TITLE: Trouble Applying Cauchy's Inequality
QUESTION [0 upvotes]: 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:
$\langle V,W \rangle \le \langle V,V \rangle ^{\frac{1}{2}}\langle W,W \rangle^{\frac{1}{2}}$

REPLY [1 votes]: 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:
$$
E[A] \cdot E[B] \ge 1
$$
{"set_name": "stack_exchange", "score": 0, "question_id": 306026}