MGMT MFE 407: Empirical Methods in Finance Homework 2: Solution
Prof. . Lochstoer TA:
January 20, 2022
Problem 1: AR(1) Process Consider an AR(1) process for rt+1 with φ0 = 0.02, and φ1 = 0.9. 1. Plot the autocorrelation function for this process for lags 0 through 20.
Copyright By PowCoder代写 加微信 powcoder
Suggested Solution:
Figure 1: Autocorrelation for lags 0 through 20 2. Is the process stationary? Explain why or why not.
Suggested solution: In order to test whether this model is stationary, we start by demeaning
it and rewrite using the lag operator, then rearrange:
xt = 0.9xt−1 + εt
xt = 0.9Bxt + εt xt(1 − 0.9B) = εt,
where xt = rt − E[rt]. The characteristic equation is 1 − 0.9z = 0,
where the root is z = 1.11 which lies outside the unit circle. Therefore, the process is stationary.
3. Give the dynamic multiplier for a shock that ocurred 6 periods ago. That is, calculate ∂rt+1−μ ∂εt
(following the notation in Lecture 5). Suggested solution: From last week’s notes:
D M j ≡ ∂ x t + j = φ j1 . ∂εt
DM6 ≡ ∂rt+j −μ = 0.96 = 0.531. ∂εt
4. If rt = −0.01, what is Et(rt+4)? (Hint: Recall that Et(rt+4) = E(rt) + Et(xt+4), where xt = rt − E(rt)).
Suggested solution: Et(rt+4) = E(rt) + Et(xt+4)
i E(rt) = 0.02 = 0.2. This means that xt = −0.01 − 0.2 = −0.21.
ii Et(xt+4) = 0.94 × (−0.21) = −0.137781
iii Et(rt+4) = 0.2 + (−0.137781) = 0.062219. Problem 2: AR(p) Processes
3. Consider an AR(2) process with φ1 = 1.1 and φ2 = −0.25
(a) Plot the autocorrelation function for this process for lags 0 through 20.
Figure 2: Autocorrelation for lags 0 through 20
Suggested Solution: We know that ρ0 = 1 we can further solve for ρ1 using
and for j ≥ 2
ρ1 = φ1ρ0 + φ2ρ1
ρj = φ1ρj−1 + φ2ρj−2
Figure 2 shows the bar plot for the autocorrelations for lags 0 through 20. (b) Is the process stationary? Explain why or why not.
Suggested Solution: We solve for the characteristic roots for the AR(2) using x= φ1 ±φ21 +4φ2,
where the characteristic roots x1, x2 are the solutions for x above. We have that x1 = 3.12 and x2 = 1.28. Both of the characteristic roots are outside the unit circle, so this AR(2) is stationary.
(c) Give the dynamic multiplier for a shock that occurred 6 periods ago. That is, calculate
∂ [rt+6 −μ] ∂εt
Suggested Solution: We can rewrite the AR(2) process as
rt −μ=φ1(rt−1 −μ)+φ2(rt−2 −μ)+εt
We then iterate this formula
rt+1 −μ=φ1(rt −μ)+φ2(rt−1 −μ)+εt+1
= (φ21 + φ2)(rt−1 − μ) + φ1φ2(rt−2 − μ) + φ1εt + εt+1
rt+2 −μ=φ1(rt+1 −μ)+φ2(rt −μ)+εt+2 = … + φ21εt + φ2εt
rt+3 −μ=φ1(rt+2 −μ)+φ2(rt+1 −μ)+εt+3 = … + (φ31 + 2φ1φ2)εt
rt+4 −μ=φ1(rt+3 −μ)+φ2(rt+2 −μ)+εt+4 = … + (φ41 + 3φ21φ2 + φ2)εt
rt+5 −μ=φ1(rt+4 −μ)+φ2(rt+3 −μ)+εt+5 = … + (φ51 + 4φ31φ2 + 3φ1φ2)εt
rt+6 −μ=φ1(rt+5 −μ)+φ2(rt+4 −μ)+εt+6 = … + (φ61 + 5φ41φ2 + 6φ21φ2 + φ32)εt
Thus the dynamic multiplier for a shock that occurred 6 periods ago is 0.380.
(d) Now, instead assume φ1 = 0.9 and φ2 = 0.8. Give the dynamic multiplier for a shock that occurred 6 periods ago. Is the process stationary? Why/why not?
Suggested Solution: The dynamic multiplier for a shock that occurred 6 periods ago is 6.778. This process is not stationary as the influence of a shock does not seem to be going away as time goes on. In fact in this case, the characteristic roots are 0.689 and −1.814. Since one of these lie inside the unit circle, we know that this process is not stationary.
(e) Instead of analytically solving for dynamic multipliers, we can easily simulate a full impulse response (that is, dynamic multipliers at all horizons). In particular, consider a positive εt shock with magnitude one standard deviation. Assume the standard deviation is 1 for simplicity. Define xt ≡ rt − μ as in class. Thus:
xt = 1.1xt−1 − 0.25xt−2 + εt.
Set the initial values equal to the unconditional mean: xt−1 = xt−2 = 0. Set all future
shock equal to their expectations, εt+j = 0 for all j > 0. As stated earlier, let εt = 1. 4
Simulate xt+j for j = 0, …, 60 given the above initial values and sequence of shocks. Plot the resulting series from xt−1 through xt+60. This is the Impulse-Response plot for a one standard deviation positive shock to εt.
