Ever wondered how your estimation of a linear function relates to the elasticities of the estimated model? I always seem to forget, especially if I have taken the logarithm on one or both sides of the equation. Here are the four cases you can have:

**Linear:**

The function has the following form (if you have more variables on the right hand side, this doesn’t change the story):

[math]Y=a + bX[/math]

The elasticity is given by:

[math]\epsilon= \frac{dY}{dX}\frac{X}{Y}=b\frac{X}{Y} [/math]

and the coefficient b is the change in Y from a unit increase in X.

**Log-linear**

[math]log(Y)=a + bX [/math]

and the elasticity is given by:

[math]\epsilon= be^{a+bX}\frac{X}{Y} = bY\frac{X}{Y} =bX[/math]

and the coefficient b is the percentage increase in Y from a unit increase in X.

**Linear-log**

[math]Y=a + b*log(X)[/math]

and the elasticity is:

[math] \epsilon= \frac{b}{X}\frac{X}{Y} =\frac{b}{Y} [/math]

and b is the change in Y caused by a 1% increase in X.

**Log-log**

[math]log(Y)=a + b *log(X)[/math]

and the elasticity is:

[math]\epsilon= \frac{bY}{X}\frac{X}{Y} =b[/math]

Depending on your regression equation the elasticity is therefore either the estimated coefficient (double log), the coefficient multiplied divided by the left-hand variable (linear-log), multiplied by the right-hand variable (log-linear) or the fraction of right-hand and left-hand variable (linear).

By the way: the formulas were written using WordPress and the Youngwhan’s Simple Latex Plug-In for writing equations in WordPress.

Theoretically, elasticity is percentage change in y over percentage change in x. log-level form is semi elasticity. Other than log-log form, in order to find elasticity, you need to multiply the beta by the initial point.

e = xdy/ydx

log-log:

d ln(y) = beta d ln(x)

dy/y = beta * dx/x

beta = xdy/ydx …which is e

log-level:

d ln(y) = beta dx

dy/y = beta dx

beta = dy/ydx is missing x in the denominator and so in order to find elasticity you need to multiply it by some value of x. This is similar for level-log form and level-level form

Great concise summary. Thank.

Great simple summary

Thanks!

I believe you need a b coefficient in your linear log function.

thanks! this really helped me!

Thanks, fixed it.

The pictures are not displaying? This looks interesting but would love to see the images that go along with the breakdown…