# $X>0, E(X)=1, E(X^2)=b, \forall a\in (0,1): P(X>a)\geq \frac{(1-a)^2}{b}$

Suppose that $$X>0, E(X)=1, E(X^2)=b$$

And We should prove for every $$a$$ such that $$0 < a < 1$$ the following statement:

$$P(X>a)\geq \frac{(1-a)^2}{b}$$

This is a preliminary course of probability and we learned only basic formulas and inequalities of Markov and Chebyshev's. I would be happy if you keep the answer simple as much as possible. Thanks.

Write $$1=\mathsf{E}X=\mathsf{E}X1\{X>a\}+\mathsf{E}X1\{X\le a\}.$$ Since $$\mathsf{E}X1\{X\le a\}\le a$$ and $$\mathsf{E}X1\{X>a\}\le \sqrt{\mathsf{E}X^2\mathsf{P}(X>a)}$$ by Cauchy-Schwartz, $$1-a\le \sqrt{\mathsf{E}X^2\mathsf{P}(X>a)},$$ which implies the result.

• why $E[X1_{x\le a}]\le a$? also, how do you prove the use of Caucy-Schwartz if you can only use the integral form of it? Mar 18, 2021 at 15:46
• @Benny Because of the indicator $1\{X\le a\}$.
– user140541
Mar 18, 2021 at 15:48
• I apperantly miss what you see. I got the same question the OP has asked in an exam, and I try to figure it out. I can understand that $E[X1_{x\le a}] = P[X\le a]$, I dont see what i miss when I can't conclude it's less than $a$. it might be trivial thing that I miss here. Mar 18, 2021 at 15:52
• @Benny $Z:=X1\{X\le a \}\le a$ because $Z=X$ on $\{X\le a\}$ and $Z=0$ on $\{X>a\}$.
– user140541
Mar 18, 2021 at 15:56
• the thing I didn't understand was that $1_{x\le a}=P(X\le a)$, even though it's pretty trivial. thanks for adding this comment for the completion of the discussion for the average undergraduate :) Mar 19, 2021 at 0:06

Let $$\mathbb{1}_A$$ be the indicator function of set $$A$$. Then for any $$a\in(0,1),$$ it is easy to see that $$X \mathbb{1}_{\{X>a\}}\ge X-a.$$ Taking the expectation of both sides yields $$E[X\mathbb{1}_{\{X>a\}}]\ge E[X]-a=1-a. \qquad(1)$$

Moreover, by the Cauchy-Schwarz inequality, $$E[X \mathbb{1}_{\{X>a\}}]\le \sqrt{E[X^2]}\sqrt{P\{X>a\}}. \qquad(2)$$

Finally, from $$(1)$$ and $$(2)$$, we deduce

$$\sqrt{E[X^2]}\sqrt{P\{X>a\}}\ge 1-a \implies P\{X>a\}\ge \frac{(1-a)^2}{E[X^2]}=\frac{(1-a)^2}{b}.$$