Log Odds in Zero Noise Limit I have the following problem: Define $p := 1 - \exp(-\exp(-\lambda^2/2\sigma^2))$ and consider the log odds:
$$\log \frac{p}{1-p} $$
I'm interested in the limiting behavior as $\sigma^2 \rightarrow 0$. Obviously the log odds go to $-\infty$, but I'd like to know how quickly. My hypothesis is that the answer will be
$-\frac{\lambda^2}{2\sigma^2}$ but I'm having trouble showing that.
This problem arises in a modified flavor of Bayesian logistic regression.

If $p=1 - \exp(-x)$
then $\log \left(\frac{p}{1-p}\right) = \log \left(\exp(x)-1\right)= \log(x) +\frac12x+\frac1{24}x^2-\frac1{2880}x^4+O\left(x^6\right)$.
Letting $x=\exp\left(-\frac{\lambda^2}{2\sigma^2}\right)$
gives $\log \left(\frac{p}{1-p}\right) = -\frac{\lambda^2}{2\sigma^2} + O\left(\exp\left(-\frac{\lambda^2}{2\sigma^2}\right)\right)$
and for decreasing $\sigma^2$, the $\exp\left(-\frac{\lambda^2}{2\sigma^2}\right)$ term quickly gets very small
