When it comes to causality tests, the typical Granger-causality test can be problematic. Testing for Granger-causality using *F*-statistics when one or both time series are non-stationary can lead to spurious causality (He & Maekawa, 1999).

Professor Giles gives an excellent example of how the TY method can be implemented.

More formal explanations can be found in the original TY (1995) paper.

In this post, I will show how Professor Giles’ example can be implemented in R.

The procedure is based on the following steps:

1. Test for integration (structural breaks need to be taken into account). Determine max order of integration (*m*). If none of the series in integrated, the usual Granger-causality test can be done.

2. Set up a VAR-model in the levels (do not difference the data).

3. Determine lag length. Let the lag length be p. The VAR model is thus VAR(p).

4. Carry out tests for misspecification, especially for residual serial correlation.

5. Add the maximum order of integration to the number of lags. This is the augmented VAR-model, VAR(p+m).

6. Carry out a Wald test for the ** first p variables only** with p degrees of freedom.

You may want to do a test of cointegration. If series are cointegrated, there must be a causality. However, Toda and Yamamoto (1995) noted that one advantage of the TY-method is that you don’t have to test for cointegration and, therefore, a pretest bias can be avoided.

The example is about causalities between prices in Robusta and Arabica coffee. The csv. file is available here.

The script below tests for causality between these two time series. The script is annotated, but let me know if I can clarify anything or if there is room for improvement.

**[embedit snippet=”bokeh-plot1″] **

list.packages <- c("fUnitRoots", "urca", "vars", "aod", "zoo", "tseries") new.packages <- list.packages[!(list.packages %in% installed.packages()[,"Package"])] if(length(new.packages)) install.packages(new.packages) library(fUnitRoots) library(urca) library(vars) library(aod) library(zoo) library(tseries) #Load data cof <- read.csv("http://christophpfeiffer.org/wp-content/uploads/2012/11/coffee_data.csv", header=T,sep=";") cof <- cof[complete.cases(cof),] names(cof)[1] <- "Date" #Adjust Date format cof["Date"]<-paste(sub("M","-",cof$Date),"-01",sep="") #Visualize plot(as.Date(cof$Date),cof$Arabica,type="l",col="black",lwd=2) lines(as.Date(cof$Date),cof$Robusta,col="blue",lty=2,lwd=1) legend("topleft",c("Arabica","Robusta"),col=c("black","blue"),lty=c(1,2),lwd=c(2,1),bty="n") #Possible structural break in 1970s. Therefore only values from 1976:01 onwards are regarded cof1<-cof[193:615,] #Visualize plot(as.Date(cof1$Date),cof1$Arabica,type="l",col="black",lwd=2,ylim=range(cof1$Robusta)) lines(as.Date(cof1$Date),cof1$Robusta,col="blue",lty=2,lwd=1) legend("topright",c("Arabica","Robusta"),col=c("black","blue"),lty=c(1,2),lwd=c(2,1),bty="n") #Test for unit roots adf.test(cof$Arabica) adf.test(cof$Robusta) kpss.test(cof$Arabica) kpss.test(cof$Arabica) adf.test(diff(cof$Arabica,1)) adf.test(diff(cof$Robusta,1)) kpss.test(diff(cof$Arabica,1)) kpss.test(diff(cof$Robusta,1)) # Since first order differencing eliminates the unit root, the maximum order of integration # is concluded to be I(1). #Set up VAR-Model #select lag order // either 2 or 6 VARselect(cof1[,2:3],lag=20,type="both") #VAR Model, lag=2 V.2<-VAR(cof1[,2:3],p=2,type="both") serial.test(V.2) #VAR-Model, lag=6 V.6<-VAR(cof1[,2:3],p=6,type="both") serial.test(V.6) #Stability analysis 1/roots(V.6)[[1]] # ">1" 1/roots(V.6)[[2]] # ">1" #Alternative stability analyis plot(stability(V.6)) ## looks fine # Model with p=6 is less likely to be serially correlated. Thus model with p=6 is selected. # Wald-test for the first 6 lags # The test can be directly done with the VAR model, however using the correct # variables is a little more tricky #VAR-Model, lag=7 (additional lag, though not tested) V.7<-VAR(cof1[,2:3],p=7,type="both") V.7$varresult summary(V.7) #Wald-test (H0: Robusta does not Granger-cause Arabica) wald.test(b=coef(V.7$varresult[[1]]), Sigma=vcov(V.7$varresult[[1]]), Terms=c(2,4,6,8,10,12)) # Could not be rejected (X2=8.6; p=0.2) #Wald.test (H0: Arabica does not Granger-cause Robusta) wald.test(b=coef(V.7$varresult[[2]]), Sigma=vcov(V.7$varresult[[2]]), Terms= c(1,3,5,7,9,11)) # Could be rejected at 10% (X2=12.3; p=0.056) # It seems that Arabica Granger-causes Robusta prices, but not the other way around.

