# $X$ is a random variable, if $\Bbb E(X^2)=1$ and $\Bbb E(X)\geq a>0$, prove that $\Bbb P(X\geq\lambda a)\geq(a-\lambda a)^2$ for $0\leq\lambda\leq 1$.

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?

• Chebyshev might be useful. Mar 28, 2019 at 3:30

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.