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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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Since the sum of the $p_i$ is $1$, identifying a probability distribution if not the same as identifying a point in $\mathbb{R}^3$, but the same as identifying a point in the simplex $\{(x_i)_{0 \leq i \leq 2} \in \mathbb{R}_+^3: \ \sum_i x_i = 1\}$. A binomial distribution correspond to a point in the intersection between the cone and the simplex. – D. Thomine Jun 28 '12 at 23:51
Bernd Sturmfels and his students have done a lot of work on applications of algebraic geometry to statistics. – Michael Hardy Jun 29 '12 at 0:24
$4xy-z^2=0$ cuts out an algebraic variety in ${\bf R}^3$, but not if $x,y,z$ are constrained to be between 0 and 1. – Gerry Myerson Jun 29 '12 at 6:33

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