Let me know if you have any suggestions.

— C

**References**

He, Z.; Maekawa, K. (1999). On spurious Granger causality. *Economic letters*, 73(3), 307–313.

Toda H.Y.; Yamamoto T. (1995). Statistical inference in vector autoregressions with possibly integrated processes. *Journal of Econometrics*, 66, 225–250.

**Data**

Christoph. This is just great! Thanks for sharing this.

Just change this for the tests.

It’s not adf.test, but ur.df. For example:

ur.df(cof$Arabica)

ur.df(cof$Robusta)

ur.kpss(cof$Arabica)

ur.kpss(cof$Arabica)

Great job! Thanks!

Claudio, thanks for pointing out.

adf.test(), kpss.test() work es well, but we need the “tseries” package loaded. Should be fine now.

Thanks for sharing this! Two questions for you.

In lm1#Wald.test (H0: Arabica does not Granger-cause Robusta)

>vcov(lm2)

>wald.test(b=coef(lm1), Sigma=vcov(lm1), Terms= c(2:7),df=6)

Should this wald.test be using lm2 instead of lm1?

Hi Christopher!

Thank you for this great piece of code. I tried it out both on regular time series and on my seasonally adjusted (in R) and wald tests differed essentially. My question is,

– Is there even a reason to wald test on seasonally adjusted time series? (I mean, you don’t!)

P.S. I used the tests on my adjusted_data so to speak, instead of Data$Variable.

Hi Cindy,

thanks for the catch, it is indeed lm2 and not lm1.

index(arab) caputres the trend index(robu) would of course also work and yields the same result.

Thanks for the quick response!

Dear Christoph,

I used the code on a data set which has 22 observations and 9 variables all variables entering into the vAR model. I determined p=k+dmax=3. this gives me 27 coefficient estimates. I have NAs in the results. Just wondering if I am doing something wrong or overfitting?

Hi,

it seems that you have relatively few observations which makes statistical analysis difficult. Any chance obtain more data? Have you considered bootstrapping?

Dear Christoph,

I am trying to run a bootstrap regression, however, the regression is producing NAs in place of coefficients for the last 2 coefficients after every bootstrap.

would appreciate any suggestion!

If you like, you can send me your R-code and the data and I’ll have a look.

christophpp@gmail.com

Dear Christoph,

Many thanks, I will email you my code and data set.

I tried to use your R-codes but in my computer it it shows “Error: could not find function “wald.test””

when i used

> wald.test(b=coef(lm1), Sigma=vcov(lm1), Terms= c(9:14),df=6)

Please help me ..

Hi Aviral,

make sure you have all packages installed. So you shouldn’t get any error messages for the first 6 lines. You can install packages with the function

install.packages(“…”)

where … is the name of the package. Specifically, for the function wald.test() you need the package “aod” to be installed. For this you need to execute the following line:

install.packages(“aod”)

once.

Hope this helps

Christoph

thankssssss it worked out …

Christoph,

The output for wald.test also gives the F-statistic, but your comments show the test is determined by the X2 results. What is the interpretation of the F output results in this case? Thanks!

