If you think this is a duplicate, please have a look at the last paragraph.

In a regression model where both dependent ($Y$) and independent ($X$) variable are in natural logs, what is the exact interpretation of the coefficient of $X$?

Take the following simple model:

$\ln Y = \beta_0 + \beta_1 \ln X + \varepsilon$

I encountered two different rules for the interpretation:

  • A $d\,\%$ increase in $X$ is associated with an $d\cdot\beta_1$ percent increase in $Y$. ("Elasticity Interpretation")
  • A $d\,\%$ increase in $X$ is associated with an $\left(\exp (\beta_1 \cdot \ln a) -1\right)\cdot 100$ percent increase in $Y$, where $a = (100 + d)/100$.
    • Source 1 (PDF) (p. 4)
    • Source 2: UCLA (last section). (The notation is different but equivalent as $a^{\beta_1} = \exp(\beta_1 \cdot \ln a)$.

To rule out that both interpretations are identical, plug in (for example) $d = 50$ and $\beta_1 = 2$:

  • Result using rule 1: $Y$ increases by $50\cdot 2 = 100\,\%$.
  • Result using rule 2: $Y$ increases by $\left( \exp (2 \cdot \ln 1.5) - 1 \right) \cdot 100 = 125\,\%$.

My guess is that the first interpretation is only approximately correct for small $d$ and $\beta_1$, but this is precisely the question: Which of the two rules is correct? (And why is the other one wrong?)

I am aware of this question, but please note that this is rather a follow-up question than a duplicate. I also know that there is that question which is very similar but unanswered. I considered improving the existing question, but there is a lot of (IMHO) superfluous information in the text that makes it unattractive to answer and I don't see how I could remove these parts without violating the original author's intent.


2 Answers 2


Yes the first is an approximation.

Using units of percent, a $d\%$ increase in X can be quantified by $cX$ where $c=1+d/100$. Then

$$\beta_0+\beta_1\ln(cX)+\epsilon=\beta_0+\beta_1\ln(c)+\beta_1 \ln(X)+\epsilon=\beta_1\ln(c)+\ln(Y)=\ln(c^{\beta_1}Y).$$

This corresponds to a $100(c^{\beta_1}-1)=100(\exp(\beta_1\ln(c))-1)$ percent increase in $Y$. If $c$ is close to 1, or equivalently, $d/100$ is close to 0, then:

$$100(\exp(\beta_1\ln c)-1)\approx 100+100\beta_1\ln(c)-100\approx d\beta_1,$$

where we are using the taylor approximation $\ln(1+x)\approx x$ for $x$ close to 0. The last expression corresponds to an approximate increase by $d\beta$.

  • $\begingroup$ But this only holds as an approximation if $d/100$ is close to 0, which is not very useful in the context of log-log regression and interpretation of a statistical model. So it is incorrect way of talking about the effect of the predictor. $\endgroup$ Mar 17, 2016 at 19:17
  • $\begingroup$ @DaltonHance: For $100\log(1+d/100)$, the error is bounded by $d^2/200$, which even if exponentiated is not completely unreasonable for $d\leq 3$. $\endgroup$
    – Alex R.
    Mar 17, 2016 at 20:18
  • $\begingroup$ We are interested in the formulating the statement "For an d% percent increase in X, the expected value of Y increases by z%" from a statistical model. An approximate method based on a taylor series is very limited for making such a statement, so limited as to be completely unusable. I see no reason to prefer it, especially when we have an exact way of determining that statement. I'm not disputing the math, just saying for the purpose of inference it's not very helpful. What if we want to talk about a 5% increase in X? $\endgroup$ Mar 17, 2016 at 20:58
  • $\begingroup$ You're asking to fit a peg in a square hole. If you want to talk about a 5% increase then you wouldn't use the approximation. There are plenty of situations I have encountered where you would be interested in a 1-3% increase and would like a quick estimate. This reasoning is extremely useful in assessing complicated nonlinear models. $\endgroup$
    – Alex R.
    Mar 17, 2016 at 21:27
  • $\begingroup$ OP did not specify that he/she was interested in an approximation that works only when considering $d<3$. Even if he/she were, I fail to see how $1.01*\beta_1$ is more useful than $1.01^{\beta_1}$. It's a trivial difference in terms of computation and the second is more accurate, more general, and is derived directly from the form of the statistical model. $\endgroup$ Mar 17, 2016 at 22:09

The second rule is correct for log-log regression, the first is incorrect.

