[svenhof]

Unfortunately, by averaging the averages you skew the results. Average of averages does not produce the same result as averaging the whole list.

[sokoloff]

Average of equal size chunks' averages does. (mathematically)

[svenhof]

Ah my bad. Thanks for that correction

[BubRoss]

It does if you weight the chunks correctly. Combine pairs of numbers and keep a weight of the remaining odd number. Either cut the weight in half of the odd number every time or use it in the average if an iteration has an even number of numbers.

[svenhof]

Thanks for that, didn't know.

[umdiff]

\left( \sum_{i=1}^m x_i/m + \sum_{i=m+1}^{2m} x_i/m \right) / 2 = \sum_{i=1}^{2m} x_i /(2m)