# 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

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