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I have developed a LSTM NN where I predict a variable multiple steps ahead. For this model, the independent variable itself is not among the predictors.

I want to compare it with the results of a VAR model benchmark. However, my understanding of the VAR is that the independent variable has to be among the predictors, in order to predict this variable ahead.

Have I misunderstood something, or is it correct that the independent variable needs to be among the predictors in a VAR model?

Thanks in advance!

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  • $\begingroup$ A serious deficiency of the VAR model is that one cannot pre-specify future values for any individual series in the model. This is why SARMAX models can be more useful where some of the X's can be pre-specified or self-predicted. $\endgroup$
    – IrishStat
    Commented Dec 5, 2019 at 13:29

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A VAR(p) model has multiple dependent variables: $$ y_t=A_1 y_{t-1}+\dots+A_p y_{t-p}+\varepsilon_t $$ where $A_1,\dots,A_p$ are coefficient matrices. The current values of the dependent variables are $y_t$; this is a vector of length $k$: $(y_{1,t},\dots,y_{k,t})$. The current values depend on past values (vectors) $y_{t-1}, \dots, y_{t-p}$. In addition, you may have independent (exogenous) variables with current values $x_t$ (a constant or a vector); then your model can be called VARX (where X stands for exogenous), something like $$ y_t=A_1 y_{t-1}+\dots+A_p y_{t-p}+B x_t+\varepsilon_t $$ where $B$ is a coefficient matrix (or a scalar in case $x_t$ is a scalar).

If there were only one dependent variable, it would be an AR(p) model. It would look the same, but $y_t,y_{t-1},\dots,y_{t-p}$s and $A_1,\dots,A_p$s would be scalars. An ARX version is also possible.

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  • $\begingroup$ Okay, so if I just need to predict a single dependent time series based on several exogenous time series, then VARX would be the one to go for among these? Would the results vary from the predicted series of this variable if the dependent variable and independent variables are used together in a VAR model? $\endgroup$
    – Kaangy
    Commented Dec 5, 2019 at 13:55
  • $\begingroup$ @Kaangy, As noted in my answer, if you have a single dependent variable, then you do not have a VAR - you have an ARX. Now, if you treat your independent variables as dependent (endogenous) and put them into a VAR, you will get forecasts of all these variables (the dependent and the "independent" ones). Whether the forecasts will be worth much depends on how the variables are interrelated, namely, whether lags of some variables are useful in predicting current values of some (the same or different) variables in the system. $\endgroup$
    – Richard Hardy
    Commented Dec 5, 2019 at 14:45
  • $\begingroup$ Sorry, meant to write ARX. Thank you for clarifying. $\endgroup$
    – Kaangy
    Commented Dec 5, 2019 at 14:50
  • $\begingroup$ @Kaangy, you are welcome! $\endgroup$
    – Richard Hardy
    Commented Dec 5, 2019 at 14:54

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