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The central moments of a probability distribution $p(x)$ are defined as:

$$\theta_n = \langle (x - \langle x \rangle)^n \rangle $$

while the non-central moments are the standard:

$$\mu_n = \langle x^n \rangle $$

By the binomial theorem, we have:

$$\theta_n = \sum_{k=0}^n \binom{n}{k}(-1)^{n-k} \mu_k \mu_1^{n-k}$$

which allows us to compute the central moments from the non-central moments. Is there an inverse to this expression, giving the non-central moments $\mu_n$ from the central moments $\theta_n$?

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1 Answer 1

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One can write:

$$\mu_n = \langle (x - \langle x \rangle + \langle x \rangle)^n \rangle$$

By the binomial theorem

$$\mu_n = \sum_{k=0}^n \binom{n}{k} \theta_k \mu^{n-k}_1$$

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  • $\begingroup$ Can this be used to calculate the variance of a probability distribution by converting a 2nd order moment centred on zero to one centered on the mean? $\endgroup$ Apr 3, 2020 at 23:49
  • $\begingroup$ Yes, and you get the standard formula $\mathrm{var}(x) = \langle x^2\rangle - \langle x\rangle^2$ @ThomasKimber $\endgroup$
    – a06e
    Jun 9, 2021 at 7:31

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