Let us assume that $Y_t$ can be described with an ARMAX process, including an exogenous covariate $X_t$, of the following form:

\begin{equation} log(Y_t)=\phi_1log(Y_{t-1})+\phi_{12}log(Y_{t-12})+\beta log(X_t) + \theta_1 \varepsilon_{t-1} + \theta_{12}\varepsilon_{t-12} + \varepsilon_t \;\;\;\; (1) \end{equation}

which can also be denoted directly as polynomials of the backshift operator B, defined by $B^aZ_t=Z_{t-a}$, such as:

\begin{equation} \phi(B) log(Y_t)= \beta log(X_t) + \theta(B)\varepsilon_{t} \;\;\;\; (2) \end{equation}

with $\phi(B)=(1-\phi_{1}B^1-\phi_{12}B^{12})$ and $\theta(B)=(1+\theta_{1}B^1+\theta_{12}B^{12})$

The question seems simple I but can't prove the answer yet:

Can we interpret $\beta$ as an elasticity, i.e. $\beta=\frac{dy/y}{dx/x}$?

In many (well published) papers, the authors interpret the $\beta$ as an elasticity but it seems not directly possible as suggested by Rob Hyndman (here) since equation $(2)$ can be rewritten as:

\begin{equation} log(Y_t)= \frac{\beta}{\phi(B)} log(X_t) + \frac{\theta(B)}{\phi(B)}\varepsilon_{t} \;\;\;\; (3) \end{equation}

from where we can see that the derivative isn't as intuitive as in a standard linear equation such as $log(Y_t)=\beta log(X_t)+\varepsilon_t $

I imagine that the answer lies in derivatives of systems including delays, but can't find the general rule to derive Equation 3. Moreover, when I estimate a similar equation on Eviews, information provided about derivatives are displayed as:

    Specification: [AR(1)=C(2),AR(12)=C(3),MA(1)=C(4),  
            MA(12)=C(5),ESTSMPL="1/01/2012 00:00 12/31/2014 23:00"] = LY -  

Variable     Derivative of Specification*

C(1)    -LX

*Note: derivative expressions do not account for ARMA components    

It seems to suggest that the derivatives is simply $C(1)$ but it also emphasizes that "derivative expressions do not account for ARMA components", what does it mean in the end?

I'd be very pleased to have an answer, and can't find any justification in the literature. Even my PhD advisor tells me that we can directly interpret $\beta$ as the elasticity but I'm really confused about that... Any clarification to suggest? Thank you.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.