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Consider a random variable $X$ that satisfies $X\geq 0\text{ a.s.}$ and $\text{Var}X<\infty$. Suppose that $0\leq \theta\leq 1$ Then show that $$P(X>\theta \mathbb{E}X) \geq (1-\theta)^2\frac{(\mathbb{E}X)^2}{\mathbb{E} X^2} $$

I tried the following chain of inequalities but it didn't work:\begin{align} P(X>\theta \mathbb{E}X)= P(X-\mathbb{E}X>(\theta-1)\mathbb{E}X)\\ =1-P(X-\mathbb{E}X\leq (\theta-1)\mathbb{E}X)\\ =1-P(\mathbb{E}X-X\ge (1-\theta)\mathbb{E}X)\\ \geq 1-P(|\mathbb{E}X-X|\ge (1-\theta)\mathbb{E}X)\\ \geq 1-\frac{\text{Var}X}{(1-\theta)^2(EX)^2}\end{align} Where in the last point I used Chebyshev's inequality, and previously markov's inequality.

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Note that $X - \theta\mathbb EX \leq X \mathbf 1_{X > \theta \mathbb EX}$, almost surely. Therefore, seeing that $$0 \leq \mathbb E[X - \theta \mathbb EX] \leq \mathbb E[X \mathbf 1_{X > \theta \mathbb E X}]$$ the Cauchy Schwarz inequality holds: $$ \mathbb E[X - \theta \mathbb EX]^2 \leq \mathbb E[X \mathbf 1_{X > \theta \mathbb E X}]^2 \leq \mathbb E[X^2]\mathbb E[1_{X > \theta \mathbb E X}]^2 = \mathbb E[X^2] P(X > \theta \mathbb E X) \\ \implies P(X > \theta \mathbb EX) \geq \frac{\mathbb E[X - \theta \mathbb E X]^2}{\mathbb E[X^2]} = \frac{(1-\theta)^2 \mathbb E[X]^2}{\mathbb E[X^2]} $$

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The proof follows if you replace $\lambda = \theta EX$ in the following post.

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