# Inequality for Expected Value of Product

Let $(\Omega, \mathbb{P}, \mathcal{F})$ be a probability space, and let $\mathbb{E}$ denote the expected value operator.

Consider the random variables $f: \Omega \rightarrow \{0,1,2\}$ and $g: \Omega \rightarrow [0,1]$, where $\Omega \subseteq \mathbb{R}^n$.

Consider the expected value of the product, namely $$\mathbb{E}\left[ f(\cdot) g(\cdot) \right] := \int_{\Omega} f(\omega) g(\omega) \mathbb{P}(d \omega).$$

Now if $f$ and $g$ are independent random variables, then $\mathbb{E}[f(\cdot) g(\cdot)] = \mathbb{E}[f(\cdot)] \mathbb{E}[g(\cdot)]$.

I am wondering about conditions under which $\mathbb{E}[f(\cdot) g(\cdot)] \geq \mathbb{E}[f(\cdot)] \mathbb{E}[g(\cdot)]$.

Additional Assumption: $\mathbb{E}[g(\cdot)] = \frac{1}{2}\mathbb{E}[f(\cdot)]$.

• What about covariance being positive? – Dilip Sarwate Jan 3 '14 at 19:05
• Of course, because $\mathbb{E}[ f g ] = \mathbb{E}[f] \mathbb{E}[g] + \text{Cov}(x,y)$. But this is just a definition, not an additional condition on $f$ and $g$. – user693 Jan 3 '14 at 19:09
• And what are $x$ and $y$ in your response to my comment? – Dilip Sarwate Jan 3 '14 at 19:39
• Sorry, I meant $f$ and $g$. – user693 Jan 3 '14 at 20:35

## 2 Answers

A sufficient condition is that, for every $\omega$ and $\omega'$ in $\Omega$, $$(f(\omega)-f(\omega'))\cdot(g(\omega)-g(\omega'))\geqslant0.$$

• Thanks for the answer. Do you think that assuming $\mathbb{E}[g] = \frac{1}{2}\mathbb{E}[f]$ may be useful? – user693 Jan 4 '14 at 21:02
• This is quite unrelated since one can always shift $f$ or $g$ by a constant without changing the direction of the inequality. – Did Jan 4 '14 at 21:35

The case when f and g are increasing functions, for which the inequality holds, is well explained in Appendix 9.9, titled "Verification of Antithetic Variable Approach When Estimating the Expected Value of Monotone Functions", of the book Simulation by Sheldon M. Ross, 5th Ed. Academic Press, 2012.