Problem set 1 - Solutions

Transcription

1 EMPIRICAL FINANCE AND FINANCIAL ECONOMETRICS - MODULE (8448) Problem set 1 - Solutions Exercise 1 -Solutions 1. The correct answer is (a). In fact, the process generating daily prices is usually assumed to be a Random Walk.. The autocovariance function for a generic AR(1) up to lag k can be easily derived as (there is no intercept since we consider de-meaned time series): 1P cov(y t ; y t k ) = E [y t y t k ] = E i e t 1P i i e t k i = k P 1 i e i = k e 1 = k y Where y is the unconditional variance of y and e is the variance of residuals (see also exercise ).The autocorrelation function is: corr(y t ; y t k ) = cov[yt;y t k] = k y = k 3. The equation for a generic AR(1) model is (there is no intercept since we consider de-meaned time series): y t = y t By an easy substitution: y t = (y t + e t 1 ) + e t Thus: y t = y t + e t If you iterate the substitution you get to: y t = n y t n + P :n 1 i e t i For a stationary process (see also exercise 5) < 1. Thus for n! 1: 1

2 P y t = 1 i e t i Which is the in nite MA representation. Exercise - Solutions 1. You can identify three steps in the Box-Jenkis approach: PRE WHITENING: make sure that the time series is stationary. If it is not, transform it in order to make it stationary (usually taking rst di erences). MODEL SELECTION: look for the best ARMA speci cation; information criteria are a useful toll in this part. MODEL CHECKING: residual tests. Make sure that residuals are not autocorrelated and check whether their distribution is normal.. Consider the AR(1) stationary process: y t = y t With residuals volatility equal to. The unconditional mean coincides with the long run expected value of the time series, which is: For n! 1, since 1 < 1: np E t [y t+n ] = 0 i 1 + i 1y t i i=i E [y] = The unconditional variance coincides with the long run expected value of the time series, which is: V t [y t+n ] = n P (i i=i 1) e For n! 1, since 1 < 1: y = e 1 1

3 3. The answer is no and we can give you an intuitive justi cation. The least square problem for equation: Y = X + e Is the solution to the linear system: X 0 X = XY This solution coincides graphically with the vector of coe cients which minimizes the sum of squared errors. The AR(1) model: y t+1 = y t + e t+1 Is compatible with the OLS assumptions, since there is no dependency in conditonal distributions. In fact: y t+1 v f(y t j e; 0 ; 1 ) And y t:t+n v Q :n 1 f(y t+i j e; 0 ; 1 ) It can be show n that the problem of tting this distribution with the most appropriate values for e; 0 ; 1 (which is the Maximum Likelihood Estimation problem) coincides with the OLS problem. However, an MA(1) model entails some form of dependency; as a matter of fact: And then y t+1 v f(e t j ) = f(y t jy t 1 ; e; 0 ; 1 ) y t:t+n v f(y t+n jy t+n 1; e; 0 ; 1 )f(y t+n 1 jy t+n 1; e; 0 ; 1 ):::f(y t+1 jy t; e; 0 ; 1 ) This more complicated form cannot be solved as an OLS problem and requires Maximum Likelihood Estimation. 3

4 Exercise 3 - Solutions In graphical representations of ACF, we can say in general that AR process show autocorrelations that decline smoothly along with lags, while in MA models autocorrelations di erent from zero are observe only in correspondence of lagged terms signi cantly di erent from zero. Time Series 1: this is clearly an AR process, and probably an AR(1) process with a root positive but rather small in absolute value Time Series : well...there is no autocorrelation in this process; we are observing the ACF of a white noise Time Series 3: only the rst two lags show an autocorrelation di erent from zero, then the ACF falls to zero. This is a MA() model. Time Series 4: Thus is again an AR process, but this time we are considering a very persistent process. For instance, it might be an AR(1) with a root very close to one. Question...how would you expect the ACF of an AR(1) process with a negative root to look like? Exercise 4 - Solutions Conside the ARMA(1,1) model: 4

