6 ms·
Unfortunately, by averaging the averages you skew the results. Average of averages does not produce the same result as averaging the whole list.
by svenhof 8y ago
Unfortunately, by averaging the averages you skew the results. Average of averages does not produce the same result as averaging the whole list.
- sokoloff 8y agoAverage of equal size chunks' averages does. (mathematically)
- svenhof 8y agoAh my bad. Thanks for that correction
- umdiff 8y ago\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)
- BubRoss 8y agoIt 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 8y agoThanks for that, didn't know.