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Continuing on the same topic as my previous post, its nice to be able to gather all the kth order moments in a single pass.

Last time I mentioned the boost/accumulators example, but you will have noticed two issues if you use that.  Firstly, moment<k> tag will give you the kth simple moment relative to zero, whereas we often want the kth central moment of a sequence relative to the mean.  Secondly, although boosts accumulator is well written it does seem to take a while to compile [~ 12 seconds for code using moment<12>].

After some playing around Ive got a faster simpler approach, where the inner loop accumulates kth powers of the element.  After you’ve run the sequence through, you can then easily extract variance, and all the kth central moments.  So in adding the more general case of kth moments, Ive made the particular variance case simpler.  That often seems to happen in programming and math!


First a bit of math and then the code.  We want to express the kth central moment in terms of the k basic moments.

First, lets define the basic moment as –

\displaystyle M_{n}^{j}= \sum_{i=1}^n {x}_i^{j}

We rearrange the binomial expansion –

\displaystyle nv_{n}^{k}= \sum_{i=1}^n({x}_{i}-\mu_{n})^k

\displaystyle = \sum_{i=1}^n \sum_{j=0}^k \binom{k}{j} {x}_{i}^j(-\mu_{n})^{k-j}

\displaystyle = \sum_{j=0}^k \binom{k}{j} (-\mu_{n})^{k-j} \sum_{i=1}^n {x}_{i}^j

So we have the kth central moment given as a weighted sum of the kth simple moments –

\displaystyle v_{n}^{k} = 1/n(\sum_{j=0}^k \binom{k}{j} (-\mu_{n})^{k-j} M_{n}^{j})

which shows that all we need to accumulate as we walk across the sequence is the kth simple powers ({x}_{i})^k .

Notice the variance is now handled as a special case where k=2.  Likewise k=0 corresponds to n, the element count and k=1 is the sum of elements.

c++ impl

Heres a basic impl of the above expression –

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