# On the interpretation of the coefficients in a log-log model as an average

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.

• 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 $\Delta\ln(.)$ ? May 8 '18 at 23:03
• @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. May 9 '18 at 8:37

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 = $\beta \Delta x$.

• 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 $E(\log(y)|X)=\beta_1+\beta_2 x$. May 9 '18 at 15:40
• 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. May 9 '18 at 15:54
• Chan, I've seen your simulation. If I do the same, but not include the error terms, then both means are actually pretty close.... May 9 '18 at 16:03
• 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. May 9 '18 at 16:09