Ok, so something that's common to the econometrics literature is that we interpret the coefficients in OLS log-linear models like such. To spell it out in the main body:

$ln(y_i)=\beta_0+\beta_1X+u_i \Rightarrow \text{if } \Delta x = 1, \text{then } \text{%}\Delta y \approx 100\beta_1 $

I think that this is a very bad approximation, although my reasoning is probably incorrect (although I do understand the derivation of why this approx holds).

Ok, so an aside:

$\frac{\dot{y(t)}}{y(t)} = g \Rightarrow ln(y_t) = gt + c $

It also follow that:

$y_{t+1} \approx y_t (1+g \Delta t) $

So here, if I plug in a change of t = 1, and let g = 1, y would be doubling with each unit change in t, and so we should approx y as 2^x instead of something of the form e^x. Of course, large-ish changes in x mess up the calculus.

However, isn't plugging in a unit change in t (in the econometrics textbook, x is t) what the econometrics textbooks are doing? A unit change in x -> 100% change in y (g and beta_1 are analogous, so g = 1 -> beta_1 =1) -> y approx doubles with each change in x -> y should be modelled as something of the form 2^x, not e^x, and there's a sizeable difference between the two, and so this contradicts that fact that the specification implies that y is in the form e^x (rather than 2^x).

I hope this makes sense.


2 Answers 2


The source of the approximation:

Given, $\ln(Y_i)=\beta_0+\beta_1X_i+u_i$, for a unit change in $X$, i.e., $X_{i+1}-X_i=1$, we have:

\begin{align} \frac{Y_{i+1}-Y_i}{Y_i} &= e^{\beta_1+\Delta u_i}-1 \end{align}

For small $x$, we use taylor expansion to say: $e^x \approx1+x$. Using this above we get:

$$\frac{\% \Delta Y}{100} \approx \beta_1+\Delta u$$

This approximation is good when $\beta_1$ is quite small. In your example, you have taken $\beta_1=1$ which makes this a bad approximation.

Consider your example with $g=0.1$.

\begin{align} \frac{y_{t+1}-y_t}{y_t} &= e^g-1 \\ &=1.1052 - 1 \tag{for $g=0.1$} \\ & \approx g \end{align}

  • $\begingroup$ @1muflon1: Do you mean that it should be $u_i$ in place of $\Delta u_i$? For the expectations part, i think then the LHS should also be changed to make it in terms of $\hat{Y_i}$, isn't it? $\endgroup$
    – Dayne
    Nov 29, 2020 at 2:09
  • 1
    $\begingroup$ right but i think that OP just omitted hats and you can interpret only coefficients that are estimated as true betas are unobservable but I guess it’s just unnecessarily pedantic I deleted my previous comment $\endgroup$
    – 1muflon1
    Nov 29, 2020 at 2:15
  • 1
    $\begingroup$ Right, it seems OP did omit hats in his/her example. But I guess the message is clear with this also. $\endgroup$
    – Dayne
    Nov 29, 2020 at 2:17
  • 1
    $\begingroup$ Thank you. I didn't stop to think that beta/g had to be approx small as well as x. And yes, I omitted hats as I thought what I was trying to convey wouldn't require them (i.e. just messing around with the true population model). $\endgroup$ Nov 29, 2020 at 14:02


$$\ln y = \beta_0 + \beta_1 x + u \implies y = \exp\{ \beta_0 + \beta_1 x + u\}$$

$$\implies \partial y / \partial x = \beta_1 y \implies \frac{\partial y / \partial x}{y} = \beta_1.$$

So we see that $\beta_1$ is the marginal change in $y$ due to infinitesimal changes in $x$ as a proportion of its level. Therefore the accuracy of the approximation

$$\beta_1 = \frac{\partial y / \partial x}{y} \approx \frac{\Delta y / \Delta x }{y}$$

$$\implies \Delta x = 1: \beta_1 \approx \frac{\Delta y }{y}$$

is nothing else than the general approximation inaccuracy issue that arises when we replace a derivative (infinitesimal change) with a disrete-interval proportional change.

In the specific case we have

$$y(x+1) - y (x) = \exp\{ \beta_0 + \beta_1 x + \beta_1 +u\} - \exp\{ \beta_0 + \beta_1 x + u\}$$

$$ = y(x)\cdot (e^{\beta_1 -1}) \implies \frac{\Delta y(x+1)}{y(x)} = (e^{\beta_1} -1).$$

We get

enter image description here

and we verify the folk wisdom that the approximation is accurate enough for purpose of economic analysis if $\beta_1 \in [-0.1,\; 0.1]$, and maybe for a larger interval, if one percentage point is not critical for the purposes of the specific research.


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.