3
$\begingroup$

Is this equation:

$$\log{(y)} = a + bx$$

semi-log or log-linear mode (or it is the same thing)?

I have two models: linear (1) and semi-log (2). The values of $R^{2}$, adjusted $R^{2}$, and Standard Error are:

  • Linear: $R^{2} = 0.6780,~\mathrm{adj.}~R^{2} = 0.6513,~~\mathrm{SE}=94.101$
  • Semi-log: $R^{2} = 0.5803,~\mathrm{adj.}~R^{2} = 0.5455,~~\mathrm{SE}=0.5493$

How to interpret this values especially from the second model?

$\endgroup$

3 Answers 3

4
$\begingroup$

This is an answer to the first part of the question regarding the description of the model: $$\log{(y)} = a + bx.......(1)$$ It is important to distinguish: i) whether a model is linear in the sense of the Classical Linear Regression Model (CLRM), and ii) whether a model has linear functional form. Model (1) is linear in the first sense because it is linear in the parameters $a$ and $b$, and this is not affected by the log of $y$. Similarly, models (2), (3) and (4) below are all linear in the CLRM sense: $$y = a + bx.......(2)$$ $$y = a + b*log(x).......(3)$$ $$log(y) = a + b*log(x).......(4)$$ However, of the above models only Model (2) has linear functional form. Models (1) and (3) could both be said to have semi-log functional form, although it is better I suggest to be more precise and indicate which variable is logged by describing (1) as semi-log (dependent) and (3) as semi-log (independent). The functional form of Model (4) is sometimes described as log-linear and sometimes as double log.

$\endgroup$
1
  • 1
    $\begingroup$ @Navi Welcome to the site. I am appreciative of your rapid acceptance of my answer - 3 minutes may be some sort of record!. What most participants do however is allow time both for feedback on answers, via votes and comments, and for any further answers, and then judge which answer is best. In this case, moreover, my answer only addressed the first part of your question. So please feel free to accept another answer, and unaccept mine, if you have second thoughts. $\endgroup$ Jul 17, 2014 at 22:14
1
$\begingroup$

I have never heard the term "semi-log regression" in 20 years. It may be in use in some substantive areas. Log linear analysis is something else - it is used when you have multiple categorical variables.

Both of your models are linear regressions. It's just that the second uses the log of y rather than y. $R^2$ has the same meaning as usual - it is the proportion of variance in y explained by the model. Adjusted $R^2$ is one way of penalizing for complexity. Since your model has only one independent variable, it is very close to the unadjusted $R^2$

EDIT answer to comment For SE of the regression see this article. It is entirely reasonable that this changes a lot when you change the scale of the dependent variable.

$\endgroup$
14
  • $\begingroup$ @ Peter Flom Thanks for you answer. How to interpret SE values i.e. such a great difference between this two? Can SE be 94.101 when R2 = 0.6780 is much lower? $\endgroup$
    – Quirik
    Jul 17, 2014 at 9:41
  • $\begingroup$ What is the SE value? What is it the standard error of? $\endgroup$
    – Peter Flom
    Jul 17, 2014 at 15:00
  • $\begingroup$ @Navi to clarify Peter's last comment: in your equation I can off the top of my head think of four standard errors: (1) the SE of $\hat{a}$, (2) the SE of $\hat{b}$ the SE of $\hat{\log(y)}$ (i.e. the predicted value of the regression line), and the SE of $\tilde{\log(y)}$ (i.e. of the predicted value of $\log(y)$ for a given value of $x$. Which SE is your question about? $\endgroup$
    – Alexis
    Jul 17, 2014 at 16:39
  • $\begingroup$ @Peter Flom Standard error of regression $\endgroup$
    – Quirik
    Jul 17, 2014 at 16:43
  • $\begingroup$ OK, I will edit my answer. $\endgroup$
    – Peter Flom
    Jul 17, 2014 at 17:37
0
$\begingroup$

Semi-log and log-linear models are terminologies used in Econometrics. They are not the same.

Log-linear models $$ Y = e^{\beta _1 }X_2 ^{\beta _2 }X_3 ^{\beta _3 }e^\epsilon $$ it is linearized as $$ \log Y = \beta_1 + \beta_2 \log X_2 +\beta_3 \log X_3 +\epsilon $$ Here $\beta$-coefficients are elasticities. For this reason, the log-linear model is also known as the "constant elasticity model".

Semilog-linear models

these are useful when coefficients are interpreted as relative variations (rate of change). The original model is $$ Y = \exp\left(\beta _1 +\beta _2 X_2 + \beta _3 X_3 + \epsilon\right) $$ and it is linearized as $$ \log Y = \beta _1 +\beta_2 X_2 + \beta_3 X_3 + \epsilon $$ In this case $$ 100 \times (\exp(\beta_k) - 1) $$ measures approximately the expected percentage change in $Y$ for a unit change in $X_k$ holding all else equal.

R squared interpretation

If you fit a linear model that is a linearized version of a non-linear model the $R^2$ must be carefully handled.

The $R^2$ is the fraction of variability of the outcome variable (or dependent variable) captured by the regression function. But the previous is referred to the linear model.

Now, for the previous log-linear model the outcome variable is $\log Y$ while the regression function is $\beta_1 + \beta_2 \log X_2 +\beta_3 \log X_3$

In the case of the semilog-linear model example, the outcome is again $\log Y$, while the regression function is $\beta _1 +\beta_2 X_2 + \beta_3 X_3$.

Side comment on the use of the $R^2$

The $R^2$ is one of the most dangerous tools. It's almost worthless and should be removed from any introductory-intermediate statistics book. People use it following silly rules of thumb, but there are serious technical issues why its misuse can be dangerous.

Intuition: suppose you run a regression and you find a spectacular $R^2 = 0.999999999999....$ approaching its "ideal maximum value". You are happy, but this means that the estimated variance of error term in the regression is close to 0... this is a violation of the assumption that your observed outcome $Y$ is a random fluctuation around the conditional mean of $Y|X$ represented by the (linear) regression function.

In the best case, a too good $R^2$ is an indication that you are fitting noise, then the question is: how large should be $R^2$ before to worry about it? Nobody knows the answer, because this depends on the bias-variance trade-off. Since of this I strongly suggest looking at other better measures of fitting.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.