Mike,

sorry for the late answer, I just got back from vacation.

Toda and Yamamoto have shown that if you add an additional lag to a correctly specified VAR-model for which at least one time-series is integrated, the parameters asymptotically follow a chi-squared distribution. Assuming these two conditions are met, looking at the F-statistic for the augmented model would simply yield a meaningless answer.

In your wald.test, why are you using Terms c(9:14) and c(2:7)? I would have thought it would be c(8:13) and (1:6) because the last lag is supposed to be exogenous and just there for the asymptotics?

Hi Rick,

for the first test we need to assess if the coefficients of Robusta are significant for the price of Arabica. For the second test we are testing if the coefficients of Arabica are significant for the price of Robusta. These are the models:

Arabica = Intercept+A*X_1+B*X_2 + Trend # lm1

No. of terms: (1) (7) (7) (1) = 16

Robusta = Intercept+A*X_1+B*X_2 + Trend # lm2

No. of terms: (1) (7) (7) (1) = 16

With A being the coefficients for the values of Arabica, B coefficients for Robusta, X_1 and X_2 the respective lagged values. For each matrix of lagged values we have 7 terms but only want to test the first 6. Then, for the first Wald test we need to look at terms 9 to 14 and for the second model at terms 2 to 7. You can look at the coefficients of each model with coef(lm1) or coef(lm2).

I’m having trouble interpreting the serial.tests. What’s the significance of the degrees of freedom and the p-values? (The p-value for serial.test(V.6) is .5.)

Any way to do this without the packages?

Bob,

of course, but why would you do it without the packages? If you want to look deeper, you can just look at the functions provided by the packages.

Martin,

sorry for the late reply. The serial test used here is a portmanteau test with the null hypothesis that residual errors are not serially correlated. A higher p-value indicates a favorable model which is why a model with 6 vs. 2 lags is chosen.

I was wondering how the method and the R code would change if one of the series has unit root and the other does not.

Yolande,

good question. The first step says:

1. Test for integration (structural breaks need to be taken into account). Determine max order of integration (m). If none of the series in integrated, the usual Granger-causality test can be done.

So, if one series has an order of integration of 1 and the other is not integrated (order of integration is 0), the maximum order of integration would be m=1.

Jumping to step 5:

5. Add the maximum order of integration to the number of lags. This is the augmented VAR-model, VAR(p+m).

This means we would still have to use the augmented model even if one series is not integrated at all. It’s the maximum order of integration that counts. The only case in which you can use the standard Granger test for non-causality is when neither of the series are integrated.

Christoph,

I was wondering how I could change the code in my blog to follow the procedure you outline above. I don’t understand the idea of including an additional lag, but not testing for it.

http://r-datameister.blogspot.com/2014_02_01_archive.html

Thank you for the prompt reply.

I do have an additional question: Where in the code and algorithm the model v.7 was used after it is set

Hi,

I was wondering if you’ve ever used the function “causality” from the package “vars”? Are the results from this function compatible with the ones from your code?

Cory,

there are some issues with differencing the data first for integrated time series and then using the standard Granger causality test (e.g. Enders, W. Applied Econometric Times Series ).

You can think of the idea to add an additional lag and then not using it for the test as a trick to get rid of the problems involved in testing integrated time series for causality.

Yolande,

you are very welcome. 🙂

Good point, model v.7 is actually not necessary for the following tests.

Hi Pablo,

as far as I know, the causality function from ‘vars’ only uses the standard Granger causality test and would not work in this context.

Hi Christoph,

This is a general question.

Why would I got -Infinity in my varselect. The lag length is 3 based on all criteria.

