Bound the variance of the product of two random varables.

Question

For two random variables $X$ and $Y$ show that the following inequality holds

$$\mathrm{Var}(XY)\leq 2\|Y\|_{\infty}^{2}\mathrm{Var}(X)+2\|X\|_{\infty}^{2}\mathrm{Var}(Y).$$

Well first I tried to show it for just indicators functions by I couldn't even show that. Any tips?

Answer 1

If $U$ and $V$ are two random variables, then the inequality $$\mathrm{Var}(U+V)\leqslant 2\mathrm{Var}(U)+2\mathrm{Var}(V)$$ takes place. Using this inequality with $U:=(X-\mathbb E[X])Y$ and $V:=\mathbb E[X]Y$, we obtain that $$\mathrm{Var}(XY)\leqslant 2\mathrm{Var}((X-\mathbb E[X])Y)+2\mathrm{Var}(\mathbb E[X]Y).$$ Since \begin{align}\mathrm{Var}((X-\mathbb E[X])Y)&=\mathbb E\left[((X-\mathbb E[X])Y)^2\right]- \left(\mathbb E\left[(X-\mathbb E[X])Y\right]\right)^2\\ &\leqslant \mathbb E\left[((X-\mathbb E[X])Y)^2\right]\\ &\leqslant \mathbb E\left[(X-\mathbb E[X])^2\right]\lVert Y\rVert_\infty^2\\ &=\mathrm{Var}(X)\lVert Y\rVert_\infty^2, \end{align} we have showed the inequality $$\mathrm{Var}(XY)\leqslant 2\mathrm{Var}(X)\lVert Y\rVert_\infty^2+2(\mathbb E[X])^2\mathrm{Var}(Y).$$