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ARMA Model - Stata Lab Session Notes

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Here are the notes from the ARMA Model Lab session. The document includes all the steps with the explanation attached. There are 8 steps. 1 - Looking to the data, 2 - Looking at the autocorrelation plot, 3 - Estimate ARMA models, 4 - Construct residuals and check if there is autocorrelation, 5 - C...

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  • November 28, 2017
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  • 2017/2018
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ARMA Model – Stata Lab Session Notes



1st Step – Looking to the data
.1
.08
.06
Density


.04
.02
0




0 20 40 60 80
VIX




- The histogram is important because if we use an ARMA model for modeling VIX you assume
a normal distribution because you use maximum likelihood.
- In our case is important to note that the series is not normally distributed.
- Formally we can test this by typing sktest VIX




- Skewness should be 0 and kurtosis should be 3  in our case we reject the H0 of normally
distributed variable.
- We already expect this because we saw the histogram.

, 2nd Step – Looking at the autocorrelation plot


Autocorrelation Partial Autocorrelation




1.00
1.00




0.80
Partial autocorrelations of VIX
0.50




0.60
0.40
0.00




0.20
0.00
-0.50




0 10 20 30 40 0 10 20 30 40
Lag Lag
Bartlett's formula for MA(q) 95% confidence bands 95% Confidence bands [se = 1/sqrt(n)]




// make a plot of the (partial) autocorrelations

ac VIX in 1/1511, name(ACplot)

pac VIX in 1/1511, name(PACplot)

- We now look at an in-sample period  (1511 days)
o You can take also half of the data and test the model on the other half.

What do we see?

- Autocorrelation declines very slowly.
- Partial autocorrelation is almost zero after lag 1 or 2 or 3 .
- This is more AR type of model than MA.
o The reason is because MA says that the autocorrelation will drop immediate to zero
after certain lag. Here it is not the case.
o AR, on the other hand, says that the partial autocorrelation drops to zero and you
can see it in the graph. However, the autocorrelation alone does not decline
exponentially to zero.
- If you see this two plots, you should think that this is more a type of AR MA model but it is
not completely AR  therefore we should go for an ARMA model.

, 3rd Step – Estimate ARMA models

//estimate various ARMA models

arima VIX in 1/1511, ar(1/1) ma(1/1)

estat acplot, name(theoACarma11)

arima VIX in 1/1511, ar(1/2) ma(1/1)

estat acplot, name(theoACarma21)



AR(2,1)
ARIMA regression

Sample: 1 - 1511 Number of obs = 1511
Wald chi2(3) = 1.27e+06
Log likelihood = -3068.989 Prob > chi2 = 0.0000



OPG
VIX Coef. Std. Err. z P>|z| [95% Conf. Interval]

VIX
_cons 20.27916 8.485587 2.39 0.017 3.647713 36.9106

ARMA
ar
L1. 1.442787 .0397349 36.31 0.000 1.364908 1.520666
L2. -.4459915 .0392521 -11.36 0.000 -.5229243 -.3690588

ma
L1. -.6463761 .03225 -20.04 0.000 -.709585 -.5831672

/sigma 1.842184 .0119603 154.03 0.000 1.818743 1.865626

Note: The test of the variance against zero is one sided, and the two-sided
confidence interval is truncated at zero.


- Coefficients = 1.44 and -0.44.
- Remember that Ф1 + Ф2 should not exceed 1  here is 0.9967955

 Immediately after you estimate the model compute estat ic, n(1511) to look at the AIC and BIC

, - You can’t do anything with the values at the moment because you must compare it with
various models.




AR(1)




- Coefficients = 0.9913489 and -0.2075646.
- Remember that Ф1 + Ф2 should not exceed 1  here is 0.7837843
o This is exactly what we previous thought because the autocorrelations looks almost
like a random walk  like an AR1 with the values of 1.



 Immediately after you estimate the model compute estat ic, n(1511) to look at the AIC and BIC

, . estat ic, n(1511)

Akaike's information criterion and Bayesian information criterion



Model Obs ll(null) ll(model) df AIC BIC

. 1,511 . -3068.989 5 6147.978 6174.58

Note: N=1511 used in calculating BIC.


- After comparing the AIC and BIC we can conclude that AR (2,1) model has lower values than
AR(1). So, AR(2,1) is better.
- The differences are not that big.
- AR(1) has higher AIC and BIC than AR(2)
o This is because the model look almost like a random walk.
o An AR(1) model with Ф = 0.99 is almost a random walk.
- What you learn from this is that if you see an AC or PAC and you see that is quite close to a
certain model then you already know that if you use other models for forecasting.

