How to interpret regression equations with logarithms, based on log difference being approximate to percentage change?

Question

$y = 4 + 2.5\,x + u$

For an increase of 1 unit of $X$ (that is, $X$ to $X+1$), we expect an increase $2.5$ units of $Y$ (that is, $Y$ to $Y+2.5$).

Is that right?

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What if there's a/an $\ln$?

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$\ln(y) = 1.7 + 0.083\,x + u$

For an increase of 1 unit of $X$ (that is, $X$ to $X+1$), we expect:

a/an $8.3\%$ increase in $Y$ (that is, $Y$ to $Y(1+8.3%)$)?

a/an $0.083\%$ increase in $Y$ (that is, $Y$ to $Y(1+0.083%)$)?

The above is supposed to make use of the fact that log difference is approximately equal to percentage change.

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$\ln(y) = 4 + 0.083\,\ln(x) + u$

For a $1\%$ increase in $X$ (that is, $X$ to $X(1+1%)$), we expect:

a/an $8.3\%$ increase in $Y$ (that is, $Y$ to $Y(1+8.3%)$)?

a/an $0.083\%$ increase in $Y$ (that is, $Y$ to $Y(1+0.083%)$)?

The above is supposed to make use of the fact that log difference is approximately equal to percentage change.

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Apologies for the confusion everyone. I am wondering about this part in Introductory Econometrics by Wooldridge. It's been a few years since I've interpreted SLR coefficients.

Why is it in the first part $0.083$ becomes $8.3%$ (instead of $0.083%$) but in the second part $0.257$ becomes $0.257%$ (instead of $25.7%$)?

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Answer 1

I think you can solve this with exponent and logarithm rules.

For the first model, $y = 4 + 2.5\,x + u$, a $1$-unit increase in $x$ will lead to a $2.5$ increase in $y$, simply because $2.5\,(x + 1) = 2.5\,x + 2.5$.

For the second model, $\ln(y) = 1.7 + 0.083\,x + u$ a $1$-unit increase in $x$ will, by the same token, increase the LHS of the equation $0.083$. But the equation renders the $\ln(y)$, not $y$. Hence, it will be the $\ln(y)$ that will be increased $0.083$. To know the increase in $y$ we will have to exponentiate: $e^{0.083} = 1.086$.

For the third model, $\ln(y) = 4 + 0.083\,\ln(x) + u$, an increase of now (as per the OP) $1\%$ in $x$ will lead to $\ln(1.01\,x)$ as opposed to $\ln(x)$ on the RHS of the equation, or $\ln(1.01\,x)=\ln(1.01)\,+\,\ln(x)$. This will increase the dependent variable by $\ln(1.01)$. But since the dependent variable is in $\log$ scale, we exponentiate to obtain the change in $y$, which is, naturally, $e^{\ln(1.01)}=1.01$.

Answer 2

For a 1-unit increase in $X$, we expect an 8.3% increase in $Y$

Not quite, because the calculation goes

$$\ln(Y') = 1.7 + 0.083(X + 1) = (1.7 + 0.083X) + 0.083 = \ln(Y) + 0.083$$

so a 1-unit increase in X changes $\ln(Y)$ to

$$\ln(Y) + 0.083 = \ln(Y) + \ln(e^{0.083}) = \ln(e^{0.083}Y) $$

So since $$e^{0.083} \approx 1.0865$$ it actually leads to about an 8.7% increase in $Y$.

It's worth noting that $e^x \approx 1 + x$ for small values of $x$, so your equation is approximately true, for small values of $0.083$.

I take it you can work out the last example from here.

Does Wooldridge have a typo? The former makes 0.083 into 8.3% and not 0.083% while the latter makes 0.257 into 0.257% and not 25.7%

I don't think so.

If

$$ \ln(Y) = 4.8 + 0.257 \ln(X) $$

then a 1% increase in $X$ gives a new $Y'$ as

$$ \ln(Y') = 4.8 + 0.257 \ln(1.01 X) = \ln(Y) + 0.257 \ln(1.01) $$

We can manipulate the right hand side

$$ \ln(Y) + 0.257 \ln(1.01) = \ln(Y) + \ln(1.01^{0.257}) = \ln(1.01^{0.257} Y) $$

So $Y' = 1.01^{0.257} Y$. Google says $1.01^{0.257} \approx 1.0026$, which is $0.26\%$, as stated.

