# concentration inequality and how to bound densities of a sequence of random variables

Let $$X_n$$ be a sequence of random Variables such that $$E(X_n)=0$$ and $$Var(X_n)=1$$ I need to bound the following:

$$P(|X_n|<\delta_n)\lesssim \delta_n$$

for a positiv sequence $$\delta_n$$ tending to zero. My question is under which asumptions this inequality holds. Lets first assume that there is a density $$f_{X_n}$$ according to the lebesgue measure for each $$X_n$$. Then it follows $$P(|X_n|<\delta_n)=\int^{\delta_n}_{-\delta_n}f_{X_n}.$$ Do i need additional asumptions to bound $$f_{X_n} for all $$n$$ in order to bound the Integral by $$2c\delta_n$$ or is it somehow enough to assume that $$Var(X_N)=1$$?

Let $$X_N$$ be $$-N, 0$$, or $$N$$ with probabilities $$\frac{1}{2 N^2}, 1-\frac{1}{N^2}, \frac{1}{2 N^2}$$.
Then, $$E[X_N]=0$$ and $$Var[X_N]=1$$.
But, for any $$\delta>0$$, $$P[|X_N|<\delta] \ge 1-\frac{1}{N^2}\rightarrow 1$$.
So, for any sequence of positive $$\delta_N$$ that converges to $$0$$, you also have $$P[|X_N|<\delta_N]\rightarrow 1$$
That shows $$Var[X_N]=1$$ is not sufficient.

You should be more precise about what the approximation means.
If $$F_N$$ is the cdf, and you want $$\lim_{N\rightarrow \infty}\frac{F_N(\delta_N)-F_N(-\delta_N)}{\delta_N}=1$$ then essentially this means (provided the derivative of $$F_N$$ exists and is denoted $$f_N$$):
$$1=2\lim_{\delta \rightarrow 0}\frac{F_N(\delta)-F_N(-\delta)}{2\delta}=2f_N(0)$$

• thank you very much. Does a integrable version of your example refutes my proof in the following answer? Mar 3 at 9:29

If i assume that $$X_n$$ is integrable, then i get by Chernoff (https://en.wikipedia.org/wiki/Concentration_inequality): $$P(|X_n|<\delta_n)\leq \delta_n E(|X_n|^{-1})$$ and by the jensen's inequality i get for all $$n$$ $$\delta_n E(|X_n|^{-1})\leq\delta_n E(X_n^{2})^{-\frac{1}{2}}$$ and by $$Var(X_n)=1$$: $$P(|X_n|<\delta_n)\leq \delta_n$$ Where did i make a mistake?

• Isn't your Jensen's inequality in the wrong direction ? Mar 3 at 11:09
• Thank you for your comment. My thought was that on $(0,\infty)$ the function $f(x)=x^{-2}$ is convex so by Jensen (ignoring the Zero cause for each n {X_n=0} is a nullset) it follows $fE(X)<E(f(X))$ ?? Mar 3 at 13:11
• @riodemarie Your proof looks correct assuming $|X_n|^{-1}$ is a valid random variable. In the counterexample I provided, $|X_n|^{-1}$ would not be a random variable because $P[X_n=0]>0$. So, I think you are OK if you add the conditions that $P[X_n=0]=0$ and $E[|X_n|^{-1}]$ exists so you can apply Markov and Jensen inequalities. Mar 3 at 13:52