Financial Econometrics Jeffrey R. Russell Midterm 2014 Suggested Solutions TA: B. B. Deng
Unless otherwise stated, e t is iid N(0,s 2 ) 1. (12 points) Consider the three series y1, y2, y3, and y4. Match each of the series up with the models the model that appears most appropriate below. r = β + β r + θε + ε. Let t 0 1 t 1 1 t 1 t a. b 0 =1, b 1 =.9, q 1 =.9, and s=5 3 b. b 0 =1, b 1 =.9, q 1 =-.9, and s=1 2 c. b 0 =1, b 1 =0, q 1 =.9, and s=1 4 d. b 0 =1, b 1 =.9, q 1 =0, and s=1 1 2
2 2. (28 points) Consider the AR(1) model y t = β 0 + β 1 y t 1 + ε t where t ~ iid N ( 0, ) a. For what values of b 0, b 1 and s is the mean reverting? ε σ. β! < 1, σ! finite Now let b 0 =.5, b 1 =.9 and s=2. b. What is the unconditional mean? β! 1 β! = 5 c. What is the unconditional variance? σ! 1 β!! = 21.05 d. Let y T =4.1. What is the conditional mean E( y y )? T+ 1 T 0.5 + 0.9 4.1 = 4.19 e. Let y T =4.1 and e T =- 1.2. What is the conditional variance Var ( y y )? T+ 1 T σ! = 2! = 4 f. What is the conditional distribution f ( y y ) T+ 1 T N 4.19, 4 3
g. Let y T =4.1 and e T =- 1.2. Find E( y y ε ) t+ t t 2, E y!!! y! =. 9! (4.1) + 1. 9! 5 = 4.27 h. Let y T =4.1 and e T =- 1.2. Find E( y y ε ) t+ t t 3, E y!!! y! =. 9! (4.1) + 1. 9! 5 = 4.34 4
3. (28 points) Here are continuously compounded returns for the DJIA from 1990 to 1/26/2012. 800 700 600 500 400 300 200 100 0-0.075-0.050-0.025 0.000 0.025 0.050 0.075 0.100 Series: RET Sample 1/03/2000 1/25/2012 Observations 3034 Mean 3.83e-05 Median 0.000449 Maximum 0.105083 Minimum -0.082005 Std. Dev. 0.012918 Skewness -0.046813 Kurtosis 10.07672 Jarque-Bera 6332.055 Probability 0.000000 For parts a. and b. of this problem, assume that the natural log of the DJIA prices follows a random walk model. The index level (not log level) at the close on 1/26/2012 was 12,756 a. Symmetry Symmetric / slightly negatively skewed b. Fatness of tails relative to Normal / fatness of tails holding variance constant Big kurtosis => fat tails c. The drift can be estimated from the sample mean and the standard deviation of the error can be estimated by the standard deviation in the above output. Find the k- step ahead forecast of the log DJIA index level as a function of k and the initial log index. Use sample estimates where they appear in the forecast equation. Random Walk Model: y! = β! + y!!! + ε! β! = 0.0000383 So, y!! = k 0.0000383 + ln 12,756 y!! = k 0.0000383 + 9.453757 5
d. Find the k- step ahead forecast error variance associated with part a. as a function of k. Use sample estimates where they appear in the forecast error variance expression. y! = β! + y!!! + ε! ε! N 0,0.012918! So, Var y!! = k 0.012918! e. Consider the following output for the DJIA continuously compounded returns. Are then predictable? Explain with the use of supporting evidence. Yes, the returns are predictable since the autocorrelations are significantly different from 0. We can tell this from the large Q- stat values and the small associated p- values. We can even see this visually since (at least) the first two autocorrelation bars are outside the dashed significance lines. Next, consider the following models fit for the DJIA returns data. Two MA models are estimated below for the DJIA returns data. The first is a pre- financial crisis data set prior to 2007. The second is a data set that primarily focuses on the period of the financial crisis, from January 2007 to present. 6