Figure 3: Impulse-Response plot
Problem 3: Applying the Box-Jenkins methodology
In PPIFGS.xls you will find quarterly data for the Producer Price Index. Our goal is to develop a quarterly model for the PPI, so we can come up with forecasts. Our boss needs forecasts of inflation, because she wants to hedge inflation exposure. There is not a single ‘correct’ answer to this problem. Well-trained econometricians can end up choosing different specifications even though they are confronted with the same sample. However, there definitely are some wrong answers.
1. We look for a covariance-stationary version of this series. Using the entire sample, make a graph with four subplots:
(a) Plot the PPI in levels. (b) Plot ∆PPI
(c) Plot logPPI (d) Plot ∆logPPI.
1947 1957 1967 1977 1987 1997 2007 2017
1947 1957 1967 1977 1987 1997 2007 2017
1947 1957 1967 1977 1987 1997 2007 2017
1947 1957 1967 1977 1987 1997 2007 2017
Figure 4: PPI Plots. The vertical shaded bars indicate NBER recessions.
log_PPI PPI
dlog_PPI delta_PPI
0 1 2 3 4 5 6 7 8 9 10 11 12
Figure 5: ACF of ∆logPPI Suggested Solution: Figure 4 presents the plots.
2. Which version of the series looks covariance-stationary to you and why? Let’s call the covari- ance stationary version yt = f(PPIt).
Suggested Solution: It seems ∆logPPI is covariance-stationary by looking at the time series graph. For ∆PPI, the unconditional variance seem to be changing. For ∆logPPI, the unconditional mean and variance seems quite stable. We’ll need to do more tests to identify stationarity.
3. Plot the ACF of yt for 12 quarters. What do you conclude? If the ACF converges very slowly, re-think whether yt really is covariance stationary.
Suggested Solution: Figure 5 shows the ACF for ∆logPPI. It converges to zero after 3 lags, though there is no clear cut off for lag 4-6. I will include 3 or 4 lags in the MA process.
4. Plot the PACF of yt for 12 quarters. What do you conclude?
Suggested Solution: Figure 6 shows PACF for ∆logPPI. I will include 2-3 lags in the
AR process.
5. On the basis of the ACF and PACF, select two different AR model specifications that seem the most reasonable to you. Explain why you chose these.
(a) Using the entire sample, estimate each one of these. Report the coefficient estimates and standard errors. Check for stationarity of the parameter estimates.
1 2 3 4 5 6 7 8 9 10 11 12
Figure 6: PACF of ∆logPPI
Suggested Solution: I will estimate the time series using AR(2) and AR (3). Table 1 gives the coefficients and standard errors for these models.
We can check whether the AR part for each model is stationary by solving for the roots for polynomials, and then make sure 1/root<1.
AR(2):1−φ1x−φ2x2 =0 AR(3):1−φ1x−φ2x2 −φ3x3 =0
The AR(2) model has 2 real roots (1.41 and −2.41) and the AR(3) has one real and two complex roots (1.32, and −1.16 ± 1.84i), respectively, all are greater than 1, thus stationary.
(b) Plot the residuals. (Note: the residuals will have conditional heteroskedasticity or ‘GARCH effects’. We will talk about this later. However, in well-specified models, the residuals should not be autocorrelated.)
Suggested Solutions: Figure 7 plots the residuals for AR(2) and AR(3) models. .
(c) Report the Q-statistic for the residuals for 8 and 12 quarters, as well as the AIC and BIC. Select a preferred model on the basis of these diagnostics. Explain your choice.
Suggested Solution: Table 2 gives the statistics for these four models. I will choose AR (3) model because the AIC is the smallest and the Q-stats are the lest significant.
Table 1: AR Estimates for ∆ log P P I Dependent variable:
0.29∗∗∗ (0.06)
dlog(PPI) AR(3)
0.27∗∗∗ (0.06)
0.16∗∗ (0.06)
0.14∗∗ (0.06)
0.01∗∗∗ (0.002)
273 824.87 0.0001 −1,639.75
ar2 0.20∗∗∗ (0.06)
Observations Log Likelihood σ2
Akaike Inf. Crit.
0.01∗∗∗ (0.001)
273 822.29 0.0001 −1,636.59
∗p<0.1; ∗∗p<0.05; ∗∗∗p<0.01 Table 2: AR Model Statistics
Q-stats(8Q) p-value
Q-stats(12Q) p-value
10.20 (0.12)
18.72∗∗ (0.04)
−1,636.59 −1622.15
5.35 (0.38)
13.83 (0.13)
−1,639.75 −1621.70
∗p<0.1; ∗∗p<0.05; ∗∗∗p<0.01 9
1947 1957 1967 1977 1987 1997 2007 2017
1947 1957 1967 1977 1987 1997 2007 2017
Figure 7: Residuals for AR(2) and AR(3) Models. The vertical shaded bars indicate NBER reces- sions.
AR(3) Residuals AR(2) Residuals
6. Re-estimate the two models using only data up to the end of 2005 and compute the MSPE (mean squared prediction error) on the remainder of the sample for one-quarter ahead fore-
where H is the length of the hold-out sample, and vi is the one-step ahead prediction error. Also report the MSPE assuming there is no predictability in yt, i.e. assuming yt follows a
random walk. What do you conclude?
Suggested Solution: The MSPE for these models are shown in Table 3. We fix the coeffi- cient estimates using data prior to Dec 2005 and fit new data sequentially to get one period ahead prediction. For the random walk, we take the real value from last period plus drift as the prediction.
Table 3: AR MSPE. This table shows the MSPE for different models. The first line does the forecast without fitting in new data. The second line shows the forecast when we fix the model but keep fitting in new data for predictions.
AR(2) without 3.40e−4
with 4.30e−4
AR(3) 3.38e−4
RW 6.38e−4
程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com