> VARselect(M,lag.max=20,type=”both”)

$selection

AIC(n) HQ(n) SC(n) FPE(n)

3 3 3 3

$criteria

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

AIC(n) 5.517709 4.919171 -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf

HQ(n) 5.139389 4.351692 -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf

SC(n) 5.693020 5.182138 -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf

FPE(n) 284.562815 237.788270 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

18 19 20

AIC(n) -Inf -Inf -Inf

HQ(n) -Inf -Inf -Inf

SC(n) -Inf -Inf -Inf

FPE(n) 0 0 0

Thanks

Hi Yolande,

this probably happens because the number of lags available in your dataset is too low to calculate the criteria for lags higher than two. If this does not explain it, you can send me the dataset and I will have a look.

Hi Christoph,

When I execute the code I get the following error. How to debug this. Can you please help me with it. I am new R software. Thank you.

> #Wald-test (H0: Robusta does not Granger-cause Arabica)

> vcov(lm1)

Error in if (is.finite(resvar) && resvar < (mean(f)^2 + var(f)) * 1e-30) warning("essentially perfect fit: summary may be unreliable") :

missing value where TRUE/FALSE needed

Hi Christoph,

Your presentation is very nice. By the way, allowing the others to ask questions and your immediate answers are very nice. I grasped the topic. I want to gift you my econometry book as soon as it is finished.

Hi Christoph,

First, thanks millions of times for the above R code.

In #29, you said: “there are some issues with differencing the data first for integrated time series and then using the standard Granger causality test (e.g. Enders, W. Applied Econometric Times Series ).”

However, what W. Enders’ say in “Applied Econometric Times Series” 3E (2010) p.321 (equivalent of above 347 in the below link) that:

“The issue of differencing is important. If the VAR can be written entirely in first differences, hypothesis tests can be performed on any equation or any set of equations using t-tests or F-tests. This follows because all of the variables are stationary. As you will see in the next chapter, it is possible to write the VAR in first differences if the variables are I(1) and are not cointegrated. If the variables in question are cointegrated, the VAR cannot be written in first differences; hence, causality tests cannot be performed using t-tests or F-tests.”

Book: http://tr.scribd.com/doc/215018030/180318659-Applied-Econometric-Time-Series-3rd-Edition-Walter-Enders-PDF

That is to say, the “issues” you mentioned are restricted to both of the variables are nonstationary AND COINTEGRATED.

When only one of the variables in a system of 2 variables is non-stationary, and the variables are not cointegrated, then Enders (2010) does not mention anything about the problematic nature of classical G-causality(1969).

Hence all in all: in a system of 2 variables:

Cases of Variables Solution Source

————— ———————– ————————

1. Both Stationary .. Classical G-causality(1969) .. Granger (1969)

2. Only one nonstationary ..(AFAIK:) Toda-Yamamoto(1995) .. ??????

3. Both nonstationary and not cointegrated .. TY(1995) .. TY(1995)

4. Both nonstationary and cointegrated (VAR–>VECM) .. (I am not sure:)TY(1995) .. ??????

I will be glad if you clarify my “AFAIK”s, “I am not sure”s and “????”s.

Best and Warm Regards,

Hi Chris, again me!

Look what Enders 3E 2010 p. 397 say (for my cases of Variables, 3rd case):

“There are three consequences if the I(1) variables are not cointegrated and you estimate the VAR in levels…:

For a VAR in levels, test for Granger causality conducted on the I(1) variables do not have a standard F-distribution. If you use first differences, you can use the standard F-distribution to test for Granger causality”.

Hence, as far as I understand, (in the case of non-cointegrated non-stationary I(1) variables) Enders2010 suggest classical G-causality(1969) on 1st differences. However, many books says, “in a 2-variable system in which AT LEAST ONE OF THE VARIABLES IS NONSTATIONARY, TY1995 is applied. Hence, for the above 3rd case, does it seem that Both G1969 (on 1st differences) and TY1995 (on levels) be applied?

Also, what do you think for my above 4th case?

