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I'm trying to prove the following:

You have random variables $A$ and $B$ over a finite probability space $\Omega$ with probability measure P. (In other words $\sum_\Omega P(s) = 1, A: \Omega \rightarrow \mathbb{R}, B: \Omega \rightarrow \mathbb{R})$
We know that $AB \ge 1$ and that $A \ge 0, B\ge 0$

How can I show:
$\text{E}[A] \cdot \text{E}[B] \ge 1 $

I specifically want to practice applications of Cauchy's Inequality for real vector spaces:
$\langle V,W \rangle \le \langle V,V \rangle ^{\frac{1}{2}}\langle W,W \rangle^{\frac{1}{2}}$

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  • $\begingroup$ It's not too hard to translate the problem into a problem about vectors where the functions A, B, and P become N-dimensional vectors and N is the size of the sample space. $\endgroup$ – Mark Feb 17 '13 at 7:47
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Let $\Omega = \{s_1, s_2, \ldots, s_n\}$, $\hat A = \left\{ \sqrt{P(s_i)A(s_i)}\right\}_{i=1}^n$ and $\hat B = \left\{ \sqrt{P(s_i)B(s_i)}\right\}_{i=1}^n$.

Apply Cauchy's equality to the vectors $\hat A, \hat B$:

\begin{align} E[A] \cdot E[B] &= \left(\sum_{i=1}^n \left(\sqrt{P(s_i)A(s_i)}\right)^2\right)\left(\sum_{i=1}^n \left(\sqrt{P(s_i)B(s_i)}\right)^2\right) \\ &\ge \left(\sum_{i=1}^n \sqrt{P(s_i)A(s_i)} \cdot \sqrt{P(s_i)B(s_i)}\right)^2 \tag{1} \end{align}

But: $$ \left(\sum_{i=1}^n \sqrt{P(s_i)A(s_i)} \cdot \sqrt{P(s_i)B(s_i)}\right)^2 = \left(\sum_{i=1}^n P(s_i)\sqrt{A(s_i)B(s_i)}\right)^2 $$

Since $\sqrt{A(s_i)B(s_i)} \ge 1$ and $\sum_{i=1}^n P(s_i) = 1$, we get: $$ \left(\sum_{i=1}^n \sqrt{P(s_i)A(s_i)} \cdot \sqrt{P(s_i)B(s_i)}\right)^2 \ge 1 \tag{2} $$

Put (1) and (2) together to get: $$ E[A] \cdot E[B] \ge 1 $$

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