = ( Q9 1 + Q9 4 + Q9 8 ) / 3 F2 = ( Q9 5 + Q9 6 + Q9 7 ) / 3 The SPSS-tables below show, in turn, the Pearson correlation between FAC1 1 and F1, and between FAC2 1 and F2. e) (4pt) i) Try to provide an explicit verbal characterization of the two rotated

factors, FAC1 1 and FAC2 1. In other words: which real-world dimensions or constructs (if any) are represented by these two factors? ii) Motivate the definition of the variables F1 and F2, and explain the order of magnitude of their correlation with FAC1 1 and FAC2 1. In case you finish the exam before the official ending time, please leave the examination hall as quietly as possible, so you will not disturb the students who are still trying to concentrate on their work! 12 Source: http://www.doksinet Answers mathematics part Question 1) a) i) The expected payoffs are: 1 2 1 1 µ1 = 3 ⋅ a + 3 ⋅ 6 = 3 a + 4 ii) The variances are: σ 12 = 1 3 {a − ( 1 3 = 3 (9 a2 − 1 4 } + {6 − ( a + 4) 16 3 2 2 µ2 = 3 ⋅ 9 + 3 ⋅ 6 = 3 + 4 = 7 2 3 1 3 } a + 4) 2 1 2 = 3 ( 3 a − 4) 2 + 23 (2 − 13 a) 2 a + 16) + 3 (4 − 3 a + 9 a 2 ) = 2 4 1 σ 22 = 13 (9 − 7) 2 + 23 (6 − 7) 2 = 3 ⋅ 4 + 1 2 6 ⋅1 = 3 3 6 27 a2 − 24 9 a+ 24 3 = 9 a2 − 3 a

+8 2 8 =2 b) The return on asset A1 is always 6, so this is effectively not a random variable. Consequently, its correlation with the return on asset A2 must be zero. Alternatively, moving from state 1 to state 2 the return on asset A2 falls, while the return on asset A1 stays the same. Hence: zero correlation. c) i) Plugging a = 6 into eq. (1), we have µ1 = 6 (of course!) and µ2 = 7 Hence: µ ( P) = µ1 λ + µ 2 (1 − λ) = 6 λ + 7(1 − λ) = 7 − λ ii) Plugging a = 6 into eq. (2), we have σ 12 = 0 (of course!) and σ 22 = 2 Hence: σ 2 ( P) = σ 12 λ2 + σ 22 (1 − λ) 2 + 2σ 1σ 2 λ(1 − λ) ρ = 0 + 2(1 − λ) 2 + 0 = 2(1 − λ) 2 d) Substituting eqs. (5) and (6) into (4), the investor wants to maximize: 1 1 U ( P) = µ ( P) − 2 σ 2 ( P) = 7 − λ − 2 ⋅ 2(1 − λ) 2 = − λ2 + λ + 6 The maximum will occur at the stationary point where the derivative w.rt λ equals zero: U = −2 λ + 1 = 0 So U attains its maximum at λ = 12 , where the optimal

portfolio consists of 50% A1 and 50% A2. e) When λ = 0, we get µ(P) = 7 and σ(P) = √2. When λ = 1, we get µ(P) = 6 and σ(P) = 0. The investor’s options are limited by the straight line which connects these points in (σ,µ)-plane. In that plane, the investor’s utility can be expressed by indifference curves which are U-shaped parabolas with their minimum at the vertical axis. Apparently, the highest possible indifference curve just touches the line when λ = 0.5, so at µ(P) = 6.5 and σ(P) = 05√2 See the graph alongside μ 7 6.5 6 f) i) Plugging a = 3 into eqs. (1) and (2), we have µ1 = 5, µ2 = 7, σ 12 = 2 and σ 22 = 2 . So asset A1 has the same variance compared to asset A2, but a lower expected payoff. ii) Moving from state 1 to state 2 the return on asset A2 falls while the return on asset A1 rises; hence a perfect negative correlation of –1. 0.5√2 σ √2 g) i) Using the results for fi), we get: µ ( P) = µ1 λ + µ 2 (1 − λ) = 5 λ + 7(1 − λ) =

