# log-odds to probabilities

For example the variance in the log-odds 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:

$$−0.65 \pm [1.96\cdot\sqrt{0.07}] = [−1.19,−0.15]$$ If I transform these to probabilities:

$$\large \frac{\exp(x)}{1+\exp(x)}=[0.23,0.46]$$

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?

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You can not transform your variance like that, because $\mathrm{Var}(f(X))\neq f(\mathrm{Var}(X))$ in general. However, $P(a\leq X\leq b)=P(f(a)\leq f(X)\leq f(b))$ if $f$ is increasing for example. – Stefan Hansen Jan 24 '13 at 14:36

## 1 Answer

Your first calculation was wrong because the variance is not log-odds, so you can't transform it to a probability. It's the variance of the log-odds, but it's not log-odds itself. It's on a different scale. You can't transform variances the same way you transform measurements.

Note in particular that since variances can't be negative, your transformed "probability variance" would always be greater than $0.5$.

Your second way of doing it gives a correct interval.

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