Impulse response: Interpreting shock and response for log-variables

Question

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))

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!

Answer 1

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.

Answer 2

Old question, but since it so far has only an incorrect answer, I'll give it a shot:

In general the interpretation of the irf is: How does a 1-standard deviation shock in the impulse affect the response (after different lags). See here

An absolute change in a logarithmized variable corresponds to a relative change in the original variable because of

$$\frac{\exp(x + \Delta x) - \exp(x)}{\exp(x)} = \exp(\Delta x) - 1$$ (the point is that this is independent of $x$).

So in your example, the correct interpretation is that a one standard deviation move in the logarithmized e corresponds to a $\exp(0.15) - 1 \approx 16\% $ move in prod at lag 3.

Or, if you want to express the impuls in non-log terms as well: A $(\exp(\sigma) - 1)\cdot 100\%$ move in the non-logarithmized e corresponds to a $16\%$ move in the non-log. prod, where $\sigma$ is the standard deviation for the logarithmized e.

The pattern here is that absolute moves in log variables correspond to relative moves in non-log-variables.

Around $0$ you approximately have $ \exp(x) = 1 + x $, which is the reason why for small values the interpretation of the prod-axis as relative change in non-log data roughly works, even though it actually shows absolut change in the log-data.