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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.

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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

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  • $\begingroup$ One follow up question: how can you justify the expansion of $\log(\exp(x) - 1)$ when $\log(x)$ is diverging to $-\infty$? I thought Taylor series expansions needed to be considered around reasonably-behaving points $\endgroup$ Commented Jan 26, 2022 at 0:09
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    $\begingroup$ Another way to see it is that, for $x \geq 0$, we have $xe^{x} \geq e^x - 1 \geq x$. So $\log(x) + x \geq \log(e^x - 1) \geq \log(x)$, skipping the Taylor series expansion. $\endgroup$ Commented Jan 26, 2022 at 1:18
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    $\begingroup$ @RylanSchaeffer Consider $\log \left(\exp(x)-1\right)- \log(x) = \log\left(\frac{\exp(x)-1}{x}\right)=\log\left(1+\frac12x+\cdots\right) = \frac12x+\cdots$ if you prefer $\endgroup$
    – Henry
    Commented Jan 26, 2022 at 2:03

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