4th Step – Construct residuals and check if there is autocorrelation

predict resARMA11 in 1/1511, residuals // make residuals

ac resARMA11 in 1/1511, name(ACresARMA11) //plot autocorrelations of the residuals

wntestq resARMA11 in 1/1511, lags(5)

- Until now we estimated the model.
- However, after you do this, you want to check if the model is ok so you check its residuals.
- You create a new variable with the code above.
0.10
0.05
Autocorrelations of AR1




0.00
-0.05
-0.10




0 10 20 30 40
Lag
Bartlett's formula for MA(q) 95% confidence bands

, - You can see that there are 1,970 missing values generated.
- Be careful always to specify you sample in the code.
- This graph does not look good because we see significantly autocorrelation in lag 1, lag 3,
and lag 5.
- Next, we run a test to confirm our outcome




- The F test is very high 41.36, and the p-value is zero.
- We can reject the H0 hypothesis that there is no autocorrelation.
o The test tells us that there is indeed autocorrelation in the residuals.

Conclusion: ARMA (1) has autocorrelation if the residuals. Therefore, let’s look at ARMA (2,1)

AR(2,1)
0.10
Autocorrelations of resARMA12




0.05
0.00
-0.05
-0.10




0 10 20 30 40
Lag
Bartlett's formula for MA(q) 95% confidence bands




- Be careful because you must estimate the model again and change the cod accordingly to fit
with ARMA (2,1)
- The graph looks slightly better than the previous one because I do not see the negative
significant autocorrelation in lag1 anymore.
- We can see something at lag3 and lag 5.
- We run the test again.

, - The F test is also high 27.83, and the p-value is zero but slightly lower than the previous one.
- Also here, we can reject the H0 hypothesis that there is no autocorrelation.
o The test tells us that there is indeed autocorrelation in the residuals.

Conclusion: ARMA (2,1) has autocorrelation if the residuals.



5th Step – Construct the fit of the model

arima VIX in 1/1511, ar(1/1) ma(1/1)

predict fitARMA11 in 1/1511, xb

twoway (line VIX time in 1/1511) (line fitARMA11 time in 1/1511)

AR(1)
80
60
40
20
0




0 500 1000 1500
time

VIX xb prediction, one-step


- The fit of the model looks amazing.
- We see the red line, but we don’t see the blue line anymore.
- The blue line is the original series and the red line is the fit of the 1150 observations.

,AR(2,1)
80
60
40
20
0




0 500 1000 1500
time

VIX xb prediction, one-step




- In terms of fit, we did an extremely good job.

Why does the fit looks so good?

- It is because these time series are full of autocorrelation.




Additional Steps – Test for homoscedasticity

gen res2ARMA11 = resARMA11^2

ac res2ARMA11 in 1/1511, name(ACres2ARMA11)

- We assume from all the ARMA models that the variance of the error terms had a mean of
zero and a variance of σ2 so it is constant.
- You can check whether is true for the VIX while computing the squared residuals (check the
code).

, 0.60
0.40
0.20
0.00
-0.20




0 10 20 30 40
Lag
Bartlett's formula for MA(q) 95% confidence bands




- Huge autocorrelation in squared residuals.
- The squared residuals of today have a lot to do with the squared residuals of yesterday.
- The graph shows that there is heteroscedasticity. The implication of heteroscedasticity is
that is harming the standard error of the coefficients. So, this is bad for the model.



Additional Steps – Log of VIX

- We saw previously that the series was not normally distributed.
- In this case we can compute the log of the VIX

gen lnVIX = ln(VIX)

arima lnVIX in 1/1511, ar(1/1) ma(1/1)

predict ehat_LN_arma11, residuals

gen ehat2_LN_arma11 = ehat_LN_arma11^2

ac ehat2_LN_arma11, name(ACres2ARMA_LOG)

graph combine ACres2ARMA11 ACres2ARMA_LOG

, Autocorrelation Partial Autocorrelation




1.00
1.00




0.80
Partial autocorrelations of VIX
0.50




0.60
0.40
0.00




0.20
0.00
-0.50




0 10 20 30 40 0 10 20 30 40
Lag Lag
Bartlett's formula for MA(q) 95% confidence bands 95% Confidence bands [se = 1/sqrt(n)]




- It looks almost the same at the first ones.



 Estimate the model but for the log of the VIX

AR(2,1)




- Coefficients = 1.509166 and -0.5116233

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