Re: Moment function
- To: mathgroup at smc.vnet.net
- Subject: [mg126978] Re: Moment function
- From: Bob Hanlon <hanlonr357 at gmail.com>
- Date: Thu, 21 Jun 2012 05:20:54 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201206200748.DAA04927@smc.vnet.net>
dist = NormalDistribution[m, s]; assume = {Element[m, Reals], Element[{n, p}, Integers], s > 0, n >= 0, p>= 0}; moment1 = Moment[dist, n*p] moment2 = Assuming[assume, Integrate[z^(n*p)*PDF[dist, z], {z, -Infinity, Infinity}]//Simplify] distp = TransformedDistribution[z^p, z \[Distributed] dist]; moment3 = Assuming[assume, FullSimplify[Moment[distp, n]]] Assuming[assume, FullSimplify[moment2 == moment3]] True {moment1, moment2, moment3} /. {n -> 1, p -> 1} {m, m, m} And@@Flatten[Table[Assuming[assume, Simplify[moment1 == moment2 == moment3]], {n, 0, 5}, {p, 0, 5}]] True The first method is very much faster. Bob Hanlon On Jun 20, 2012, at 3:48 AM, paul <paulvonhippel at yahoo.com> wrote: > I would like a formula for the mth moment of Z^p, where Z~Normal(mu,sigma) and p is a positive integer. With m=p=1, the answer should be mu -- i.e., the first moment of a normal variable is the mean mu. But that's not what I get when I do it this way: > > momentPN = > FullSimplify[ > Moment[TransformedDistribution[Z^p, > Z \[Distributed] NormalDistribution[\[Mu], \[Sigma]]], m], {m > 0, > p > 0, Element[p, Integers], Element[m, Integers]}] > > FullSimplify[ > momentPN /. {p -> 1, m -> 1}, {\[Sigma] > 0, Element[\[Mu], Reals]}] > > I know other ways to get the right answer, but why doesn't this method work? >
- References:
- Moment function
- From: paul <paulvonhippel@yahoo.com>
- Moment function