(no subject)

[eirias]

Sox!

(You have to admit, baseball is a lot more fun than trying to take the limit of [n!/(x!y!(n-x-y)!)]*[(p1)^x]*[(p2)^y]*[(1-p1-p2)^(n-x-y)] as n->inf.)

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[dolson.livejournal.com]

You rang?

This is probably too late to be useful, but I'll go ahead and post it since the Sox aren't playing tonight and I need some other way of procrastinating before my exam on Wednesday.

What you want to do is take a limit of the trinomial probability distribution to end up with another probability distribution. For simplicity, I'll discuss the binomial case. There are two important limits of this distribution.

If you take the limit of the binomial distribution P(x) = [n!/(x!*(n-x)!)]*p^x*(1-p)^(n-x) where n->inf and p->0 such that for all n, np=a, where a is constant, you end up with the Poisson distribution p(x) = a^x*e^(-a)/x!. This requires knowing the limit lim t->inf [1-(u/t)]^(t) = e^(-u). The key here is replacing n and p with the constant quantity a.

If you consider a limit where both n and np are much larger than 1 but finite, you can expand x about the mean in a "small" parameter u << np and get a normal approximation with mean np and variance np(1-p). (Doing this expansion requires Stirling's formula: n! ~ sqrt(2*pi*n)*(n/e)^n.) Here, if you take the limit n->inf and switch variables, you end up with a continuous distribution which is normal.

If you consider the large n limit without specifying behavior for p, a normal distribution is the natural answer. Thinking about the binomial distribution as resulting from summing n coin flip results, the central limit theorem states that for large n the distribution should approach a normal distribution with mean np and variance np(1-p).

In the multivariate case, taking the corresponding limits results in distributions which are products of the one-variable distributions.