Regards…

As if it seems no matter (stationary/non-stationary, integrated/cointegrated) what the variables in 2-variable system, TY1995 correctly finds G-causality:

Eiji Kurozumi, Khashbaatar Dashtseren 2011: “Statistical Inference in Possibly Integrated/Cointegrated Vector Autoregressions: Application to Testing for Structural Changes”:

“…..Toda and Yamamoto (1995) propose to estimate a model with intentionally augmented lags, and show that the estimated parameter of interest has a limiting normal distribution IRRESPECTIVE OF WHETHER THE VARIABLES ARE (TREND) STATIONARY, INTEGRATED, OR COINTEGRATED….”

Anyway, Chris, I wanna learn what you think as well.

An offer to the above code: The revealing of why the relevant packages was loaded; that is to say, for which function… would be better:

library(fUnitRoots)

library(urca)

library(vars) # VARselect, serial.test are in vars

library(aod) # wald.test is in aod

library(zoo)

library(tseries) #adf.test, kpss.test are in tseries

Another offer to the above code: I looked at the above code and observed the following code snippet:

# Model with additional lag is set up.

V.71″

1/roots(V.7)[[2]] # result became “>1”

# Since the above two value exceeds (in absolute value) 1, V.7 is stable.

# Fact: VAR is stable if the eigenvalues of RHO part of VAR exceeds 1

equivalently you can use:

roots(V.7)[[1]] # result became “<1"

roots(V.7)[[2]] # result became "<1"

and get the same stability of the VAR(7).

Best and Warm Regards,

Erdogan CEVHER

Third Offer to the above code:

Though p values of serial.test’s are enough to decide, it would be nicer to show the ACFs of the residuals of the VARs as well. Hence, I offer to add the followings just after the relevant tests:

plot(serial.test(V.2))

plot(serial.test(V.6))

In my #39,

Instead of:

library(vars) # VARselect, serial.test are in vars

….etc……

the following is nicer:

library(vars) # for VARselect, serial.test

….etc……

Hi Erdogan,

thanks for making my argument more precise. The issues in differencing integrated time series first and then applying the standard Granger-causality test are that the following two conditions have to be met:

1. The time-series have to be integrated by the same order.

2. The time-series must not be conintegrated.

Personally, I would simply stick to the TY-method as soon as one series is integrated.

The TY-method also works when time series are cointegrated. There is no need to test for cointegration first (see TY 1995).

Thanks for the stability analysis.

All the best and good luck for your book!

Lohit,

make sure you have installed all the packages (you only need to due this once):

install.packages(“fUnitRoots”)

install.packages(“…”)

Let me know if this helped.

Hi Christoph,

I installed all packages but I still get the error below

>vcov(lm1)

Error in if (is.finite(resvar) && resvar < (mean(f)^2 + var(f)) * 1e-30) warning("essentially perfect fit: summary may be unreliable") :

missing value where TRUE/FALSE needed

best regards

Agapi

Agapi,

if you just copy and paste the entire code and successfully download the data this should not happen.

A few things to look at:

1. Have you copied the entire code?

2. Do you get any previous warnings?

3. Have you installed R correctly?

Looking at the warning, it seems that there is something wrong with the data you were supposed to download (‘essentially perfect fit’).

#Load data

cof <- read.csv("http://www.christophpfeiffer.org/app/download/6938079586/coffee_data.csv", header=T,sep=";")

Hi Christopher,

Thanks for the great post. I’m testing for Granger Causality under the TY approach you outline here but when using the function wald.test with d.o.f=p the response when the function is run is a Chi-squared test with df=p+m.

If you look up the help for the package wald.test {aod}, under the arguments for df it says:

“A numeric vector giving the degrees of freedom to be used in an F test, i.e. the degrees of freedom of the residuals of the model from which b and Sigma were fitted”

Given that we want an asymptotically chi-square statistic distributed with p df under the null, does the package wald.test have the ability to specify the degrees of freedom for a Chi-squared test, or does df only work for an F test?

If this is the case do you know any other Wald test package for R that would allow you to specify the degrees of freedom specifically for the Chi-Squared statistic?