5 With y t 1 = 3:4, u t 1 = 1:3. y t = 0: :69y t 1 + 0:4u t 1 + u t 1. Forecasts can be easily derived as: Remember that: ^y t = 0: :69(3:4) + 0:4( 1:3) = 1:836 ^y t+1 = 0: :69( 0:03) + 0:4( 1:804) = 0:7438 MSE = P t=1:n e t n Another very useful metrix which is used in model testing is the mean absolute percentage error: MAP E = P t=1:n j et y t j 1 n In order to solve the exercise you just need: MSE = ( 0:03 1:836) +(0:961+0:7438) = 3: The exponential smoothing model will have the form: Thus: And: ^y t = ^y t 1 + (y t 1 ^y t 1 ) ^y t = 3 + 0:15(3:4 3) = 3:06 ^y t+1 = 3:06 + 0:15( 0:03 3:06) = :596 MSE = ( 0:03 3:06) +(0:961 :596) = 6: The model with the best t to data is the ARMA; we can say that beacause it has a smaller MSE than the exponential smoothing model. 4. Exponential smoothing gives you a point forecast for the underlying stochastic process, but it is NOT a probabilistic model. This means that ir makes no assumption on the properties of errors; with ES it is not possible to de ne con dence intervals (this is why ES is also called a NON PARAMETRIC method). On the other hand, ARMA models entail distributional assumptions and are probabilistic models. 5

6 Exercise 5 - Solutions 1. A process with a deterministic trend is: y t = t + e t While a process with a stochastic trend is: y t = 0 + y t This process is actually a Random Walk.. Tstats of a regression between integrated series are upward biased; this happens since any couple of trends is by de nition highly correlated. However, this apparently high signi cance might not be supported by any actual relationships between the data generating processes. Exercise 6 - Solutions 1. Consider the demeaned series y 1:t ; t an AR(1) model such that: Now rewrite it as: y t = y t y t = y t Where = ( 1). Testing for the null = 1 is the same as testing for = 0. Dickey-Fuller test this null using the statistic: DF = ^ S:E:(^) This statistic follows a non-standard distribution, whose quantiles have been derived trough simulation exercises. See Section of Brooks for more details.. The power of a statistical test is the probability of rejecting the null hypothesis when the alternative is true (i.e. the probability NOT to make a II type error). On the other hand, a low power test has an high probability of making a type II error, that is of failing to reject the null hypothesis when the alternative is true. For what concern Dickey-Fuller, this means that you might accept the existence of a unit root even if the process is stationary. The power of test decreases the closer the root of the process is to unity. This means that while for processes with low persistence (ex. = 0:7) or highly explosive (ex. = 1:3) DF can be useful for highly persistent but stationary processes (ex. = 0:96) the test turns out to be quite unreliable. 3. You should implement exactly the same scheme of Dickey-Fuller, but on di erenced time series. Thus you should consider: 6

7 y t = y t And rewrite it as: y t = y t Then the test will be exactly identical to the standard DF. Exercise 7 - Solutions 1. Cointegration is a way to model persistent processes together. Unit root processes can not be t with a meaningful probabilistic model, and it is impossible to forecast their behavior since their variances have an explosive behavior. However, it is sometimes possible to identify a long run error correction model linking two series; in this way it is possible to study the behavior of their di erences and to build forecasting frameworks. Take the example of log prices and log dividends. These series contain a unit root, and it is not feasible to t a reliable univariate model for them. However, a wide stream of literature focuses on the properties of residuals obtained as a di erence between the two series (also called price dividend ratio); this residuals are useful in order to predict di erences in prices, or, in other words, returns.. Remember that if you two series such that the rst one is I(p) and the second one is I(q), then the residuals of the cointegrating vector is I(p-q). In this speci ca case since both the series are I(1), the cointegrating vector will be I(0). The residuals will therefore be stationary. For a formal de nition of cointegration see Section Brooks. 3. Consider the two integrated series y 1:t and x 1:t. The rst step of the Engle-Granger method consists in running the OLS regression: y t = x t + e t You then collect the residuals (note that in this regression all the stats are biased by the presence of unit roots). Then run a second regression, with the form of an error correction model: y t = x t + (y t x t ) + u t Which is a representation of the cointegration relationship. 7