Linear Regression with a Dependent Variable that is a Ratio

I'm doing linear regressions where the dependent variable is a ratio that can range from 0.01 to 100.

Is it ok to take the log of the dependent variable and the regression on that? I'm matching the results of a study and that is what they did.

What is the difference of taking the log versus using the ratio as-is?

• I am looking an assessment of house prices. My independent variable is Assessed House Price divided by Sales Price. My dependent variables are several race categories (percent black, white, hispanic, and asian) and median household income. I'm finding that census tracts with a larger percent of blacks have a higher assessment to sales price ratio than other areas. May 15, 2013 at 20:57
• Why don't you use logistic regression? You can define the ratio as your dependent variable is many statistical packages. Jun 11, 2014 at 23:54
• Logistic regression is typically for binary values, or proportions (between 0 and 1). It's not applicable here since the ratio can exceed 1. Sep 7, 2017 at 22:11

When you take the log of the ratio, keep in mind what what that is: $log({a \over b}) = log(a) - log(b)$
Does using this value as a dependent variable make sense in your problem?

Now, as to using the raw ratio - this can be problematic. Kronmal 1993 makes the argument that a regression with a ratio as the dependent variable:
${Y \over Z} = \alpha_0 + \alpha_XX + \epsilon$
which can be described as
$Y = Z1_n\alpha_0 + ZX\alpha_X + Z^{-1}\epsilon$
is a submodel of
$Y = \beta_0 + \beta_XX + Z1_n\alpha_0 + ZX\alpha_X + Z^{-1}\epsilon$

aka...

• Regress numerator by original independent variables, denominator, and denominator times the original variables
• Weight regression by (inverse) denominator

Only in the case where $\beta_0$ and $\beta_X$ were zero would the original regression model be valid.

Caveat - I'm not convinced I have a complete understanding of ratios either.

• Are you assuming a fixed Z? I've got 27,000 cases (aka properties) and the Y (assessment price) and Z (sales price) differ for each one. May 15, 2013 at 21:03
• No, Z is a nxn diagonal matrix, with the diagonals being your sales price. My notation in the first equation might be confusing since it isn't using matrix notation. $Z^{-1}Y = \alpha_0 + \alpha_XX + \epsilon$ would be in line with the rest. May 16, 2013 at 3:36
• Ok. I looked at the suggested approaches in your linked question and they make sense. I do not know enough about this to actually recommend them one way or the other, but if someone confirms your suggestions than I could try them out. May 16, 2013 at 15:46