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Discrete distributions

Suppose our random variable $ X$ can take only values $ v_1, v_2,
\ldots, v_N$. We insist that $ 0\leq f_X(v_i) \leq 1$, and

$\displaystyle \sum_{i=1}^N f_X(v_i) = 1.
$

If each $ v_i$ is equally likely to occur, then these conditions lead to a uniform distribution with $ f_X(v_i) == 1/N$. A binomial distribution expresses how often we expect a particular value to occur out of $ n$ events. If $ f_X(e)$ = $ p$ (the probability that our chosen event occurs once is $ p$), then the probability of exactly $ k$ occurrences is:

$\displaystyle f_X(k) = {n \choose k} p^k(1-p)^{n-k}.
$

The reasoning is that in a sequence of $ n$ events there are $ n$   choose$ k$ ways to select exactly $ k$ event $ e$s, and the probability of each selection is the product of the $ p$s with the product of the $ (1-p)$s (the probability that $ e$ doesn't occur).

This is a discrete normal distribution.



Danny Heap 2002-12-16