Hi Christoph, thanks for the post. Question – looking at these two lines:

lm1<-lm(arab~arab.l[,2:8]+robu.l[,2:8]+index(arab))

lm2<-lm(robu~arab.l[,2:8]+robu.l[,2:8]+index(arab))

In the expression for lm2, shouldn't it be index(robu)? Seems that comment #18 above suggests that also?

Thanks!

Hi Alexei,

it does not matter for the end result. index(arab) and index(robu) should be identical.

Hi Christopher,

One of your points were if R was correctly installed.

I had to re-installed 32bit version of R.

Many thanks!

Agapi

Would you be able to point me in the direction of how to build a calibration dataset for this kind of modeling in general? E.g., generate two dataseries with specific lags, coefficients, amounts of noise, etc?

Hello Alexei,

sure. But I need some more specifics on your dataset.

What kind of data do you have and what is your outcome?

Hi Nicholas,

the df parameter in the Wald test only refers to the F-distribtuion and is actually not necessary for the TY-test.

I am not aware of any other wald test function in R. But in any case have a look here:

http://www.statlect.com/Wald_test.htm

and here:

http://www.utstat.toronto.edu/~brunner/oldclass/appliedf12/lectures/2101f12WaldWith

Hi everyone,

I made some changes to the code which should take care of the NA error you were getting. Let me know if it works.

Greg,

thanks for the input and sorry for my delay in responding. Using the existing VAR model is indeed more straightforward than the manual computation.

We only need to change the V.6 model to a V.7 model and use the correct terms which can be a bit tedious.

Thanks a lot… Your new code is working alright…

Thanks a lot Sir for giving such codes to estimate TY causality . Thank You Very much.

First I want to thank you for this code of TY Granger causality. I have a problem while trying to set the VAR model through VARselect, I get no output “Error in y – z$residuals ” What should I do please? Thanks in advance

Marwa,

are you copying & pasting the above code or are you working on your own dataset? If you work with your own code, please share it.

Dear Christoph it has worked thank you!

But I want to know whether the adf.test and the kpss.test are including the trend and the drift and how can I test their significance!

thank you

Thanks for this nice code example!

However, I have two questons:

1) Why do we separately check for serial correlation in the error terms. Could this point not be addressed using a Newy-West estimator for the covariance matrix in the WaldTest?

2) How would it be possible to test whether the cumulative impact is zero (i.e. the sum of the coefficients is zero).

Thanks in advance for your reply!

Renard,

apologies for my late reply.

1) I am sure there are other methods that work as well.

2) The Wald test is pretty close to the cumulative test. At least for the no-impact version. If all coefficients are zero, the cumulative impact is zero as well. For the other case it may be that effects cancel each other out. To assess this, you could have a look at the coefficients and see if they can possibly cancel each other out.

Thank you very much for sharing the code Christoph.

One quick question: why is “cof” and not “cof1” used in the UR tests?

Hi Basile,

you are right, cof1 probably makes more sense, since that is the data we are building the model for. The results are probably not affected, but still a good point.

Hi Christoph:

Thanks for the code in R! I appreciate it greatly. However, (1) could you share a code snippet on how to do the Wald Test with the VAR? You comment: ” # The test can be directly done with the VAR model, however using the correct & # variables is a little more tricky” and (2) why is it that you do the Wald Test with one additional lag?

I have three variables: GDP, CO2 emissions, and Energy Consumption for Puerto Rico. I have chosen the VAR model (it has 9 lags, don’t know if having that many lags is affects my analysis), but since I have three variables I am a bit confused on how to run the Wald Test. Especially on how to choose the “Terms” argument for each test. I hope that you can help me with my question.

what if the two time series that you have used namely ” arabica” and “robusta ” have different order of integration (d1 and d2 respectively )then which one should be used for the augmented VAR model ..

