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Topic: Moment function
Replies: 1   Last Post: Jun 21, 2012 5:24 AM

 Bob Hanlon Posts: 726 Registered: 10/29/11
Re: Moment function
Posted: Jun 21, 2012 5:24 AM

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@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?
>