<p>The source of the approximation:</p>
<p>Given, <span class="math-container">$\ln(Y_i)=\beta_0+\beta_1X_i+u_i$</span>, for a unit change in <span class="math-container">$X$</span>, i.e., <span class="math-container">$X_{i+1}-X_i=1$</span>, we have:</p>
<p><span class="math-container">\begin{align}
\frac{Y_{i+1}-Y_i}{Y_i} &amp;= e^{\beta_1+\Delta u_i}-1
\end{align}</span></p>
<p>For small <span class="math-container">$x$</span>, we use taylor expansion to say: <span class="math-container">$e^x \approx1+x$</span>. Using this above we get:</p>
<p><span class="math-container">$$\frac{\% \Delta Y}{100} \approx \beta_1+\Delta u$$</span></p>
<p>This approximation is good when <span class="math-container">$\beta_1$</span> is quite small. In your example, you have taken <span class="math-container">$\beta_1=1$</span> which makes this a bad approximation.</p>
<p>Consider your example with <span class="math-container">$g=0.1$</span>.</p>
<p><span class="math-container">\begin{align}
\frac{y_{t+1}-y_t}{y_t} &amp;= e^g-1 \\
&amp;=1.1052 - 1 \tag{for $g=0.1$} \\
&amp; \approx g
\end{align}</span></p>

CC BY-SA 4.0
2020-11-29T02:09:40.240Z
@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?
Dayne
CC BY-SA 4.0
2020-11-29T02:15:15.893Z
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
1muflon1
CC BY-SA 4.0
2020-11-29T02:17:52.203Z
Right, it seems OP did omit hats in his/her example. But I guess the message is clear with this also.
Dayne
CC BY-SA 4.0
2020-11-29T14:02:22.373Z
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).
threelinewhip
CC BY-SA 4.0
2020-11-29T01:47:20.760Z
2020-11-29T01:47:20.760Z
Dayne
<p>When</p>
<p><span class="math-container">$$\ln y = \beta_0 + \beta_1 x + u \implies y = \exp\{ \beta_0 + \beta_1 x + u\}$$</span></p>
<p><span class="math-container">$$\implies \partial y / \partial x = \beta_1 y \implies \frac{\partial y / \partial x}{y} = \beta_1.$$</span></p>
<p>So we see that <span class="math-container">$\beta_1$</span> is the marginal change in <span class="math-container">$y$</span> due to infinitesimal changes in <span class="math-container">$x$</span> as a proportion of its level. Therefore the accuracy of the approximation</p>
<p><span class="math-container">$$\beta_1 = \frac{\partial y / \partial x}{y} \approx \frac{\Delta y / \Delta x }{y}$$</span></p>
<p><span class="math-container">$$\implies \Delta x = 1: \beta_1 \approx \frac{\Delta y  }{y}$$</span></p>
<p>is nothing else than the general approximation inaccuracy issue that arises when we replace a derivative (infinitesimal change) with a disrete-interval proportional change.</p>
<p>In the specific case we have</p>
<p><span class="math-container">$$y(x+1) - y (x) = \exp\{ \beta_0 + \beta_1 x + \beta_1 +u\} - \exp\{ \beta_0 + \beta_1 x + u\}$$</span></p>
<p><span class="math-container">$$ = y(x)\cdot (e^{\beta_1 -1}) \implies \frac{\Delta y(x+1)}{y(x)} = (e^{\beta_1} -1).$$</span></p>
<p>We get</p>
<p><a href="https://i.sstatic.net/TxriF.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/TxriF.png" alt="enter image description here" /></a></p>
<p>and we verify the folk wisdom that the approximation is accurate enough for purpose of economic analysis if <span class="math-container">$\beta_1 \in [-0.1,\; 0.1]$</span>, and maybe for a larger interval, if one percentage point is not critical for the purposes of the specific research.</p>

CC BY-SA 4.0
2020-11-29T17:54:15.783Z
2020-11-29T21:16:47.987Z
2020-11-29T21:16:47.987Z
Alecos Papadopoulos
<p><a href="https://i.sstatic.net/PI6RJ.png" rel="nofollow noreferrer">Ok, so something that's common to the econometrics literature is that we interpret the coefficients in OLS log-linear models like such.</a>
To spell it out in the main body:</p>
<p><span class="math-container">$ln(y_i)=\beta_0+\beta_1X+u_i \Rightarrow \text{if } \Delta x = 1, \text{then } \text{%}\Delta y \approx 100\beta_1 $</span></p>
<p>I think that this is a very bad approximation, although my reasoning is probably incorrect (although I do understand the derivation of why this approx holds).</p>
<p>Ok, so an aside:</p>
<p><span class="math-container">$\frac{\dot{y(t)}}{y(t)} = g \Rightarrow ln(y_t) = gt + c $</span></p>
<p>It also follow that:</p>
<p><span class="math-container">$y_{t+1} \approx y_t (1+g \Delta t) $</span></p>
<p>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.</p>
<p>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 -&gt; 100% change in y (g and beta_1 are analogous, so g = 1 -&gt; beta_1 =1) -&gt; y approx doubles with each change in x -&gt; 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).</p>
<p>I hope this makes sense.</p>

<econometrics>
Log-linear ordinary least squares specification coefficient loses its interpretation when we change an explanatory variable by a unit?
threelinewhip
# Log-linear ordinary least squares specification coefficient loses its interpretation when we change an explanatory variable by a unit?

(Asked by: username_0 on 2020-11-28)

[Ok, so something that's common to the econometrics literature is that we interpret the coefficients in OLS log-linear models like such.](https://i.sstatic.net/PI6RJ.png)
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.

**Tags:** econometrics

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## 2 answers

### Answer by username_1 on 2020-11-29. Score: 2 | Accepted answer

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}

> **Comments:**
> **username_1**: @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?  
> **username_2**: 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  
> **username_1**: Right, it seems OP did omit hats in his/her example. But I guess the message is clear with this also.  
> **username_0**: 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).  

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### Answer by username_3 on 2020-11-29. Score: 4

When

$$\ln y = \beta\_0 + \beta\_1 x + u \implies y = \exp\{ \beta\_0 + \beta\_1 x + u\}$$

$$\implies \partial y / \partial x = \beta\_1 y \implies \frac{\partial y / \partial x}{y} = \beta\_1.$$

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](https://i.sstatic.net/TxriF.png)](https://i.sstatic.net/TxriF.png)

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.


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