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 $f_{X_n}<c$ 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 $f_N$):
$$1=2\lim_{\delta \rightarrow 0}\frac{F_N(\delta)-F_N(-\delta)}{2\delta}=2f_N(0)$$

  • $\begingroup$ thank you very much. Does a integrable version of your example refutes my proof in the following answer? $\endgroup$
    – riodemarie
    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?

  • $\begingroup$ Isn't your Jensen's inequality in the wrong direction ? $\endgroup$ Mar 3 at 11:09
  • $\begingroup$ 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 $fE(X)<E(f(X))$ ?? $\endgroup$
    – riodemarie
    Mar 3 at 13:11
  • $\begingroup$ @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. $\endgroup$
    – John L
    Mar 3 at 13:52

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