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.
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