<p>The short answer is no, this is not so when the dependent variable is in log, unless you <em>define</em> the $\%\Delta$ as (100 times) log-difference. The reason is that $E(\log y) \ne \log E(y)$.</p>

<p>We don't need a multiple regression model to show this so consider $\log(y)=\beta_0 + \beta_1 x + u$. It is the <strong>dependent variable in log</strong> that matters, so let us not take log to $x$ for notational brevity.</p>

<p>It is true that $\%\Delta y \approx 100 \beta_1 \Delta x$ if $\Delta u = 0$ (<em>ceteris paribus</em>) when $\beta_1 \Delta x \approx 0$. But we cannot interpret it in terms of $E(y|x)$. The reason follows.</p>

<p>As you can see, we have $y = \exp(\beta_0 + \beta_1 x + u)$. When $x$ increases from $x_1$ to $x_1+\Delta x$, $y$ increases from $y_1 = \exp(\beta_0 + \beta_1 x_1 + u_1)$ to $y_2 = \exp \{ \beta_0 + \beta_1 (x_1+\Delta x)+u_2 \}$. <strong>Importantly,</strong> $u$ can also change because $u$ is not held fixed, which is why I wrote $u_1$ and $u_2$.</p>

<p>Now, what is the change rate of $y$? We have
$$
\frac{y_2}{y_1} -1 = \exp(\beta \Delta x) \exp(\Delta u) -1 \approx (1+\beta \Delta x) \cdot \exp(\Delta u) -1,
$$
where $\Delta u = u_2-u_1$. Here, the left-hand side is the change rate of $y$. You can verify that the change rate of $y$ is approximately $\beta \Delta x$ if $u_1 = u_2$, i.e., if $u$ is held fixed so that $\exp(\Delta u) = 1$. But the problem is that $u$ is <strong>not held fixed</strong> so $\exp(\Delta u)$ can be <strong>very different from 1</strong>. When $x$ and $u$ are independent, the average growth rate is $\exp(\beta \Delta x) \cdot E(e^{\Delta u}) -1$.</p>

<p>How different would this be from $\beta \Delta x$? If $u\sim N(0,\sigma^2)$ and $u_1$ and $u_2$ are mutually independent, then $u_2-u_1 \sim N(0,2 \sigma^2)$, and thus $E(e^{u_2-u_1}) = \exp(\sigma^2)$. For example, if $\beta_1 \Delta x_1 = 0.01$ and $u\sim N(0,1)$, then the average growth rate of $y$ is $\exp(0.01) \times e - 1 \approx 1.75$, that is, approximately 175%, which is hugely different from 1%. Interesting, isn't it?</p>

<p><strong>Simulations:</strong> Use R to do the following simulation.</p>

<pre><code>set.seed(1)
n &lt;- 1e6
x1 &lt;- rnorm(n)
x2 &lt;- x1+1
y1 &lt;- exp(1+0.01*x1 + rnorm(n))
y2 &lt;- exp(1+0.01*x2 + rnorm(n))
100*mean(y2/y1-1)
# 175.2406
mean(log(y2)-log(y1))
# 0.01004419
</code></pre>

<p>When $\beta_1 \Delta x = 0.01 \times 1 = 0.01$, the average growth rate is not 1% but 175%. But $\Delta E(\log y)$ is approximately 0.01.</p>

<p>Note 1: The above discussion is for the log-level model. Little is changed for the log-log model.</p>

<p>Note 2: If you define the growth rate as the log-difference, the expected growth rate equals the expected log-difference = $\beta \Delta x$.</p>

CC BY-SA 4.0
2018-05-09T15:40:02.233Z
Chan, thanks for your answer. However, I'm not sure I agree with you on whether we should include the error terms in the computations, since $E(\log(y)|X)=\beta_1+\beta_2 x$.
An old man in the sea.
CC BY-SA 4.0
2018-05-09T15:54:14.300Z
Your equation is true, and that’s the case of Note 2 in my answer. With the usual definition of % change, you have $E(y_2/y_1-1)$ for the expected % change (divided by 100), which inevitably involves the error distribution as I derived. Please see the simulation part which is more revealing.
chan1142
CC BY-SA 4.0
2018-05-09T16:03:38.633Z
Chan, I've seen your simulation. If I do the same, but not include the error terms, then both means are actually pretty close....
An old man in the sea.
CC BY-SA 4.0
2018-05-09T16:09:13.200Z
That’s for $\Delta u = 0$ I mentioned in the third paragraph. In that case they are close to each other as you correctly found.
chan1142
CC BY-SA 4.0
2018-05-09T11:02:02.720Z
2018-05-09T11:02:02.720Z
chan1142
<p>In the linear regression model $E(y|X)=x'\beta$, and so we're capable of interpreting a $\beta_i$ as $\Delta E(y|\text{all fixed except } x_i)=\beta_i \Delta x_i$. I'm thinking in discrete changes.
Hence, we're able to interpret the estimator of $\beta_i$ as an estimation of what happens on average.</p>

<p>I was wondering if this interpretation of the coefficients' estimators is also preserved when we have a log-log model... $ E(\%\Delta y|\text{all fixed except } x_i)\approx \beta_i \%\Delta x_i$</p>

<p>Any help would be appreciated.</p>

CC BY-SA 4.0
2018-05-08T23:03:50.467Z
To me, without any $\Delta$, a log-log model could already be considered as  the continuous-variation interpretation version of that you give, which, that being said, is a kind of partial derivative. Do you precisely mean $\Delta\ln(.)$ ?
keepAlive
CC BY-SA 4.0
2018-05-09T08:37:56.313Z
@Kanak I'm not sure I understand you... Yes, we can have $\Delta \log(.)$, but then we wouldn't have the interpretation of the coefficient as an average for sure. I was wondering if we could say "on average" when interpreting the coefficient of a log-log model.
An old man in the sea.
<econometrics>
On the interpretation of the coefficients in a log-log model as an average
An old man in the sea.
# On the interpretation of the coefficients in a log-log model as an average

(Asked by: username_0 on 2018-05-08)

In the linear regression model $E(y|X)=x'\beta$, and so we're capable of interpreting a $\beta\_i$ as $\Delta E(y|\text{all fixed except } x\_i)=\beta\_i \Delta x\_i$. I'm thinking in discrete changes.
Hence, we're able to interpret the estimator of $\beta\_i$ as an estimation of what happens on average.

