Getting From Concentration Inequality to Interval Length

Question

I've seen this used some times and I would like to ask what steps are taken on the way to getting there:

E.g. assuming bounded variance, we can use Chebyshev concentration inequality: for any $t>0$,

$$ P(\lvert X - \mu \rvert > t) \leq \frac{\sigma^2}{t^2} $$

What are the steps I would need to take to find the bound for the length of some interval around $\mu$ that contains probability mass $1-\eta$ of the distribution of $X$, for $0 \leq \eta \leq 1$?

For example

Given bounded fourth moment:

$$ E[(X-\mu)^4] \leq C_4 \bigg(E[(X-\mu)^2]\bigg)^2 $$

the following concentration inequality applies

$$ P(\lvert X - E[X] \rvert \geq t \sigma) \leq \frac{C_4}{t^4} $$

Then let $I_{1-\eta}$ be the interval around $\mu$ containing $1-\eta$ probability mass, then using the concentration inequality we know that:

$$ \text{length}(I_{1-\eta}) \leq \frac{C_4^{\frac{1}{4}}\sigma}{\eta^\frac{1}{4}} $$ What are the steps taken to arrive at this length bound? Taken from this paper, lemma 11 (concentration) and p43 (length)

Answer 1

I think this might be how to think of it :

Rewrite the concentration inequality as

$$ \mathbb{P} \bigg( \lvert X - E[X] \rvert \geq t \sigma C_4^\frac{1}{4} \bigg) \leq \frac{1}{t^4} $$

then

$$ \mathbb{P} \bigg( \lvert X - E[X] \rvert \geq \eta^{-\frac{1}{4}} \sigma C_4^\frac{1}{4} \bigg) \leq \eta $$

then if you removed the $\eta$-quantile, the remaining $1-\eta$ mass of points would lie within

$$ \lvert X - E[X] \rvert \leq \frac{C_4^\frac{1}{4}}{\eta^\frac{1}{4}} \sigma $$