0
$\begingroup$

I have the following model: $y=a_0+a_1(x_1/z_1)+a_2(x_2)+e $, where

$y$ = the average log of variable y,

$x_1$ = the average ratio of $ x_1/z_1 $,

$x_2$ = is the averae log of the variable $x_2$.

The values of the coefficients are :

$a_1=0.012$,

$a_2=0.049$.

The data are cross-sectional macroeconomic variables.

I wonder how can I interpret these coefficients since i am confused from some similar papers that i have seen they inerpret it with different ways. I mean in one paper says:

" 1% increase of x1 increases y by 1.2% but another paper says 0.012%" and the same with the x2 which is log " the first says 4.9% and the second 0.049%"

which of them are correct?

$\endgroup$
1
  • $\begingroup$ I wonder whether your definition of $x_1$ is correct, since it appears on both sides of the equation? $\endgroup$
    – statchrist
    Commented Apr 14, 2015 at 22:35

1 Answer 1

2
$\begingroup$

It appears you have estimated a relation

$$ \ln Y = a_0 + a_1(X_1/Z_1) + a_2 \ln X_2$$

so

$$a_2 = \frac {\partial \ln Y}{\partial \ln X_2}$$

By treating partial differentials as quantities (I won't tell if you don't)

We have

$$\frac {\partial \ln Y}{\partial \ln X_2} = \frac {\partial \ln Y}{\partial \ln X_2}\cdot \frac {1}{\partial X_2/\partial X_2} = \frac {\partial \ln Y}{\partial X_2}\cdot \frac {1}{\partial \ln X_2/\partial X_2}$$

$$=\frac {1}{Y}\frac {\partial Y}{\partial X_2} \cdot \frac {1}{1/X_2} = \frac {\partial Y}{\partial X_2} \cdot \frac {X_2}{Y} = \frac {\partial Y/\partial X_2}{Y/X_2}$$

The last expression is the definition of point elasticity (the ratio of "marginal over the average"), and for small changes in $X_2$, it approximates the more general definition of elasticity expressed as

$$\frac {\text {% Δ in Y}}{\text{% Δ in}\, X_2}$$

In your case therefore

$$\frac {\text {% Δ in Y}}{\text{% Δ in}\, X_2} = a_2 = 0.049$$

then, if $\text{% Δ in}\, X_2 =0.01$ we have

$$\text {% Δ in Y} = 0.049 \times 0.01 = 0.00049 = 0.049\%$$

So the second paper is correct (the change in $Y$ is ~"five basis points", as a banker would say).

$\endgroup$
2
  • $\begingroup$ Tiny remark: in your second equation, I think you were going for $a_2$, not $a_1$. $\endgroup$
    – hejseb
    Commented Apr 15, 2015 at 5:31
  • $\begingroup$ @Alecos Papadopoulos I really thank you about your analytical answer. Now it is totaly clear to me. $\endgroup$
    – Ant
    Commented Apr 15, 2015 at 12:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.