Subject: Declaring function as probability distribution?
From: Stavros Macrakis
Date: Sat, 27 Jul 2013 12:29:54 -0400
I don't know of any way to inform Maxima that integrate(p(x),x,minf,inf) =
1.
Why is declare(p(x),posfun) not quite good enough? When does it matter
that p(x) can be exactly 0? I suppose you could work with q(x)^2 instead
of p(x)?
All that being said, if you tell us what integrals and sums over functions
of p(x) you want to evaluate, perhaps we can help out.
-s
On Mon, Jul 22, 2013 at 9:03 PM, Mike Valenzuela <mickle.mouse at gmail.com>wrote:
> Hello,
>
> I've been looking for a way to encode some information to make Maxima make
> certain simplifications automatically.
>
> Specifically I need to encode the following two properties of probability
> distributions:
> (1) integrate(p(x),x,minf,inf) = 1, (when working with distributions from
> the continuous domain)
> (2) p(x)>=0 for all x
>
> declare(p(x), posfun) is almost good enough, but not quite.
> I'm unsure how to tell integrate (and sum) that the sum over all values is
> known.
>
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