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I am not sure if this is possible, plus I am running on no sleep, so here is what I am asking. Assume I have a list of random numbers. In this instance we will have a list containing $[5, 4, 3, 2, 1]$. I want to retain as little data about these numbers as possible in memory and update the average as new numbers get introduced.

So the average of this list of numbers is $\frac{5+4+3+2+1}{5}=3$. If I add another random number to the list, say $6$, it becomes $[6, 5, 4, 3, 2, 1]$. However, let's say we don't know the previous numbers in the list, just the count of them and their average before we append $6$ to the list. Given these details, I can calculate the new average based on just the count and old average.

$\frac{Count * Avg + NewNumber} {Count + 1} = NewAverage$

or $\frac{5 * 3 + 6} {5 + 1} = 3.5$

Simple enough. Now, is it possible to get the average of the past $X$ numbers without actually having a reference to them? I can't think of a better way to express that without it sounding odd or misleading. Assume after calculating the average and getting the count, and any other scalar numerical values, the list of numbers is not retrievable.

Also I am new to this StackExchange so forgive my etiquette and feel free to correct me on it.

TheKobra
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    Is X variable? Like your list keeps growing, and at any point you want to be able to calculate the average of the past 5 numbers, or the past 3, or whatever past number you want? If so, then no, I don't think there's any way without remembering all the numbers (or a more-or-less equivalent amount of other information). – Duncan Jul 24 '14 at 22:22
  • Yes, that's what I was concerned about. Thanks. – TheKobra Jul 31 '14 at 20:20

1 Answers1

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I think you need to keep $X$ numbers in memory if you want to do that.

An alternative is to gradually forget old numbers, so $$New Average = \frac{(X-1)OldAverage+NewNumber}{X}$$

This weights the new number by $1/X$, and earlier ones by $(X-1)^a/X^{a+1}$ so they lose $1/e$ of their weight every $X$ numbers.

Empy2
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    Also there is the Kalman filter which involves a lot more calculations. – Empy2 Jul 24 '14 at 19:44
  • This seems like it would work, just have to store one extra number. Thank you. – TheKobra Jul 24 '14 at 19:48
  • My apologies. I spoke too soon. Sleep deprivation sucks. I can't retrieve the last X number use in this equation because that would require a list of the numbers to be stored. Thus I don't think this will work without storing that list in memory which is the problem I am trying to avoid. – TheKobra Jul 24 '14 at 19:57
  • I modified the question to make it more clear. – TheKobra Jul 24 '14 at 20:02