Logarithms and regression

Question

I have a regression where I am trying to predict Y, using several variables. My variables are A, B, C.

I'm using R for my analysis. My equation is as follows:

mod <-lm(  log(Y/B) ~ log(A/B) + log(A/C), data = dataframe)

and a summary of the model is as follows:

Coefficients:
                      Estimate Std. Error t value Pr(>|t|)    
(Intercept)          -1.326698   0.080683  -16.44   <2e-16 ***
log(A/B)              0.756779   0.007507  100.81   <2e-16 ***
log(A/C)             -0.292373   0.012762  -22.91   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 0.6925 on 17063 degrees of freedom
Multiple R-squared: 0.4043, Adjusted R-squared: 0.4042 
F-statistic:  5790 on 2 and 17063 DF,  p-value: < 2.2e-16 

I understand that log(j/k) = log(j) - log(k); so log(Y/B) can be written similarly. Is there a way that I can include the B variable in the dependent part of my regression equation?

Also, it is acceptable to use a variable for normalizing, on both dependent and independent variables? I have seen other examples where the same variable is used in this way, but I was not certain if this makes sense.

Answer 1

By virtue of properties of logarithms, the original model,

$$\log(Y/B) = \beta_0 + \beta_1\log(A/B) + \beta_2\log(A/C) + \varepsilon$$

($\varepsilon$ is a zero - mean random variable) can be written

$$\log(Y) - \log(B) = \beta_0 + \beta_1 (\log(A)-\log(B)) + \beta_2(\log(A)-\log(C)) + \varepsilon,$$

which algebraically is identical to

$$\log(Y) = \beta_0 + (\beta_1+\beta_2)\log(A) + (1-\beta_1)\log(B) - \beta_2\log(C) + \varepsilon.$$

Note particularly that nothing funny has happened to the random term $\varepsilon$, so that fitting one (via least squares or maximum likelihood) will give the same results as fitting the other.

Conversely, if you start with the general model

$$\log(Y) = \gamma_0 + \gamma_1\log(A) + \gamma_2\log(B) + \gamma_3\log(C) + \delta,$$

comparing coefficients shows that $\gamma_0 = \beta_0$, $\delta = \varepsilon$, and

$$\gamma_1 + \gamma_2 + \gamma_3 = (\beta_1+\beta_2) + (1-\beta_1) + (-\beta_2) = 1.$$

These are the only relations among the coefficients, as evidenced by the fact you can recover the betas from the gammas in many ways such as

$$\beta_1 = 1 - \gamma_2, \quad \beta_2 = -\gamma_3 \quad \text{or}$$

$$\beta_1 = \gamma_1 + \gamma_3, \quad \beta_2 = \gamma_1 + \gamma_2 -1,$$

for example.

Thus the original model is the general model with a single linear constraint. Either method will result in equivalent fits and the same parameter estimates. Therefore,

1. Yes, there is a way to include $B$ as an independent variable, as shown; and

2. The original model makes sense.

Answer 2

Just doing

$$ \log(Y) \sim \log(A) + \log(B) + \log(C) $$

would be equivalent to the model you're fitting, under the constraint that the coefficients sum to 1 (thanks whuber for making this clear). The coefficients would be changed to compensate for the redundancies in your current formulation, but the fit would be the same.

Edit: If you ignore the constraint and fit the model above then note that the $R^2$ value will change (and typically be much lower) when compared to your original formulation. This is because $B$ appears on both sides of the regression function, so there may appear to be an impressive $R^2$ regardless of whether $Y$ relates to $A,B,C$. See

A=runif(100)
B=runif(100)
C=runif(100)
Y=runif(100)

g1 = lm( log(Y/B) ~ log(A/B) + log(A/C))
g2 = lm( log(Y) ~ log(A)+log(B)+log(C))

as an example. Clearly none of the variables are related to each other, but model g1 shows a pretty impressive $R^2$ (around $45\%$ in my simulation).