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I have a question related to the interpretation of Impulse Response Function (IRF) functions. Assume we do have two time-series that have been both log-transformed and are stationary. When applying a IRF in the vars package, how do we "read" the x and y-axis correctly?

Example:

# Load data and apply VAR
library("vars")
data(Canada)
data <- Canada
data <- data.frame(data[,1:2])
var <- VAR(data, p=3, type = "both")
plot(irf(var, impulse = "e", response = "prod", boot = T, cumulative = FALSE, n.ahead = 20, ci=0.95))

irf-example

Which interpretation is correct?

  1. A 1% log-increase of e causes a 15% increase of prod at lag 3?
  2. A 1% increase of e causes a 15% log-increase of prod at lag 3?
  3. Both is wrong, correct is:

Thanks for your help!

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The VAR(p) model is defined as

$$y_t = \Phi_0 + \Phi_1 y_{t-1} + ... + \Phi_p y_{t-p} + u_t$$

In your case $p=3$ and $y_t \in \mathbb R^2$ with the two variables $y_t =(prod_t,e_t)$. The infinite MA($\infty$) representation is

$$y_{t+h} = \sum_{i=0}^\infty \Psi_i u_{t+h-i}$$

and the impulse response function can be read of from the coefficients

$$\{\Psi_h\}_{ij} = \frac{\partial y_{it +h}}{\partial u_{jt}}$$

How does a shock in the $j$-proces proces at time $t$ affect the $i$-proces at time $t+h$. A 1-unit change in the $u_{jt}$ is an unexpected increase in $y_{jt}$ by 1 full unit. Since both your processes are in logs I would therefore say that

A 1% unexpected increase in e three periods back is a 15% increase in prod today.

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  • $\begingroup$ Thank you. That's roughly what I also had in mind. But please have a bit of patience, I like to wait for a bit more confirmation, or alternative explanations. $\endgroup$ – RaymondD Dec 21 '18 at 15:25

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