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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.

Yakir
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2 Answers2

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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? – Bernard Mar 18 '21 at 15:46
  • @Benny Because of the indicator $1{X\le a}$. –  Mar 18 '21 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. – Bernard Mar 18 '21 at 15:52
  • @Benny $Z:=X1{X\le a }\le a$ because $Z=X$ on ${X\le a}$ and $Z=0$ on ${X>a}$. –  Mar 18 '21 at 15:56
  • @Benny Also $E[X1_{x\le a}] \ne P[X\le a]$. However, the expectation can be bounded by the square root of ... by the Cauchy-Schwartz inequality which says that for square integrable random variables $Y$ and $Z$, one has $\mathsf{E}|YZ|\le \sqrt{\mathsf{E}Y^2\mathsf{E}Z^2}$ –  Mar 19 '21 at 00:02
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    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 :) – Bernard Mar 19 '21 at 00:06
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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

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

peaceful breeze
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