An upper bound on the expected value of the square of random variable

Question

Let $X$ and $Y$ be two random variables such that:

1. $Y$ is a geometric random variable with the success probability $p$ (the expected value of $Y$ is $1/p$).
2. $X \geq 0$.
3. $\mathbb{E}(X) \leq \mathbb{E}(Y)$.

I would be grateful for any help of how one could upperbound $\mathbb{E}(X^2)$ in terms of $p$.

Answer 1

You can't. Let $X = 0$ with probability $1-\varepsilon$ and $\mathbb E[Y]/\varepsilon$ with probability $\varepsilon$.