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I've seen this in Probability theory $\sum p_ix_i$. What $p_i$ stands for?

Can some one prove that $\sum^3 p_i.x_i^2$ is bigger than $(\sum^3 p_i.x_i)^2$

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closed as not a real question by Jasper Loy, Davide Giraudo, tomasz, Alexander Gruber, Nameless Dec 27 '12 at 12:36

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1 Answer

up vote 3 down vote accepted

$p_i$ may means $P(X = x_i)$. As example, let $X$ be some discrete random variable taking values in the set $\left\{x_1,\ldots,x_n\right\}$. Then its expectation is given by $$ E(X) = \sum_{i=1}^nx_iP(X=x_i) = \sum_{i=1}^nx_ip_i. $$

Regard your second question:

You want to show that

$$ (\sum_{i=1}^3p_ix_i)^2\leq\sum_{i=1}^3p_ix_i^2. $$ Assuming that we have a random variable $X$ taking values $\left\{x_1,x_2,x_3\right\}$, then, what you actually want to show is that $$ (E(X))^2\leq E(X^2), $$ which readily follows from the fact that the $\text{var}(X)$ is non-negative.

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that means to me $\sum_{i=1}^{n=3}$ –  probability Dec 27 '12 at 12:13
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