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This is a problem in KaiLai Chung's A Course in Probability Theory.

Given a nonnegative random variable $X$ defined on $\Omega$, if $\mathbb{E}(X^2)=1$ and $\mathbb{E}(X)\geq a >0$, prove that $$\mathbb{P}(X\geq \lambda a)\geq (a-\lambda a)^2$$ for $0\leq\lambda \leq 1$.

Let $A=\{x\in \Omega:X(x)\geq \lambda a\}$, we get $$\int_A (X-\lambda a)\geq a-\int_A\lambda a -\int_{A^c}X$$ and $$\int_A (X^2-\lambda^2 a^2)=1-\int_A\lambda^2a^2-\int_{A^c}X^2$$ I want to contrast $\int_A (X-\lambda a)$ and $\int_A (X^2-\lambda^2 a^2)$, but I don't know how to do it, could anyone gives me some hints?

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  • $\begingroup$ Chebyshev might be useful. $\endgroup$
    – copper.hat
    Mar 28, 2019 at 3:30

1 Answer 1

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You have $$ a\le\mathbb E(X) = \int_{X\le\lambda a}X\,dP + \int_{X\ge\lambda a}X\,dP\,\le\,\lambda a + \int_{X\ge\lambda a}X\,dP. $$ Hence, $$ a(1-\lambda)\,\le\,\int_{X\ge\lambda a}X\,dP\,\le\,\left(\int_{X\ge\lambda a}X^2\,dP\right)^{1/2}\cdot P(X\ge\lambda a)^{1/2}\,\le\,P(X\ge\lambda a)^{1/2}. $$ Square this and you're done.

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  • $\begingroup$ Thank you very much! $\endgroup$
    – Xin Fu
    Mar 28, 2019 at 3:44

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