An upper bound on the expected value of the square of random variable dominated by a geometric random variable

Question

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

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

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

Answer 1

If $0 \leq X \leq Y$ almost surely, then $0 \leq X^2 \leq Y^2$ almost surely, as $x \mapsto x^2$ is monotonous on $\mathbb{R}_+$. We also know, that if $X \leq Y$ almost surely, then $EX \leq EY$. Thus $0 \leq E(X^2) \leq E(Y^2) = (EY)^2 + Var(Y) = \frac{1}{p} + \frac{1 - p}{p^2} = \frac{1}{p^2}$. Thus $E(X^2) \in [0; \frac{1}{p^2}]$. And the is the best possible bound.