1

I am wondering how I could express the variance of log-odds into understandable terms.

For example the variance in the log-odds of crime being reported to the police between neighbourhoods is 0.07 (0.01), and the mean log-odds is -0.65 (0.03). Just transforming this variance to probability given me 0.51. So if the probabilities have a variation of 0.51, the standard deviation is then 0.7, which would make the 95% coverage bound wider than 1, which is impossible for probabilities.

Another idea was to calculate the 95% coverage bounds of my log-odds:

$(-0.65) +/- [1.96\sqrt{0.07}] = [-1.19, -0.15]$

If I transform these to probabilities:

$\exp(x)/[1+\exp(x)] = [0.23,0.46]$

So the chance of a crime being reported to the police can vary from 23% to 46%.

So the 95% range is actually only 0.23 or 23%.

Could someone explain me why this first calculation was wrong, and whether the second is correct? Would you have any other suggestions on how to make this variance more graspable?

Marloes
  • 790
  • 1
    You can approximate the variance of a transformed variable via Taylor expansions. – Glen_b Jan 24 '13 at 14:28
  • 2
    If I've correctly parsed this question, you are asking why $f( {\rm var}(X) ) \neq {\rm var}(f(X))$, where $f$ is the inverse-logit function. For some intuition on why this inequality exists, you may want to read about Jensen's Inequality. – Macro Jan 24 '13 at 14:56

0 Answers0