7 − 2 λ σ 2 ( P ) = σ 12 λ2 + σ 22 (1 − λ) 2 + 2σ 1 σ 2 λ(1 − λ) ρ = = 2 λ2 + 2(1 − λ) 2 + 2 ⋅ 2 λ(1 − λ)(−1) = 2 λ2 + 2 − 4 λ + 2 λ2 − 4 λ + 4 λ2 = 8 λ2 − 8 λ + 2 Plugging this into eq. (4), the investor wants to maximize . 1 U ( P) = µ ( P ) − 12 σ 2 ( P) = 7 − 2 λ − 2 (8 λ2 − 8 λ + 2) = −4 λ2 + 2 λ + 6 The maximum will occur at the stationary point where the derivative w.rt λ equals zero: U = −8 λ + 2 = 0 which solves for λ = 14 . ii) As the assets are negatively correlated, it is beneficial to diversify even though one of the assets is inferior, because this will reduce the overall variance of the portfolio at a modest cost in terms of the expected return. 13 Source: http://www.doksinet Question 2) a) - Imagine that a = 0, 1, 2, 3 or 4. In this case, no matter how you vote, you won’t reach the 6 “ayes” needed to get a positive payment. Since voting “aye” would cost you € 150, your unique best

response is to vote “nay” - Imagine that a = 6, 7, 8 or 9. In this case, no matter how you vote, the 6 “ayes” required for a positive payment have already been achieved. If you vote “aye” too, your payment increases by € 1, but you have to pay the € 1.50, so you’re worse off by € 050 To illustrate: if a = 7, you get € 8 – € 150 = € 650 by voting “aye”, but € 7 by voting “nay”. Clearly, your unique best response is again to vote “nay” - Imagine that a = 5. If you vote “nay”, your net payoff is simply € 0 But if you vote “aye”, your pivotal vote implies that the hurdle of 6 students is reached, so you will receive € 6. Factoring in the € 150 cost of voting “aye”, your net payoff is € 4.50 So in this case, your unique best response is to vote “aye” Clearly, your best response is not always the same: typically it’s “nay”, but in the pivotal scenario it’s “aye”. Consequently, you (or any other student) don’t

have a dominant strategy. b) A strategy profile is a Nash equilibrium when the students are playing best response to each other. So let’s evaluate systematically whether this is the case for the eleven distinct profiles. - Imagine that k = 0; all students vote “nay”, no payments are made, and no one pays the € 1.50 fee Clearly, each “nay”-voter is playing best response here: given the “nay”-votes of the other 9 students, voting “aye” would make him worse off, because he would have to pay the € 1.50 fee, for no gain - Imagine that k = 1, 2, 3 or 4, so that no payments are made, but the “aye”-voters do have to pay the € 1.50 fee Each “nay”-voter is again playing his best response here, for the same reason as for the previous scenario. But an “aye”-voter is not playing best response: by voting “nay”, he could avoid the € 1.50 fee - Imagine that k = 5; again no payments are made, but the “aye”-voters do have to pay the € 1.50 fee Any

“aye”-voter is now not playing best response, for exactly same reason as in the previous scenario. Moreover, a “nay”-voter is also not playing best response: after all, by voting the pivotal “aye”, he would receive € 6, implying a net payoff of € 4.50 which is better than the € 0 he gets now - Imagine that k = 6; all “nay” students receive € 6 now, while the “aye”-voters pocket only € 4.50 Any “aye”-voter is now playing best response; after all, given the votes of the others, him voting “nay” instead would reduce his net payoff from € 4.50 down to € 0 Any “nay”-voter is also playing best response: voting “aye” instead would reduce his net payoff by € 0.50 - Imagine that k = 7, 8, 9 or 10. You can easily verify that a “nay”-voter is now playing best response, but an “aye”-voter isn’t: given the votes of the others, he could increase his net payoff by € 0.50 by voting “nay” Conclusion: the strategy profiles with k = 0

and k = 6 constitute the Nash equilibria. c) Let k denote the total number of “ayes”. It will be clear immediately that the Pareto optimal strategy profile must involve k ≥ 6, because only such profiles yield a positive total net payoff. In particular, with k “ayes”, each student will receive € k, for a total payment of € 10k. With each “aye” paying € 150 for a total fee of € 15k, the net total payoff is simply € 8.5k This amount is maximized by choosing k as large as possible, ie for k = 10 In words: the Pareto optimal strategy profile involves all students voting “aye”. d) You should easily be able to convince yourself, and the corrector, that the same considerations apply as for item a). So if a = 5, your vote is pivotal and your (unique) best response is to vote “aye”; for any other value of a, your (unique) best response is “nay”. e) Consider the simplified version of the game as proposed in the hint. Paraphrasing item d), student 3 will vote