Thanks

I went through your code and looked very promising , i really appreciate the amount of time that you are spending in replying to everybody’s query its laudatory . I tried to run your code but is showing a problem ,

#Wald-test (H0: Robusta does not Granger-cause Arabica)

> vcov(lm1)

Error in if (is.finite(resvar) && resvar wald.test(b=coef(lm1), Sigma=vcov(lm1), Terms= c(9:14),df=6)

Error in if (is.finite(resvar) && resvar # Could not be rejected (X2=8.6; p=0.2)

>

> #Wald.test (H0: Arabica does not Granger-cause Robusta)

> vcov(lm2)

Error in if (is.finite(resvar) && resvar wald.test(b=coef(lm2), Sigma=vcov(lm2), Terms= c(2:7),df=6)

Error in if (is.finite(resvar) && resvar < (mean(f)^2 + var(f)) * 1e-30) warning("essentially perfect fit: summary may be unreliable") :

missing value where TRUE/FALSE needed ….

I went through all the questions and answers posted up there and came across people who had faced similar kind of problem and you have provided the remedy in the form of changes in the code but I am still facing that problem … I have installed R properly and have loaded all the required packages …

One more question is regarding the values that you have taken

#####

lm1<-lm(arab~arab.l[,2:8]+robu.l[,2:8]+index(arab))

lm2<-lm(robu~arab.l[,2:8]+robu.l[,2:8]+index(arab))

Why you have taken arab[,2:8] and robu too ??

#Wald-test (H0: Robusta does not Granger-cause Arabica)

vcov(lm1)

wald.test(b=coef(lm1), Sigma=vcov(lm1), Terms= c(9:14),df=6)

In Terms=c(9:14) , what's the reason behind taking 9:14 value , would you elaborate it please

# Could not be rejected (X2=8.6; p=0.2)

######

Thank you

Do you have script for var model using exogenous variables for scenario anlysis?

Hi Braulio,

the alternative code for the Wald test is:

#Wald-test (H0: Robusta does not Granger-cause Arabica)

vcov(lm1)

wald.test(b=coef(lm1), Sigma=vcov(lm1), Terms= c(9:14),df=6)

#Wald.test (H0: Arabica does not Granger-cause Robusta)

vcov(lm2)

wald.test(b=coef(lm2), Sigma=vcov(lm2), Terms= c(2:7),df=6)

Using the additional lag in the VAR model but not testing for it is the trick of the TY-method.

Hi Sam,

you always use the highest order of integration for the augmented var model.

Hi VARS Newby,

you can include exogenous variables directly in the definition of the VAR model.

for example: http://stackoverflow.com/questions/22513336/var-with-exogenous-variables

Sam,

I just tested the code. It seems to be working fine for me. Have you figured out a solution in the meanwhile? If not you can send me the code you are actually using and I will have a look.

Looks like we are getting close to a hundred comments. Seems about time to organize all comments into a FAQ to make them more accessible. If anyone is willing to help out with this, let me know!

Is there a way to calculate the terms range?

Christoph,

Why did you use the following term?

Terms=c(2,4,6,8,10,12), Terms= c(1,3,5,7,9,11)

#Wald-test (H0: Robusta does not Granger-cause Arabica)

wald.test(b=coef(V.7$varresult[[1]]), Sigma=vcov(V.7$varresult[[1]]), Terms=c(2,4,6,8,10,12))

# Could not be rejected (X2=8.6; p=0.2)

#Wald.test (H0: Arabica does not Granger-cause Robusta)

wald.test(b=coef(V.7$varresult[[2]]), Sigma=vcov(V.7$varresult[[2]]), Terms= c(1,3,5,7,9,11))

# Could be rejected at 10% (X2=12.3; p=0.056)

Hello Christoph,

could you explain what you mean by:

#Alternative stability analyis

plot(stability(V.6)) ## looks fine

Is a rather flat line the best of rather a high fluctuation?

Hey,

Thanks a lot for this awesome article.

I have similar data. A bit bigger; several time series calculated from the polynomial fit of my experimental data.

Every time I try to run wald.test() it returns: “Error in L %*% V : non-conformable arguments”

Any suggestion where I can find info?

