Probability inequalities with indicator variables

Question

Consider a non-negative random variable $X$ such that $\mathbb{E}(X^2)>0$ is finite, and let $t\in(0,1)$. I need to prove that $\mathbb{E}\left[X\mathbb{I}_{X>t\mathbb{E}(X)}\right]\geq(1-t)\mathbb{E}(X)$ and $\mathbb{P}\left[X>t\mathbb{E}(X)\right]\geq(1-t)^2\frac{\mathbb{E}(X)^2}{\mathbb{E}(X^2)}$. I know that $\mathbb{E}\left[\mathbb{I}_{X>t\mathbb{E}(X)}\right]=\mathbb{P}[X>t\mathbb{E}(X)]$, and I feel like I should use the tower law somehow to deal with the product with the indicator variable, but I am not sure what to condition on. For the second inequality, I suspect Cauchy-Schwarz will come into play, but I can't really see a path to success. Could someone please guide me on this? Thanks!

Answer 1

For the first inequality, write $X = X \cdot \mathbb{I}_{X > t\mathbb{E}(X)} + X \cdot \mathbb{I}_{X \leq t \mathbb{E}(X)}$ and take expectations of both sides: $$ \mathbb{E}(X) = \mathbb{E}(X \cdot \mathbb{I}_{X > t \mathbb{E}(X)}) + \mathbb{E}(X \cdot \mathbb{I}_{X \leq t \mathbb{E}(X)}) \leq \mathbb{E}(X \cdot \mathbb{I}_{X > t \mathbb{E}(X)}) + t \mathbb{E}(X). $$ Then rearranging gives $\mathbb{E}(X \cdot \mathbb{I}_{X > t \mathbb{E}(X)}) \geq (1-t) \mathbb{E}(X)$.

Now for the second inequality, use the first part and apply Cauchy-Schwarz to $\mathbb{E}(X \cdot \mathbb{I}_{X > t \mathbb{E}(X)})$: $$ (1-t)\mathbb{E}(X) \leq \mathbb{E}(X \cdot \mathbb{I}_{X > t \mathbb{E}(X)}) \leq \mathbb{E}(X^2)^{1/2} \cdot \mathbb{E}(\mathbb{I}^2_{X > t\mathbb{E}(X)})^{1/2} = \mathbb{E}(X^2)^{1/2} \cdot \mathbb{P}(X > t\mathbb{E}(X))^{1/2}. $$ Now square both sides and rearrange and you get the second inequality.