General formula for normal moments

Yesterday I wrote about how to find moments of the normal distribution using Sage. Then GlennF left a comment saying it’s not too hard to work out the moments analytically and outlined a proof. I’ll fill in a few details here.

First, start with a standard normal distribution Z. That is, Z has mean 0 and variance 1. By symmetry, the odd moments of Z are 0. For the even moments, integration by parts shows that E(Z2m) = (2m – 1) E(Z2m – 2). Apply this relation recursively until you get E(Z2m) = (2m – 1)!!. (See this post if you’re unfamiliar with double factorial. Note that (-1)!! is defined to be 1.)

For a general normal random variable X with mean μ and variance σ2, define Z = (X – μ)/σ. Then Z is a standard normal and X = σZ + μ. Apply the binomial theorem and note that the odd terms are zero.

 E{(sigma Z + mu)^n} = sum_{i=0}^n {nchoose i}sigma^i E(Z^i) mu^{n-i}  = sum_{j=0}^{lfloor n/2 rfloor} {n choose 2j} (2j - 1)!!, sigma^{2j}mu^{n - 2j}

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3 comments on “General formula for normal moments
  1. GlennF says:

    I like your statement of it better than mine. For some reason I can’t explain today, I chose to re-express the double factorial in terms of ordinary factorials. It looks much cleaner *not* doing that, though.

  2. SteveBrooklineMA says:

    Very nice. Is there an application? I thought that high moments were interesting for theoretical questions, but not really useful when applied to “real” data.

  3. John says:

    I’ve haven’t had much need for higher moments. But I’ve been working with Edgeworth expansions lately, and higher moments pop up. In this tech report I needed up to 5th moments of a normal. With more terms in my Edgeworth approximations, I’d need higher moments.

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