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Results for i 1 i 0 var vecm
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1 vote
1 answer
20 views

Stationarity Conditions VECM

process is convert it to a VAR model $$ y_{t} = \left(I+\Pi+\Gamma_{1}\right)y_{t-1}+\cdot\cdot\cdot+\left(\Gamma_{p-1}-\Gamma_{p-2}\right)y_{t-p+1}-\Gamma_{p-1}y_{t-p}+u_{t} $$ find the eigenvalues of … the companion matrix $$ \left[\begin{array}{ccccc} I+\Pi+\Gamma_{1} & \Gamma_{2}-\Gamma_{1} & \cdot\cdot\cdot & \Gamma_{p-1}-\Gamma_{p-2} & -\Gamma_{p-1}\\ I & 0 & 0 & 0 & 0\\ 0 & I & 0 & 0 & 0\\ 0 & …
John's user avatar
  • 2,297
4 votes
Accepted

Eigenvalues of Johansen Trace Test

First, we will revisit the Vector Autoregressive (VAR) model and its Vector Error Correction Model (VECM) representation: VAR and VECM Models For an $m$-dimensional VAR($p$) model: \begin{equation} Y_t … I still need to read up on how canonical correlation analysis works to understand, but I can see that you will can get $m$ values from performing CCA on $\Delta Y$ and $Y_{t-1}$. …
Robert Long's user avatar
  • 65.9k
3 votes
Accepted

Non-stationary time series: what are the advantages of doing analysis in levels instead of d...

Building on Why do we need a VECM specification if the I(1) processes are cointegrated? … \end{equation} $$ Here, $z_t'=(y_t,x_t)$, $a'=(\alpha,0)$ and $b'=(1,-\beta)$. I estimate a VAR(1) for the first differences and a VECM, and report the forecasts for $y_t$ for the two models. …
Christoph Hanck's user avatar
0 votes

VAR regression between I(1) and I(0)

I suggest running at least KPSS in addition to ADF. Not all tests are created equal and some might be more powerful than others, especially in the case of near-unit-root processes. … As regards cointegrated VAR, it makes no sense if you have an I(1) variable and an I(0) variable, but given that you are not sure whether or not this is indeed the case, a Johansen VECM might throw light …
Pavel Filip's user avatar
3 votes

VAR regression between I(1) and I(0)

If you are fairly sure that $Y_1$ is I(1) and $Y_2$ is I(0), take the first difference of $Y_1$ and model $(\Delta Y_1,Y_2)$ using a VAR. … There is no point in using a VECM, as there is no cointegration and thus no error correction term. …
Richard Hardy's user avatar
2 votes
2 answers
57 views

VAR regression between I(1) and I(0)

I am considering two time series and I would like to to a VAR regression between them. The ADF test rejected stationarity in only one of them, so the time series would be I(0) and I(1). … Should I consider doing a VAR of the I(0) in levels and the differences in the I(1)? Should I still consider a VECM? …
dleal's user avatar
  • 243
0 votes
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Unit root test and cointegration

If $d=1$ and there is a single series that is I(1) while the other ones are I(0), you take first differences of that variable and model that together with the other variables using a VAR. … If, on the other hand, the I(1) variables are cointegrated, you model them using VECM; you also include the I(0) variables on the side as in this answer. …
Richard Hardy's user avatar
4 votes
1 answer
104 views

Eigenvalues of Johansen Trace Test

Suppose we have an m-dimensional VAR(p) model $$Y_t = \Pi_1 Y_{t-1} +\Pi_2 Y_{t-1} +\dots + \Pi_p Y_{t-p} +\varepsilon_t$$ Then the VECM corresponding to this model is $$\Delta Y_t = \Pi Y_{t-1} +\sum_ … But I'm a little confused how the eigenvalues lie between 0 and 1. I have heard that they are the square of the canonical correlations between $\Delta Y_t$ and $Y_{t-1}$. …
Joseph Basford's user avatar
0 votes
1 answer
523 views

Johansen Cointegration Test at Levels or First Differences in a VAR Model

Having taken the first difference, they do all become stationary at I(1). I was wondering, do I perform the Johansen cointegration test on the I(0) variables or the I(1) variables. … 2) If I have cointegration present, and my I(0) variables are non-stationary, I should switch to a VECM model? ) …
Tom's user avatar
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1 vote
Accepted

Can second order differences stationary series be estimated via VECM model without consideri...

It looks like you have $x_1$ and $x_3$ being I(1) and $x_2$ being I(2). This is what I would do. Take the first difference of $x_2$ to make it I(1) and call that $y$: $y:=\Delta x_2$. … Second, VECM will not be appropriate as there can be no cointegration between variables that, after the differencing, are at most I(0). …
Richard Hardy's user avatar
1 vote
Accepted

What is the best time series model to find "time lag" between 2 sets of data?

You have two options (probably a few more, but these 2 are these, I have in my mind): 1.) … With a VAR you can capture season, trend levels. If a VAR in levels is not possible (which would be the case with sales and spendings) use VECM or VAR in differences. …
Patrick Bormann's user avatar
1 vote
0 answers
37 views

If all endogenous variables are I(1), and if just two of them are cointegrated, can I incorp...

Suppose a VAR(p) model containing $s$ endogenous variables such that $Y_{t} = (y_{1t}, \ \ldots, y_{st})$. It was verified that, for all $i \in \{1, \ \ldots, s\}$, $y_{it} \sim I(1)$. … 0)$. …
Aquinate244's user avatar
1 vote
0 answers
162 views

VAR or VECM for I(0) e I(1)

i did johansen (x1,x2,x3,x4) and Trace test indicated 1 cointegrating eqn(s) at the 0.05 level What should i follow now? … I saw this thread- VAR or VECM for a mix of stationary and nonstationary variables , but couldnt understand what should i do in eviews …
Henrique couto's user avatar
6 votes

Why Are Impulse Responses in VECM Permanent?

What would be a correct interpretation of an impulse response that does not go back to 0 in a VECM? … Here's how I fed in the unit impulse from x: T <- 50 x0 <- 100 shock <- 1 offset <- 0 beta <- 1.3 # y_t - beta * x_t is stationary Alpha <- matrix(c(.3, .4), ncol=1) Beta_T <- matrix(c(1, -beta), ncol …
Ben Ogorek's user avatar
  • 5,447
2 votes
1 answer
230 views

MLE estimation of mean-adjusted state-space model

(maturities)) i <-1 for (i in 1:nrow(yield)) { y <- as.data.frame(yield[i,]) data <- cbind(y,X) names(data)<- c("y", "c1","c2","c3") reg <- lm(formula = y~c1+c2+c3-1, data = data) beta[i,]<- … reg$coefficients eps[i,] <- reg$residuals } colnames(beta)<- c("b0","b1","b2") library(vars) VAR(beta,p = 1, type = "const")->varbeta varbeta$varresult varlagbeta <- matrix(c(varbeta[["varresult"]][ …
PK1998's user avatar
  • 151

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