Depending on your interpretation, it looks like you've actually encountered three different rules for interpretation of the $\beta_1$ term in log-log regression. Source 2 UCLA, under your second "rule" is the correct rule for log-log regression, but it is not mathematically equivalent to what you have given as the second "rule": $a^{\beta_1} = e^{\beta_1 \mathrm{ln}(a)} \neq e^{\beta_1 \mathrm{ln}(a)} - 1$.

But I think this demonstrates that your question is more a confusion of language than a confusion of math and so I want to spend some time to explain that.

To see this consider, $$\mathrm{ln}(Y_0) = \beta_0 + \beta_1 \mathrm{ln}(X_0)$$ and $$\mathrm{ln}(Y_1) = \beta_0 + \beta_1 \mathrm{ln}(X_0*d)$$

Note that we have ignored the error term here and so properly what we are considering is the expected value of $Y$ at $X_0$ and some new value $X_0*d$. Note also that if you want to talk about a percent increase then $d$ needs to be some number greater than 1. That is, a 50% increase is equivalent to $d = 1.5$. I think you may be stumbling in your thinking in the translation between ratios and percentages. So if you want to translate the English language phrase "a z% increase" to the ratio $d$, you would find $d = (100 + z)/100$. It can be a little trickier to think about decreases, but the math is similar: a z% decrease is equivalent to $d = (100 - z)/100$. For example, a 25% decrease in $X$ means that $X$ is 75% of its original value, and 0.75 is the number you need to use to translate that English language phrase to the correct formula.

So after that tricky business of translating English to math it is easy to see that 50% increase in $X$ results in a $1.5^{\beta_1}$ increase in $Y$. Returning to our example see that:

$$ \mathrm{ln}(Y_1) - \mathrm{ln}(Y_0) = (\beta_0 + \beta_1 \mathrm{ln}(X_0*d)) - (\beta_0 + \beta_1 \mathrm{ln}(X_0)) $$

$$\mathrm{ln}(Y_1) - \mathrm{ln}(Y_0) = \beta_1 (\mathrm{ln}(X_0) + \mathrm{ln}(d)) - \beta_1 \mathrm{ln}(X_0)$$

$$\mathrm{ln}(Y_1) - \mathrm{ln}(Y_0) = \beta_1 \mathrm{ln}(d) $$

$$\mathrm{ln}(\frac{Y_1}{Y_0}) = \beta_1 \mathrm{ln}(d) $$

$$\frac{Y_1}{Y_0} = d^{\beta_1} $$

So notice that the ratio of our new $Y$ to our old $Y$ is equal to ratio of the new $X$ to our old $X$ raised to the $\beta_1$ power. In terms of your numerical example a 50% increase in $X$ with $\beta_1 = 2$ means that $d= 1.5$ and so $\frac{Y_1}{Y_0} = 1.5^2 = 2.25$. This is where language comes back in to the problem. $Y_1$ is 225% of it's original value, but that may not mean the same thing in English as a 225% increase. If we take a 50% increase to mean a ratio of 1.5, then it make sense that we should return to our formula for translating percentages to ratio and subtract 1 from our ratio to talk about a percentage increase, i.e. $2.25 - 1 = 1.25$ or a 125% increase.

  • $\begingroup$ Thank you for your illustrative explanation - but "translating English to math" was not my problem (I think). The reason why I listed UCLA under "Rule 2" was that I already realized that the two sources under "Rule 2" lead to the same result, although using different notations/wordings. $\endgroup$
    – CL.
    Mar 21, 2016 at 18:13

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