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I have a list of random numbers with unknown distribution. This list is divided in N chunks with equal length. Then I have the average of every chunk, and the average of these averages.

Is it possible to derive the difference (min,max) between the average of the averages of these chucks and the average of the whole list, in case of unknown distribution?

Edit: I've missed the most important part - it is about integer arithmetics and therefore in this case the averaging introduces rounding.

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  • $\begingroup$ average of the averages of these chucks is the same as the average of the whole list $\endgroup$
    – Markoff Chainz
    Commented Jun 24, 2019 at 8:43
  • $\begingroup$ @markoff , it is about rounded integers, I forgot to mention . $\endgroup$
    – IvanDi
    Commented Jun 24, 2019 at 8:57
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    $\begingroup$ A simple conservative bound can be obtained by assuming that one option always gets rounded down and the other gets rounded up, or vice versa. $\endgroup$
    – user856
    Commented Jun 24, 2019 at 9:23

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Maybe I didn't really understand your question, but average of the averages of these chunks is the same as the average of the whole list.

For example, suppose that your list is $[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]$

And you divide it into 5 chunks $[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15]$.

Average of each chunk : $[2],[5],[8],[11],[14]$

Average of average of chunks : $\frac{2+5+8+11+14}{5} = 8$

Average of the whole list : $\frac{1+2+...+15}{15} = 8$

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