Question 1

Consider the regression of Morgan Stanley daily stock prices, MSt , on the daily SP500 index, SPt :

MSt  = β1 + β2 SPt + ut  = β\ xt + ut , t=1,...,T,              (1) where xt  is a 2x1 vector with elements 1 and SPt . Assume that E[xtut] = 0.

1. (6 points) Suppose you obtain OLS estimates βˆ1   = −46 and βˆ2   = .2. with HAC standard errors respectively equal to 2.18 and 0.01. Discuss how/whether you would use this regression output to test the hypothesis H0  : β2  = 0, and mention any other analysis you would carry out to support your decision.

A: The standard errors may not be valid if the regression is spurious (which is likely, as we are considering stock prices). You can run a unit root test on the residuals. If reject, can use output to do a t-test=  > 1.96 so reject.

2. (6 points) Answer question 1.1 again, in light of the fact that the time series of estimated residuals from regression (1), uˆt , looks as follows:

A: The errors look non-stationary (the are smooth and have a trend), suggesting we are in the presence of a spurious regression and so cannot use output. Again, you can run a unit root test on the residuals to confirm

3. (6 points) Explain how you would test whether TetuTns on the SP500 index Granger-cause Morgan Stanley stock TetuTns.

A: Compute returns on both SP500 and MS by taking log price di↵erences from the previous day, then regress MS returns on lags of MS returns and lags of the SP500 returns. Test if lags of SP500 are jointly significant. If they are, conclude there is Granger causality.

4.  (7 points) Discuss whether you think an Error Correction Model would be the right model for the vector (MSt ,SPt )\ , based on your answer to question 1.2.

A: An ECM is the right model only if MSt  and SPt  are cointegrated.  If in question 1.2 we found a unit root (as is likely), the two are not cointegrated so the ECM is not the right model.  If we reject a unit root, this is evidence of cointegration so an ECM is appropriate

Question 2

Consider the model

yt  = −c1 ⇤ 1(t < T/2) + c2 ⇤ 1(t > T/2) + "t + ✓1"t−1 + ✓2"t−2, t = 1, ...,T,   (2)

where "t  ⇠ i.i.d.N(0, 1) and 1( · ) equals 1 when the statement within parentheses is true and equals zero otherwise.

1.  (6 points) Under which conditions on the parameters is the model stationary? Find the (unconditional) mean and variance of yt  under these conditions.

A: When c1  = c2  = c, in which case we have an MA(2) with mean c and variance 1+✓1(2)+✓2(2)

2.  (6 points) Under which conditions is the model invertible?

A: Roots of equation 1 + ✓1 x + ✓2 x2  = 0 greater than 1 in absolute value

3.  (6 points) Given a quadratic loss function, what is the optimal 1-step ahead forecast at time T based on model (2)?

A: 1-step ahead forecast is c2 + ✓1"T + ✓2"T−1

4.  (7 points) What are the parameters of model (2) and how would you estimate them?

A: Parameters are c1 ,c2 , ✓ 1 , ✓2 . You can estimate them by MLE

Question 3

Consider the model

yt    = σt尸{zzt}, t = 1, ...,T,                         (3)

"t

σt(2)  = w + ↵1"t(2)−1 + ↵2"t(2)−2                                                                  (4)

where zt ⇠ i.i.d.N(0, 1).

1.  (5 points) Derive the unconditional mean and unconditional variance of yt . A: E[yt] = 0 and Var[yt] =

2.  (6 points) Explain how you would estimate the parameters by OLS (i.e., by running a regression).

A: This is an ARCH(2), which is an AR(2) for "t(2) . "t(2) can be approximated here (due to the lack of a conditional mean) with yt(2), so estimate by regressing yt(2) on yt(2)−1  and yt(2)−2

3.  (7 points) Suppose that, after running the regression in question 3.2, you find that the residuals of the regression are autocorrelated. Explain how you would modify the model for yt and how you would estimate the new model.

A: You need to increase the order of the ARCH (e.g., start with ARCH(3) and increase progressively until you don’t find any residual autocorrelation).  You can still estimate these models by OLS.

4.  (7 points) Explain how you would construct an estimate of the standardized residual, zˆt , using results from the regression in question 3.2.

A: zˆt =  where σˆt = ^ + ↵ˆ1yt(2)−1 + ↵ˆ2yt(2)−2