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 $\theta_n$?


1 Answer 1


One can write:

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

By the binomial theorem

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.