“aye” if and only if there is a single previous “aye”, so that mr. 3 is the pivotal voter Student 2 understands and anticipates this. So he will vote “aye” only if there is no previous “aye” (ie by student 1). If there was a previous “aye”, he’d vote “nay” and be confident that mr 3 would provide the second “aye”, so that mr. 2 could earn the €2 payment without the €150 fee Student 1 understands and anticipates all of this. In particular, he can be confident that if he votes “nay”, both students 2 and 3 will vote “aye”, so that he will get the €2 payment without paying the €1.50 fee Altogether, student 1 will vote “nay”, students 2 and 3 will vote “aye”. By extension: in the original version of our game, the first 4 students will all vote “nay”, and the last 6 students will all vote “aye”. 14 Source: http://www.doksinet Answers statistics part Question 1) a) i) In period 88 i.e April 1999, the return on Akzo was 2106%

higher than predicted (by model 1) on the basis of the market return. ii) The criterion for an influential observation is that its in- or exclusion will make an appreciable difference to the estimated slope coefficient of the model. An outlier doesn’t have to be influential: in particular, if its value for the predictor is close to the average, then its in- or exclusion would mainly shift the line, i.e change the intercept, but not tilt it. But in the case at hand, our outlier lies more or less in the North-East of our scatterplot In particular, the value for the predictor Market is clearly above average. This makes it likely that this observation will tilt the fitted line a bit counter-clockwise. Alternatively: visualizing what would happen without the outlier, we should expect a somewhat flatter line. b) i) First-order autocorrelation means that the error of our model (1) is linearly related to the previous error, either positively or negatively. ii) The residual regression is

run on the residuals of model (1); after all, the question is whether model (1) displays first-order autocorrelation. So we need the Durbin-Watson statistic for model (1), ie d = 2110, not the Durbin-Watson statistic of our residual regression (which should have been unchecked under the “Statistics” option). iii) The null hypothesis implies that there is no linear relation between the successive errors of model (1): this corresponds with a zero slope in our residual regression. Since theory implies the absence of autocorrelation, the appropriate alternative is two-sided. This test is performed by default by SPSS; we only need to check the p-value of the slope coefficient of our residual regression. Since it is as high as 0395, we cannot reject the null at 10%. c) A test at the 10% significance level implies that there is 10% chance of making a Type I error. That means: rejecting the null when it is actually true, just as a result of sampling chance, “by accident” so to say.

Now imagine that the CAPM is correct and that our null hypothesis is true across the board, i.e alpha is zero for all 16 stocks If so, we would expect 10% out of 16 i.e 16 of such “false alarms”, say 1 or 2 So two significant alphas out of 16 is perfectly normal if indeed all true alphas are zero; this does not constitute real evidence against the CAPM. d) The null cannot be tested directly by looking at the output. We’ll have to calculate the relevant test statistic by hand: 1.093 − 1 t= = 1.39 0.067 With 194 ≈ 180 df, this is in between the 5% and 10% critical t-values of 1.653 and 1286 Consequently, the twosided p-value is somewhere between 10% and 20% e) i) In the first subperiod, we have Breakt = 0, so the model reduces to: ( Rit − R0t ) = αi + βi ( RMt − R0t ) + εit Consequently, in the first subperiod, the estimated intercept (alpha) is simply –0.705 and the estimated slope (beta) is 1234 In the second subperiod, we have Breakt = 1, hence: ~ ( Rit − R0t ) =

αi + α~i + ( βi + βi )( RMt − R0t ) + εit Consequently, in the second subperiod, the estimated intercept (alpha) is –0.705 + 1383 = 0678, and the estimated slope (beta) is 1.234 – 0227 = 1007 ii) Collecting all this information, we get the sketch alongside. Akzo-Nobel 1.234 1.007 0.678 6.093 After Market -0.705 Before f) i) “No parameter break”, “same line in both periods”, “no alpha- and no beta-break”, etc. ii) A joint hypothesis like this requires a partial F-test. Model (2) is the Complete Model with SSEC = 5389927 and k = 3; model (1) is the Reduced Model with SSER = 5537.414 and g = 1 Hence: (5537.414 − 5389927) /(3 − 1) F= = 2.63 5389.927 /(196 − 3 − 1) With 2 numerator-df and 192 ≈ 180 denominator-df, the 10% critical F-value is 2.33 With 263 > 233, we can reject the null at the 10% significance level. iii) The only feature of the dummy-interaction model which we need for the test is its SSE; but this number can be found by adding

