# Some probability inequalities

Let $x$ be random variable, such that $E(x)=0,E(x^2)=1$ and $P(x^2\geq s^2)\geq\displaystyle\frac{C}{s^t}$, where $C>0,s\geq 1 , t>0$. Let $m<n$ and $m,n$ are natural numbers very big. Let also $L\geq 1$ . Consider $1-(1-\frac{n}{2}P(x^2\geq Ln))^m$.

Assume (*) $\frac{n}{2}P(x^2\geq Ln)\leq \frac{2c_0}{m}$, where $0 <c_0 <0.6$;

using inequality $(1-y)^m\leq 1-\displaystyle\frac{my}{2}$, valid for $y\in [0, c_0]$ , with natural $n$, we get

$$1-(1-\frac{n}{2}P(x^2\geq Ln))^m\geq\displaystyle\frac{nm}{4}P(x^2\geq Ln),$$

using assumption $P(x^2\geq s^2)\geq\displaystyle\frac{C}{s^t}$ with $s^2=Ln$ , we get $$1-(1-\frac{n}{2}P(x^2\geq Ln))^m\geq \displaystyle\frac{nm}{4}P(x^2\geq Ln)\geq \frac{mnC'}{(Ln)^{\frac{t}{2}}}.$$

How to show, that if $t\geq 4$, then (*) holds?

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You need to put dollar signs around your LaTeX to get it to compile right. –  Nate Eldredge Dec 1 '11 at 5:10
@David: You need to enclose your math within $$. At the moment it is unreadable. – user17762 Dec 1 '11 at 5:11 ## 2 Answers Markov inequality: let Z denote a nonnegative integrable random variable. Then, for every positive z,$$ \mathrm P(Z\geqslant z)\leqslant z^{-1}\mathrm E(Z). $$Application: consider Z=x^2 for a square integrable random variable x such that \mathrm E(x^2)=1, and z=n for any positive integer n. Then \mathrm P(x^2\geqslant n)\leqslant n^{-1}. Note that this proof does not use the hypothesis that \mathrm E(x)=0 nor the lower bound on the tail of the distribution of x^2. Now, what you seem to be asking for is a proof that \mathrm P(x^2\geqslant n)\leqslant\Theta(n^{-2}) under the additional hypotheses that \mathrm E(x)=0 and that \mathrm P(x^2\geqslant s)\geqslant Cs^{-t/2} for every s\geqslant1, for some t\geqslant4. But there is no hope such a result could hold, is there? Consider a symmetric random variable such that \mathrm P(x^2\geqslant s)=\Theta(s^{-u}) when s\to\infty, for some positive u. Then your hypothesis may hold as soon as u\leqslant t/2 (for the tail estimate) and u\gt1 (for the square integrability) and your conclusion asks for u\geqslant2. These simply do not fit. - Yhank you. But what about the proof I've posted. At it seems it warks for n>n_0? – David Dec 6 '11 at 15:37 I have no idea what you are talking about. Is the part of my post which reads But there is no hope such a result could hold not clear enough? – Did Dec 9 '11 at 11:47 Probably it is better reformulate the question I've posted. Please tell me if my solution is correct and where I have mistakes. Thank you for advice. Let x be random variable with mean zero and variance 1. And such that for C>0, t>0 and natural n, P(x^2\ge n)\geq \frac{C}{n^t}. We want to show that if t\ge 2, then P(x^2\ge n)\geq \frac{2}{n^2}. First, we show that for any n\geq \widetilde{n},$$ \begin{align} P(x^2\geq n)\leq \frac{2}{n^2}. \end{align} Suppose to the contrary that \begin{align} P(x^2 \geq n_i) > \frac{2}{n_i^2} \end{align} for some infinite sequence n_1, n_2, \ldots, n_i, \ldots, such that \begin{align} P(x^2 > n_i^2) > 2P(x^2 > n_{i+1}^2). \end{align} By assumption 1=E x^2 = \int_{0}^\infty x^2 d \mu. $$Here \mu is a probability measure. We can break up this integral into the sum:$$ E x^2 = \sum_{i=0}^\infty \int_{n_i}^{n_{i+1}}x^2 d \mu. $$Consider now$$ \begin{align} \int_{n_i}^{n_{i+1}} x^2 d\mu \geq n_i^2 \int_{n_i}^{n_{i+1}} d\mu &= n_i^2(P(x^2\geq n_i^2)-P(x^2 \geq n_{i+1}^2)) &\gt n_i^2\frac 12 P(x^2\geq n_i^2) &\gt 1. \end{align}  This is a contradiction to $E x^2=1$.

Thus, with $\widetilde{n}\leq n$ \begin{align} P(w^2\geq Kn)\leq \frac {2}{n^2} . \end{align} But, within condition \begin{align} \frac{C}{n^t}\leq P(w^2\geq Kn)\leq \frac{2}{n^2}, \end{align} and $\displaystyle{n\geq \left(\frac{C}{2}\right)^{\frac{1}{t-2}}=\widetilde{\widetilde{n}}}$. Thus, for any $n\geq \max\{\widetilde{n},\widetilde{\widetilde{n}}\}$, $\frac{C}{n^t}\leq P(w^2\geq Kn)\leq \frac{2}{n^2}$.

If $t\geq 2$, then for any $n \in N$ and ,$C=2$ $P(w^2\geq Kn)=\frac{2}{n^2}$. Thus, $\frac{C}{n^t}\leq P(w^2\geq Kn)\leq \frac{2}{n^2}$ for any natural n.

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