Editing Central moment (section)

===Relation to moments about the origin===
Sometimes it is convenient to convert moments about the origin to moments about the mean. The general equation for converting the {{mvar|n}}-th-order moment about the origin to the moment about the mean is

<math display="block">
\mu_n = \operatorname{E}\left[\left(X - \operatorname{E}[X]\right)^n\right]
= \sum_{j=0}^n \binom{n}{j} {\left(-1\right)}^{n-j} \mu'_j \mu^{n-j},
</math>

where {{mvar|μ}} is the mean of the distribution, and the moment about the origin is given by

<math display="block">
\mu'_m = \int_{-\infty}^{+\infty} x^m f(x)\,dx = \operatorname{E}[X^m] = \sum_{j=0}^m \binom{m}{j} \mu_j \mu^{m-j}.
</math>

For the cases {{math|1=''n'' = 2, 3, 4}} — which are of most interest because of the relations to [[variance]], [[skewness]], and [[kurtosis]], respectively — this formula becomes (noting that <math>\mu = \mu'_1</math> and <math>\mu'_0=1</math>):

<math display="block">\mu_2 = \mu'_2 - \mu^2\,</math>
which is commonly referred to as <math> \operatorname{Var}(X) = \operatorname{E}[X^2] - \left(\operatorname{E}[X]\right)^2</math>

<math display="block">\begin{align}
\mu_3 &= \mu'_3 - 3 \mu \mu'_2 +2 \mu^3 \\
\mu_4 &= \mu'_4 - 4 \mu \mu'_3 + 6 \mu^2 \mu'_2 - 3 \mu^4.
\end{align}</math>

... and so on,<ref>{{Cite web|url=http://mathworld.wolfram.com/CentralMoment.html|title = Central Moment}}</ref> following [[Pascal's triangle]], i.e.

<math display="block">\mu_5 = \mu'_5 - 5 \mu \mu'_4 + 10 \mu^2 \mu'_3 - 10 \mu^3 \mu'_2 + 4 \mu^5.\,</math>

because {{nowrap|<math> 5\mu^4\mu'_1 - \mu^5 \mu'_0 = 5\mu^4\mu - \mu^5 = 5 \mu^5 - \mu^5 = 4 \mu^5</math>.}}

The following sum is a stochastic variable having a '''''compound distribution'''''

<math display="block">W = \sum_{i=1}^M Y_i, </math>

where the <math>Y_i</math> are mutually independent random variables sharing the same common distribution and <math>M</math> a random integer variable independent of the <math>Y_k</math> with its own distribution. The moments of <math>W</math> are obtained as

<math display="block">\operatorname{E}[W^n]= \sum_{i=0}^n\operatorname{E}\left[\binom{M}{i}\right] \sum_{j=0}^i \binom{i}{j} {\left(-1\right)}^{i-j} \operatorname{E} \left[ \left(\sum_{k=1}^j Y_k\right)^n \right], </math>

where <math display="inline">\operatorname{E} \left[ {\left(\sum_{k=1}^j Y_k\right)}^n\right] </math> is defined as zero for <math>j = 0</math>.