the separate SSE’s for the subperiods, without actually estimating model (2), along the lines of proposal I) in the basis for questions e-f). In fact, this is how the “real” Chow test works 15 Source: http://www.doksinet Question 2) a) i) The key thing is that these linear combinations are defined in such a way that their variance (“eigenvalue”) is maximized, so that they represent the largest possible fraction of the information that is contained in our dataset. The first factor (with the factor loadings reported in the first column of the “Component Matrix”) is defined precisely in this way. It has a variance of 1981, ie almost 25% of all the variance The second factor is again a linear combination of Q9 1 to Q9 8 with the largest possible variance, but subject to the restriction that it must be uncorrelated with (“orthogonal to”) the first factor. Logically, it has a lower variance than this first factor. Similarly, the third factor is the maximum-variance

linear combination that is uncorrelated with both the first two factors, and it necessarily has a lower eigenvalue than the second factor. Etcetera ii) It’s the sum of the squared factor loadings for the first factor or component, i.e 06842 + + 06712 = 1981 iii) The process stops once all the information contained in the eight original variables, i.e 100% of the variance of their standardized versions, has been captured. Logically, this takes 8 factors: eg with only 7 linear combinations, it would not be possible to fully replicate the original 8 variables and capture all their variance. b) The eigenvalue for factor 3, i.e 1129, is only marginally above 1; this may reflect sampling chance Moreover, the scree plot indicates a clear discontinuity after two factors; there is a sharp drop in eigenvalue from factor 2 to 3. Both these considerations may lead us to settle for 2 rather than 3 retained factors. On the downside, if you retain only 2 factors, then only 46.760% of the total

variation in the dataset will be captured by them c) i) It’s the sum of the squared factor loadings for either motive, e.g (–0276)2 + (–0046)2 = 0079 ii) Most motives have a reasonably high correlation, typically 0.4 or higher, with at least one other motive This is what allows them to be grouped together into a factor. But for motives Q9 2 and Q9 3 (and to a lesser extent also Q9 5), this is not the case: their correlations with the other motives are very low. iii) If you retain more factors, you’d expect a generic increase in the communalities, but disproportionately so for the two motives which are currently underrepresented. The issue is: does it take one or two extra retained factors to achieve a substantial increase in these two communalities? Now the key thing to see is that motives Q9 2 and Q9 3 do not only have low correlations with the remaining six motives; their mutual correlation of 0.029 is also extremely low. So if you retain a third factor, it is very unlikely

that it will be able to capture these two motives in one swoop. On the basis of the available information, it seems much more likely that it will take two extra factors to achieve the desired effect, where one of them will have a relatively big loading on Q9 2, and the other a relatively big loading on Q9 3. d) i) In the “Component Plot”, each dot represents one of the eight motives, and the coordinates along the “Component 1” and “Component 2” axes are equal to the factor loadings that are reported in the “Component Matrix” table. Eg the first motive, Q9 1, has coordinates (0684, 0308) There is an equivalent relation between the “Component Plot in Rotated Space” and the “Pattern Matrix”. ii) The ultimate goal is to end up with factors which have a clear interpretation. So ideally, we would like to associate each original variable more or less uniquely with a single factor. Visually, this would mean that the dots representing the original variables would bunch

up along the ends of the axes that correspond with the factors. But in the original solution presented by the “Component Plot”, this is clearly not the case: for several goals, it holds that both factors have rather high loadings. The idea is to rotate the axes of the plot in such a way, that the ideal picture is more closely approximated. The “Component Plot in Rotated Space” shows that this has been achieved to some degree: motives 1, 4, 6, 7 and 8 “bunch up” at the end of an axis, while motive 5 clearly has the largest loading on one axis. iii) Comparing the rotated with the original plot, it seems that the entire system of axes has been rotated counterclockwise by something like 30 degrees. In other words: our oblique rotation, which in principle allows for nonorthogonality, actually looks very orthogonal This would suggest that an orthogonal rotation like Varimax would have been appropriate here. N.B: This conclusion is supported by the fact that the correlation

between the rotated factors as reported in the “Component Correlation Matrix” (which is deliberately not printed here) is as small as –0.025 e) i) Using the standard rules described in the New Reading, the “Pattern Matrix” tells us that: 1) the first factor mainly consists of motives 1, 4 and 8; 2) the second factor mainly consists of motives 5, 6 and 7. So the contrast seems to be something like “university” vs. “city”, “study” vs “fun”; in fact, any labels are acceptable, as long as they intuitively match the underlying three motives for both factors. ii) The definitions of F1 and F2 are the “extremistic” versions of the allocations discussed at item i): 1) The first factor mainly consists of motives 1, 4 and 8, but with different weights. F1 only contains motives 1, 4 and 8, with identical weights. Consequently, the behavior of F1 will be very similar, but not totally identical, to that of FAC1 1; hence the very high correlation. 2) Same idea for F2

versus F2 1. 16