Best wishes,

thiago

Hi Chrostoph,

this is really great work! One question. What if

#Stability analysis

1/roots(V.6)[[1]] # “>1”

1/roots(V.6)[[2]] # “>1”

the test always turns our pairs where at least one of the values is <1. Still choosing the lag with the lowest distortion or giving up the analysis?

Chris,

I would also look at the stability analysis from

plot(stability(V.6))

This can show you when the results are stable and when they are not. You may then want to exclude certain periods from your analysis to account for that and redo the test after the adjustment.

Thiago,

I realize my response is quite delayed to say the least. Shoot me an email, if the problem persists.

Dear Christoph,

thank you for your quick reply! Yes, I already did that and that’s exactly how I would also approach it now. I concluded that I could solve the problem by excluding one very extreme outlier from the dataset.

One last question: This paper here http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.198.9108&rep=rep1&type=pdf mentions a bootstrap simulations for the MWALD test. Do you have any recommendation how to do this? That one didn’t really help: http://stackoverflow.com/questions/32057573/wald-testing-bootstrapped-estimates-in-r

Dear Christoph,

great work! Only one question: Why is there a constant+trend (“both”) considered for the VAR model, but not for the Wald test?

Thank you Manuel. I see what you are getting at.

The trend is implicitly included in the Wald test since that is the model that is used for the Wald test, however for the test we only want to focus on whether there is a causality between Robusta and Arabica. So only these variables are included.

Hello Christoph,

In case that I have more than 2 variables how I can check if x is caused by y and z.

I have tried to adj a bit your code from the aor library but it does not work.

If you have hints that would be helpful.

All the best

Nick

Hi Christoph,

Thank you very much for the amazing script! I would like to ask you that in the case of more than 1 structural break, let´s say 3 or 4 how should we proceed with the following process?

Thank you very much.

Jozefina

Hi Jozefina,

I would recommend finding the structural breaks using Bai-Perron and then testing for each phase separately. You can find the Bai-Perron test in the strucchange package.

Thank you

Christoph

Hi Mr Christophor. I hve time series and İ could not apply you code toda yamomoto. my wald test result is wald.test(b=coef(V.4$varresult[[1]]), Sigma=vcov(V.4$varresult[[1]]), Terms=c(2,4,6,8,10,12))

Error in qr.solve(L %*% V %*% t(L)) : singular matrix ‘a’ in solve

> wald.test(b=coef(V.4$varresult[[2]]), Sigma=vcov(V.4$varresult[[2]]), Terms=c(2,4,6,8,10,12))

Error in qr.solve(L %*% V %*% t(L)) : singular matrix ‘a’ in solve

I sent may data you please help ne get right code.Tank you.

Buen día Christoph,

Tengo algunas preguntas:

1) En este parte “VAR-Model, lag = 7 (lag adicional, aunque no probado)” tenemos que realizar las pruebas de que este VAR este bien especificado (No tenga prroblemas de autocolorrelacion), o no es necesario?

2) Cual resultado de causalidad es el que debo utilizar cuando aplique este método a una investigación, los que arroja el test de wald para el V.6, ó el test de wald para el VAR aumentado; es decir, el V.7?

Thanks so much for this, it’s great.

Is there a way to control for confounding variables? Or do you know where I could find out how to do this? I’m new to this type of analysis and R so examples such as yours with the code are incredibly helpful.

Thanks,

Ruth

Hello Ruth,

I don’t know if Christoph answer in 2021 but his script is amazing. I really want to thanks him for that.

Ruth, do you know why he did that :

wald.test(b=coef(V.4$varresult[[1]]), Sigma=vcov(V.4$varresult[[1]]), Terms=c(2,4,6,8,10,12))

wald.test(b=coef(V.4$varresult[[1]]), Sigma=vcov(V.4$varresult[[1]]), Terms=c(1,3,5,7,9,11))

I understand the code but not the elements in the c vectors in Terms .

Thanks in advance.

Tita

useful code, for which thanks. But I think you are missing the last entries in the Term values for the calls to wald.test. Should these not include, respectively, 14 and 13?