My Math Forum Ref: Proof of non-central moments of the normal distribution

 February 13th, 2014, 11:37 AM #1 Newbie   Joined: Feb 2014 Posts: 3 Thanks: 0 Ref: Proof of non-central moments of the normal distribution Hi, I am currently reading through an article where they using the noncentral moments for the normal distribution (and actually also the chi-squared distribution), but what i need to know is how these non-central moments has been determined. I have tried googling the answer without luck. In most places, i.e wikipedia they just state the moments - so i have the answer. But what i am looking for is a proof of how to archieve the answer. Can anybody help me with an reference? Preferable online, but a book would also suffice. ---- So for clarification, if $X\sim N(\alpha,\sigma^{2})$ when i want to determine what $E[X^{k}]=\underset{-\infty}{\overset{\infty}{\int}}x^{k}\frac{1}{\sqrt {2\pi\sigma^{2}}}e^{\frac{-(x-\alpha)^{2}}{2\sigma^{2}}}dx$ In particular I'm interested in the 4. moment., so a proof only for the absolute moments would be fine - if it makes a difference.
 February 13th, 2014, 02:39 PM #2 Global Moderator   Joined: May 2007 Posts: 6,684 Thanks: 658 Re: Ref: Proof of non-central moments of the normal distribu (Answered in another forum). y = x - ? gives the non-central moment as a finite linear combination of central moments.
 February 17th, 2014, 06:30 AM #3 Newbie   Joined: Feb 2014 Posts: 3 Thanks: 0 Re: Ref: Proof of non-central moments of the normal distribu Thanks man, I also found another solution. I solved it with taylorexpansion of e^tx like this: $E(e^{tX})=E\left[1+\frac{tE(X)}{1!}+\frac{t^{2}E(X^{2})}{2!}+\frac{ t^{3}E(X^{3})}{3!}+...\right]$ And using that on the normal distribution i first calculated the integral $M(t)=\frac{1}{\sqrt{2\pi\sigma^{2}}}\underset{-\infty}{\overset{\infty}{\int}}e^{tx-\frac{(x-\mu)^{2}}{2\sigma^{2}}}d = e^{\mu t+\frac{1}{2}\sigma^{2}t^{2}}$ And then it is possible to differentiate n times (to find the n'th moment) and evaluate at t=0, which yields that $\frac{M^{(n)}(t)}{\partial^{n}t}|_{t=0}=E(X^{n})$

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