Dependent Variable: RET Method: Least Squares Date: 02/08/10 Time: 22:16 Sample: 1/01/1990 1/26/2007 Included observations: 4436 Convergence achieved after 6 iterations MA Backcast: 12/27/1989 12/29/1989 Variable Coefficient Std. Error t- Statistic Prob. C 0.000353 0.000141 2.505348 0.0123 MA(1) 0.001926 0.015028 0.128172 0.8980 MA(2) - 0.024203 0.015032-1.610151 0.1074 MA(3) - 0.014765 0.015037-0.981913 0.3262 R- squared :.000766 Dependent Variable: RET Method: Least Squares Date: 02/08/10 Time: 22:15 Sample (adjusted): 1/01/2007 6/03/2009 Included observations: 633 after adjustments Convergence achieved after 7 iterations MA Backcast: 12/27/2006 12/29/2006 Variable Coefficient Std. Error t- Statistic Prob. C - 0.000457 0.000608-0.751799 0.4525 MA(1) - 0.143672 0.039738-3.615450 0.0003 MA(2) - 0.106021 0.039924-2.655579 0.0081 MA(3) 0.090011 0.039781 2.262692 0.0240 R- squared :.041178 f. For the crisis sample, find the forecast of the return E( r ε, ε, ε,...) as a t+ k t t 1 t 2 function of εt, εt 1, εt 2,..., and the estimated parameters above. You should have forecasts for all k>0. E r!!! F! = c + θ! ε! + θ! ε!!! + θ! ε!!! E r!!! F! = c + θ! ε! + θ! ε!!! E r!!! F! = c + θ! ε! E r!!! F! = c for k > 3 7
g. Explain in plain English what the difference is between the pre- crisis and the crisis return dynamics. In the pre- crisis sample, the MA coefficients were NOT statistically significant, so we can infer LITTLE dependence relation using these. Hence we could not use previous errors to predict future returns. In the crisis sample, we do see significant MA coefficients. The values of these MA coefficients imply that we might see returns overshoot on surprises and then oscillate to a stable value. Notice the signs of the MA coefficients in the crisis sample suggest reversal behavior. 4. (20 points) Below is the estimated GARCH model for the DJIA return series in problem 3. Variance Equation C 1.06E- 06 1.37E- 07 7.757034 0.0000 RESID(- 1)^2 0.070310 0.004392 16.01029 0.0000 GARCH(- 1) 0.921003 0.005118 179.9711 0.0000 a. The last return in the sample is -.0104 and the value of h T associated with the last observation in the sample is.000125. Write the one- step ahead forecast of the variance. h!!! = 0.00000106 + 0.070310 0.0104! + 0.921003 0.000125 = 0.00012379 b. Find the expected variance for the cumulative return over the next 5 days. Report the annualized volatility. where So!!!! σ! = E h!!! F! = σ! + α + β!!! h!!! σ! ω 1 α β = 0.00000106 1 0.070310 0.921003 = 0.00012202 σ! + α + β!!! h!!! σ! 8
! = 0.00012202 + 0.991313!!! 0.00012379 0.00012202!!! = 0.00061880 5 day cumulative variance So, annualized volatility is: 0.00061880 252 5 = 0.1766 c. What number does the forecast for the variance of the cumulative return get close to (in an annualized basis) when the number of days gets large? Cumulative Variance: 252 σ! = 252 0.00012202 = 0.0307 Cumulative Standard Deviation: 0.0307 = 0.1754 Now consider the following TARCH (asymmetric GARCH model). d. What is the one- step ahead conditional variance if the last return in the sample is -.0104 and the value of h T associated with the last observation in the sample is.000125. h!!! = 1.44 10!! +.013761 0.0104! +. 160407.0104! + (.922455). 000125 9
Now consider the following output obtained from the symmetric GARCH(1,1) model. e. What do the autocorrelations tell you about whether or not the GARCH(1,1) model appears to fit the volatility dynamics well? Be specific. Yes, the model does seem to fit reasonably well. From eyeballing the autocorrelation bars, we don t see much significance for the first few lags. Also, the autocorrelations only start to meet a non- negligible p- value after the 4 th lag. 5. (12 points) Let r t denote the continuously compounded returns on an asset and let d t denote the price dividend ratio of the asset. You use the price dividends to forecast future returns over different time horizons. Suppose that rt = β1dt 1+ ε and t dt = φdt 1 + η where e t t and h t are iid, mean zero and independent of each other. Both r t and d t are mean reverting and stationary. 10
a. Find E( r + r d ) t+ 1 t t 1 E r!!! + r! d!!! = β! d!!! φ + 1 b. Find E( r + r + r d ) t+ 2 t+ 1 t t 1 E r!!! + r!!! + r! d!!! = β! d!!! φ! + φ + 1 c. What is E( r +... + r + r d ) t+ k t+ 1 t t 1! β! d!!! φ!!!! = β! d!!! 1 φ!!! 1 φ d. What does the forecast converge to as k gets large? β! d!!! 1 1 φ 11