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I came across the following in explaining the log-linear regression model.

Given the model $\log(Y_i) = β_0 + β_1X_i + u_i$

The expected value of $\log(Y)$ given $X$ is $β_0 + β_1X$.

So far, so good. But then it says:

'When $X$ is $X+ΔX$, the expected value is given by $\log(Y+ΔY)$'.

I don't see why this is necessarily the case. Could someone explain why $\log(Y+ΔY) = β_0 + β_1(X+ΔX)$?

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  • $\begingroup$ I edited slightly the formatting, if you feel it's wrong feel free to revert the changes $\endgroup$
    – Firebug
    Commented May 24, 2022 at 11:59
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    $\begingroup$ can you add the full quote. they haven't defined $\Delta Y$, so can't you solve for $\Delta Y$ $\endgroup$
    – seanv507
    Commented May 24, 2022 at 12:21
  • $\begingroup$ I would warmly advise consulting other sources about log-linear regressions. The reason is that the quoted statement is silly insofar as it simultaneously asks us to suppose $\log Y$ is a linear function of $X$ (that's the model) and to think of changes in $Y$ as being additive (that's what "$Y+\Delta Y$" means). These are contradictory approaches. The problem is that changes in $Y$ should be thought of and expressed multiplicatively rather than additively. Anyone creating such confusion is not going to be a good source for insight or correct information. $\endgroup$
    – whuber ♦
    Commented May 24, 2022 at 18:56

2 Answers 2

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Your question is complete without defining what $\Delta Y$ is. Regardless, a tick thing about log-linear regression is that if you want to get the expected value after anti-log transformation, a bias has to be corrected. Details are given in many classical papers (e.g., Beauchamp, J.J. and Olson, J.S., 1973. Corrections for bias in regression estimates after logarithmic transformation. Ecology, 54(6), pp.1403-1407; Sprugel, D. G. "Correctiong for bias in log-transformed allometric equations." Ecology 64 (1983): 209-210; Newman MC. Regression analysis of log‐transformed data: Statistical bias and its correction. Environmental Toxicology and Chemistry: An International Journal. 1993 Jun;12(6):1129-33.)

Here is the gist: suppose that $ Z=log(Y)=\beta_0+\beta_1*X+u$. The expected value of Z or log )Y) is simply $E[Z]=E[log(Y)]=\beta_0+\beta_1*X$, but the expectation of Y is not $exp(\beta_0+\beta_1*X)$, but a more correct one is $E(Y)=exp(\beta_0+\beta_1*X+{\sigma}^2/2)$. Whatever you are trying to derive, if it involves anti-log transformation and taking expectation, this bias correction should not be ignored.

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    $\begingroup$ Because the question is about log-log regression, rather than linear regression in the original variables, the issue is not about estimating the expectation of the original variables. The question is explicitly about the conditional expectation of $\log Y$ rather than of $Y.$ $\endgroup$
    – whuber ♦
    Commented May 24, 2022 at 18:40
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IF the model is $\log(Y_i) = β_0 + β_1X_i + u_i$ then it follows that for another $X_j$

$$\log(Y_j) = β_0 + β_1X_j + u_j$$

We can define then $Y_j= Y_i + \Delta Y $, with $ (Y_j > 0, Y_i > 0)$, and $X_j=X_i + \Delta X$. This implies that

$$\log(Y_i + \Delta Y) = β_0 + β_1(X_i + \Delta X) + u_j$$

Notice, however, the error term in this form: it's $u_j$, not $u_i$. You can make arguments regarding the expected value from here, given $X_i, \Delta X$ and model assumptions.

As @whuber rightfully pointed in the comments, this is a general result that is valid under linearity (of the independent variables). You can see this because the solution is completely agnostic to the fact that the link function is a logarithm. It would've worked the same under any function.

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    $\begingroup$ It's unnecessary to introduce the exponential: the statement in the question is a completely general interpretation of any model that is linear in the explanatory variable. Taking a round trip through two nonlinear transformations only obscures this basic fact. $\endgroup$
    – whuber ♦
    Commented May 24, 2022 at 18:06
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    $\begingroup$ I see your point, we can directly jump from $Y_j=Y_i+\Delta Y$ to the result. I didn't put much thought into this tbh @whuber. Since I knew the result was right, then it was only a matter of brute-forcing mathematically. But I'll edit this $\endgroup$
    – Firebug
    Commented May 24, 2022 at 18:18
  • $\begingroup$ @Firebug What do you mean 'we can directly jump from Yj=Yi+ΔY to the result.' Are you saying there is a simpler derivation than what you wrote? $\endgroup$
    – R.S.
    Commented May 25, 2022 at 20:16
  • $\begingroup$ @R.S. No, the current derivation is the one whuber alluded to. You can check the revisions of the answer to see the previous one $\endgroup$
    – Firebug
    Commented May 25, 2022 at 23:16

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