# Logit Variable Interpretation

If I have a rate variable, for example unemployment rate, that has been transformed such that $\text{unemployment new} = \ln(\frac{\text{unemployment}}{1-\text{unemployment}})$, how do I interpret this variable? I know the log distributes, so this is actually the $\ln(\text{unemployment}) - \ln(1-\text{unemployment})$ - is this the rate of change in the unemployment rate? What about if I then difference $\text{unemployment new}$ across years? I ran a quick experiment and saw that as unemployment goes from $0.05$ to $0.1$, $\text{unemployment new}$ goes from $-2.94$ to $-2.19$, and the difference is $0.75$. But this isn't a 75% increase, so I'm not sure how to interpret it.

• If it helps, your new variable is on the log odds scale. So not a rate of change.
– Dan
Commented Sep 3, 2014 at 20:02
• I am also assuming that unemployment is a percentage.
– Dan
Commented Sep 3, 2014 at 20:03

The probabilistic interpretation is that logits are log odds. Odds are ratios of probabilities: the odds of event $A$ are given by $\frac{\Pr(A)}{\Pr(\neg{A}) }=\frac{\Pr(A)}{1-\Pr(A)}$ for $\Pr(A)\in(0,1)$. Clearly, then, odds must always be a positive number. Thus, taking the logarithm maps the odds to $\mathbb{R}$. This property, $\text{logit}:(0,1)\to\mathbb{R}$ is very helpful across the field of statistics, but is not directly relevant to your interpretation problem.
You're not alone in finding logits somewhat opaque to interpret, though, because it's two transformations removed from ordinary probability. First, let's talk about odds. Odds describe relative chance of an outcome: how much more likely is $A$ than $\neg{A}$. In your particular context, we can say that you are less likely to randomly sample an unemployed person from the large population of interest than you are to sample an employed person. We know this because the sign is negative. We also know the magnitude is fairly large from its absolute value. Moreover, the logarithm makes the log odds symmetric about $0$. What this means is that $\text{logit}(\Pr(A))=-\text{logit}(\Pr(\neg{A}))$.