I was wondering if this interpretation of the coefficients' estimators is also preserved when we have a log-log model... $ E(\%\Delta y|\text{all fixed except } x\_i)\approx \beta\_i \%\Delta x\_i$

Any help would be appreciated.

**Tags:** econometrics

> **Comments:**
> **username_1**: To me, without any $\Delta$, a log-log model could already be considered as  the continuous-variation interpretation version of that you give, which, that being said, is a kind of partial derivative. Do you precisely mean $\Delta\ln(.)$ ?  
> **username_0**: @Kanak I'm not sure I understand you... Yes, we can have $\Delta \log(.)$, but then we wouldn't have the interpretation of the coefficient as an average for sure. I was wondering if we could say "on average" when interpreting the coefficient of a log-log model.  

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## 1 answers

### Answer by username_2 on 2018-05-09. Score: 1

The short answer is no, this is not so when the dependent variable is in log, unless you define the $\%\Delta$ as (100 times) log-difference. The reason is that $E(\log y) \ne \log E(y)$.
We don't need a multiple regression model to show this so consider $\log(y)=\beta_0 + \beta_1 x + u$. It is the dependent variable in log that matters, so let us not take log to $x$ for notational brevity.
It is true that $\%\Delta y \approx 100 \beta_1 \Delta x$ if $\Delta u = 0$ (ceteris paribus) when $\beta_1 \Delta x \approx 0$. But we cannot interpret it in terms of $E(y|x)$. The reason follows.
As you can see, we have $y = \exp(\beta_0 + \beta_1 x + u)$. When $x$ increases from $x_1$ to $x_1+\Delta x$, $y$ increases from $y_1 = \exp(\beta_0 + \beta_1 x_1 + u_1)$ to $y_2 = \exp \{ \beta_0 + \beta_1 (x_1+\Delta x)+u_2 \}$. Importantly, $u$ can also change because $u$ is not held fixed, which is why I wrote $u_1$ and $u_2$.
Now, what is the change rate of $y$? We have
$$
\frac{y_2}{y_1} -1 = \exp(\beta \Delta x) \exp(\Delta u) -1 \approx (1+\beta \Delta x) \cdot \exp(\Delta u) -1,
$$
where $\Delta u = u_2-u_1$. Here, the left-hand side is the change rate of $y$. You can verify that the change rate of $y$ is approximately $\beta \Delta x$ if $u_1 = u_2$, i.e., if $u$ is held fixed so that $\exp(\Delta u) = 1$. But the problem is that $u$ is not held fixed so $\exp(\Delta u)$ can be very different from 1. When $x$ and $u$ are independent, the average growth rate is $\exp(\beta \Delta x) \cdot E(e^{\Delta u}) -1$.
How different would this be from $\beta \Delta x$? If $u\sim N(0,\sigma^2)$ and $u_1$ and $u_2$ are mutually independent, then $u_2-u_1 \sim N(0,2 \sigma^2)$, and thus $E(e^{u_2-u_1}) = \exp(\sigma^2)$. For example, if $\beta_1 \Delta x_1 = 0.01$ and $u\sim N(0,1)$, then the average growth rate of $y$ is $\exp(0.01) \times e - 1 \approx 1.75$, that is, approximately 175%, which is hugely different from 1%. Interesting, isn't it?
Simulations: Use R to do the following simulation.
set.seed(1)
n <- 1e6
x1 <- rnorm(n)
x2 <- x1+1
y1 <- exp(1+0.01*x1 + rnorm(n))
y2 <- exp(1+0.01*x2 + rnorm(n))
100*mean(y2/y1-1)
# 175.2406
mean(log(y2)-log(y1))
# 0.01004419

When $\beta_1 \Delta x = 0.01 \times 1 = 0.01$, the average growth rate is not 1% but 175%. But $\Delta E(\log y)$ is approximately 0.01.
Note 1: The above discussion is for the log-level model. Little is changed for the log-log model.
Note 2: If you define the growth rate as the log-difference, the expected growth rate equals the expected log-difference = $\beta \Delta x$.

> **Comments:**
> **username_0**: Chan, thanks for your answer. However, I'm not sure I agree with you on whether we should include the error terms in the computations, since $E(\log(y)|X)=\beta_1+\beta_2 x$.  
> **username_2**: Your equation is true, and that’s the case of Note 2 in my answer. With the usual definition of % change, you have $E(y_2/y_1-1)$ for the expected % change (divided by 100), which inevitably involves the error distribution as I derived. Please see the simulation part which is more revealing.  
> **username_0**: Chan, I've seen your simulation. If I do the same, but not include the error terms, then both means are actually pretty close....  
> **username_2**: That’s for $\Delta u = 0$ I mentioned in the third paragraph. In that case they are close to each other as you correctly found.  

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