Answer 3

You can derive all answers yourself by using a simple intuition. Here how it goes. We start with the definition of the derivative: $$y(x)'\approx\frac{\Delta y(x)}{\Delta x}$$

Next we differentiate both sides of the linear model $y=b_0+b_x x$ and get this: $$\frac{\Delta y(x)}{\Delta x}\approx b_x $$ $$\Delta y(x)\approx b_x \Delta x$$

Then we say the magic phrase "per unit increase in x", which is the same as saying $\Delta x=1$, and proceed to the simplified formula $$\Delta y(x)\approx b_x $$

For the log we notice the following: $$\left(\ln y(x)\right)'=\frac{y'}{y}\approx\frac{\Delta y(x)}{y(x)\Delta x}$$

Then for a model $\ln y=b_0+b_1 x$ we get the following by differentiating both sides of the equation: $$\frac{\Delta y}{y}\approx b_x \Delta x$$ or for a unit increase in x: $$\frac{\Delta y}{y}\approx b_x $$ The left hand side can be interpreted as "percent change" but expressed in decimals, so 0.1 would mean 10%.

I'm sure you can pick up from here. It's easy to see what happens when the log is on the right hand side with $\ln x$. It's very similar. Your confusion was in the meaning of the "percent change", which I already explained and shown that it corresponds to terms like $\frac{\Delta s}{s}$

Answer 4

You asked a number of questions. I will paraphrase each, and then answer them.

First question: If $f(x) = 4 + 2.5 x + u$, what is $f(x+1)-f(x)$? The answer is 2.5, for all $x$.

Second question: If $f(x) = \exp(1.7+0.083 x+u)$, what is $f(x+1)-f(x)$?

$$ \begin{eqnarray} f(x+1)-f(x) & = & \exp(1.7+0.083(x+1)+u) - \exp(1.7+0.083 x+u) \\ & = & \exp(1.7+0.083 x+u+0.083) - \exp(1.7+0.083 x+u) \\ & = & \exp(1.7+0.083 x+u) \cdot \exp(0.083) - \exp(1.7+0.083 x+u) \\ & = & [\exp(0.083)-1] \cdot \exp(1.7+0.083 x+u) \\ &\approx& 0.083 \cdot \exp(1.7+0.083 x+u) \\ &=& 0.083 \cdot f(x) \\ &=& 8.3\% \cdot f(x) \end{eqnarray} $$

Where in the $\approx$ step I used the approximation (which comes from the Taylor series) that for small $x$, $\exp(x) \approx 1+x$. So if $x$ increases by 1, $f(x)$ increases by 8.3%.

Third question: Here your question is a little unclear (there's at least one typo), but I think what you want to ask is: If $f(x) = \exp(4 + 0.083 \ln(x) + u)$, what is $f(1.01 \cdot x)-f(x)$? This would correspond to the question of how much $f(x)$ changes for a 1% increase in $x$.

$$ \begin{eqnarray} f(1.01\cdot x)-f(x) &=& \exp(4 + 0.083 \ln(1.01 x) + u) - \exp(4 + 0.083 \ln(x) + u) \\ &=& \exp(4 + 0.083 (\ln(1.01)+\ln(x)) + u) - \exp(4 + 0.083 \ln(x) + u) \\ &=& \exp(4 + 0.083 \ln(x) + u) \cdot \exp(0.083 \ln(1.01)) - \exp(4 + 0.083 \ln(x) + u) \\ &=& [\exp(0.083 \ln(1.01))-1] \cdot \exp(4 + 0.083 ln(x) + u) \\ &\approx& [\exp(0.083 \cdot 0.01)- 1] \cdot f(x) \\ &\approx& (1 + 0.083 \cdot 0.01 - 1) \cdot f(x) \\ &=& 0.083 \cdot 0.01 \cdot f(x) \\ &=& 0.083\% \cdot f(x) \end{eqnarray} $$

Where at the first $\approx$ I used the approximation that $\ln(1+x) \approx x$ for small $x$, and at the second $\approx$ I again used the approximation that $\exp(x) \approx 1+x$ for small $x$. (Of course, these are really the same approximation, but one stated in terms of $\exp()$ and one in terms of $\ln()$.)

So the answer is that a 1% increase in $x$ will cause a 0.083% increase in $f(x)$.

Fourth question: Does Wooldridge have a typo? No. If you compare the answers to your second and third question, it should now be clear why this is.