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I have a random variable $f(x) = \frac{1}{1 + e^{-x}}$ which is a logistic transformation of a Gaussian random variable $x \sim \mathcal{N}(\mu, \sigma^2)$.

I want to bound $|f(x) - f(\mu)|$. Are there any useful bounds in this case?

E.g. for Gaussian distribution $y \sim \mathcal{N}(\mu, \sigma^2)$ with probability at least $1 - \delta$ it holds that: $$ |y - \mu| \leq \sqrt{2 \log(1 / \delta)} \sigma. $$

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    $\begingroup$ Using the same letter, $\sigma,$ to refer to two different things is potentially confusing. $\endgroup$ – Michael Hardy May 29 '18 at 17:18
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    $\begingroup$ Aitchison's book on compositional data analysis has quite a bit on this distribution. $\endgroup$ – Michael Hardy May 29 '18 at 17:35
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Since you have $\sigma$ denoting both variance and the function, I will change the function to $f(x) = \frac{1}{1+e^{-x}}$.

Let $z \sim N(0,1)$. The function $z \mapsto f(\sigma z + \mu)$ is Lipschitz [with constant $\sigma/4$]. By Gaussian concentration for Lipschitz functions (see Theorem 8 here), you have $|f(\sigma z + \mu) - f(\mu)| \le t$ with probability $\ge 1 - C \exp(-c t^2 / (\sigma/4)^2)$. Equivalently, $|f(\sigma z + \mu) - f(\mu)| \le \frac{\sigma}{4}\sqrt{c^{-1} \log(C/\delta)}$ with probability $\ge 1 - \delta$. I believe the exact constants are $C=c=2$.

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  • $\begingroup$ Can we provide a better bound if we take into account small derivatives of $f(x)$ wrt $x$ for large absolute values of $x$? For Gaussian variable the bound in my example is rather tight, but I suppose that for logit-normal variable it is not the case for a large part of real line. $\endgroup$ – Alexey Zaytsev May 29 '18 at 19:41

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