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Consider a random variable $X$ that can take on the values $0,1$ and $2$. So we have $p_i = P(X=i), \ i = 0,1,2$ $\sum_{i=0}^{2} p_i = 1$ and $0 \leq p_i \leq 1$ So identifying a random variable $X$ is the same thing as identifying a point $(p_0,p_1,p_2) \in \mathbb{R}^3$. Now suppose $X$ is a binomial random variable. Then $p_i = P(X = i) = \binom{2}{i} q^{i}(1-q)^{2-i}$ and $4p_{0}p_{2}-p_{1}^{2} = 0$

How is this related to algebraic geometry? The above equation is an algebraic variety in $\mathbb{R}^3$?

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    $4xy-z^2=0$ cuts out an algebraic variety in ${\bf R}^3$, but not if $x,y,z$ are constrained to be between 0 and 1.2012-06-29

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