# Bound the variance of the product of two random varables.

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?

• No here I am not assuming they are independent. – Stone Nov 29 '14 at 1:51
• In the general case, $$Var(XY)= 2 \mathbb{E}(X) \mathbb{E}(Y) \text{Cov}(X,Y)-\text{Cov}(X,Y)^2+\mu _{2,2}+2 \mu _{1,2} \mathbb{E}(X)+2 \mu _{2,1} \mathbb{E}(Y)+\text{Var}(Y) (\mathbb{E}[X])^2+\text{Var}(X) (\mathbb{E}[Y])^2$$ ... may serve as a useful starting point. – wolfies Nov 29 '14 at 1:54
• sorry but what are $\mu_{2,2}$ and $\mu_{2,1}$ suppose to be? – Stone Nov 29 '14 at 2:02
• Sorry - I should have specified. They denote the product central moments: $$\mu _{r,s}=E\left[(X-E[X])^r (Y-E[Y])^s\right]$$ – wolfies Nov 29 '14 at 2:04
• @Eupraxis1981 Well I don't think you need to assume that they are bounded since if say $X$ is unbounded and if $Y\neq 0$ then the right hand side is infinite and the inequality still holds and in the case that $Y=0$ then both sides are zero and the result still holds. So yes it is fair to assume that they are both bounded a.s – Stone Nov 29 '14 at 6:42

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 $$\mathrm{Var}(XY)\leqslant 2\mathrm{Var}(X)\lVert Y\rVert_\infty^2+2(\mathbb E[X])^2\mathrm{Var}(Y).$$