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say I have a regression of the sort:

fraction on men aged 18+ incarcerated = a + blog(income) +e, where the unit of observation is a county.

would the correct interpretation be, an increase in 1 log point of income (100% increase in income) leds to a b*100 percentage point change in the fraction of men incarcerated? or is it a b percent change?

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    $\begingroup$ This sounds like a logistic regression instead of straight linear regression. Please give some additional details about your data. $\endgroup$ – Dave Dec 7 '19 at 21:31
  • $\begingroup$ Yes I understand it may fit a logistic regression, but I am only interested in an association and some degree of magnitude, not prediction, and whether I understand what coefficient means in words. the variable for the fraction is just the number of men 18+ incarcerated divided by the total number of men age 18+, so it is a proportion of men of that category incarcerated $\endgroup$ – Steve Dec 7 '19 at 21:38
  • $\begingroup$ So no biggie if your predicted value is 106%? $\endgroup$ – Dave Dec 7 '19 at 23:20
  • $\begingroup$ no if my goal is just statistical association (and eventually causal inference) in which case I am not trying to impose any additional distributional assumptinos to get at a sign and magnitude. and once again, I am just trying to make sure I understand what an interpretation in this case is.. $\endgroup$ – Steve Dec 7 '19 at 23:22
  • $\begingroup$ I am not sure if off-the-shelf logistic regression will work here since the outcome is not binary. Fractional response regression may perform better. See Papke, L. E., and J. M. Wooldridge. 1996. Econometric methods for fractional response variables with an application to 401(k) plan participation rates. Journal of Applied Econometrics 11: 619–632 for an intro. $\endgroup$ – Dimitriy V. Masterov Dec 9 '19 at 22:36
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Given a linear regression model $$E[y \vert x] = \alpha + \beta \cdot \ln(x),$$ the partial derivative with respect to $x$ is $$\frac{\partial y }{\partial x} = \beta \cdot \frac{1}{x}.$$ Solving for $\beta$, you get $$ \beta = \frac{\partial y }{\partial x} \cdot x \approx \frac{\Delta y}{\Delta x/x}.$$

I often find it helpful to multiply the denominator by 100 to convert it to percent, which gets you

$$\frac{\beta}{100} \approx \frac{\Delta y}{100 \cdot\frac{\Delta x}{x}}.$$

The numerator is just a change in $y$ in levels, and the denominator is percent change in $x$.

To given an example, if $\hat \beta=-0.5,$ that means that a 1% change in income is associated with a $-.5/100=-.005$ decrease in the fraction of adult men incarcerated. This is a half of a percentage point decline in the incarceration fraction.

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  • $\begingroup$ so this is equivalent of scaling the independet variably by 100, so then in your example where Beta is -.5, it would show up in the output as -0.005, correct? so I could equivalently multiply the dependent variable by 100, so then I just have a 1 percent increase in x leads to a Beta_hat=.5 percentage point decline in incarceration fraction $\endgroup$ – Steve Dec 10 '19 at 1:08
  • $\begingroup$ The output would show .5, which you would have to divide by 100 yourself (or have your software do it). $\endgroup$ – Dimitriy V. Masterov Dec 10 '19 at 1:14
  • $\begingroup$ ah ok. so you mean if I left the regression as is (y on logx), the it would tell me a 100% increase in x leads to a .5 decrease, so I can scale it down by 100 to make it a 1 percent decrease $\endgroup$ – Steve Dec 10 '19 at 1:19
  • $\begingroup$ That is correct. $\endgroup$ – Dimitriy V. Masterov Dec 10 '19 at 1:54

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