Is there any advantage of using "central moments" over "moments" when approximating a distribution to a known distribution using moment matching? I have noticed that in lot of papers.
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2$\begingroup$ Since central moments and regular moments are related via linear transformations, there is no theoretical advantage of one form over the other. In some cases, there might be a small computational advantage in beginning with one form versus the other. $\endgroup$– Dilip SarwateCommented Mar 17, 2012 at 19:21
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$\begingroup$ @DilipSarwate, that looks like it addresses the OP's question pretty directly. I wonder if you'd like to switch it to an answer from a comment. $\endgroup$– gung - Reinstate MonicaCommented Mar 18, 2012 at 18:08
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Following the suggestion by gung:
Central moments $\mu_n$ are related to moments $m_n$ through linear transformations:
$$\mu_n = E[(X-\mu)^n] = \sum_{i=0}^n \binom{n}{i}(-\mu)^{n-i}E[X^i] = \sum_{i=0}^n \binom{n}{i}(-\mu)^{n-i}m_i.$$
An adaptation, emendation, and incorporation of a proposal by @Flav for an improved statement:
One theoretical advantage to central moments is that they are invariant to translations along the $x$ axis: the central moments of $Y = X + a$ are the same as the central moments of $X$. From a computational perspective, in some cases, (non-central) moments are slightly easier to calculate than central moments, while in other cases, the central moments are slightly easier to calculate.
answered Mar 18, 2012 at 18:33
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$\begingroup$ Thanks for the reply. I solved a moment matching problem and have seen that expression for central moment is much simpler than that of moment $\endgroup$ Commented Mar 19, 2012 at 16:35
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$\begingroup$ @prasenjit: Don't forget to designate this as your answer if it turns out to be the best. $\endgroup$– WayneCommented Jun 21, 2012 at 20:54