# Getting From Concentration Inequality to Interval Length

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)

• No general bound exists, apart from the trivial lower bound of $0$ and upper bound of $+\infty:$ you need to specify more about how the interval is constructed or about the distribution of $X.$ Would you like to clarify your question? – whuber Mar 30 '19 at 21:16
• @whuber thank you, I have added an example that led to this question – eeek Mar 31 '19 at 11:41

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

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