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is it possible to calculate the regular average of a sequence of numbers when i dont know everything of the sequence, but just everytime i get a new number i know the total count of numbers and the average for the numbers - 1.

for example: 2 3 10 the average is of course: 5

but in the last step to calculate i only have access to the previous average of 2 and 3: 2.5 the next number: 10 and the count of numbers: 3

if this is possible, how?

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up vote 14 down vote accepted

Yes, and you can derive it from the expression for the average. Let the average of the first $n$ numbers be $\mu_n$. The formula for it is

$$\mu_n = \frac{1}{n} \sum_{i=1}^n x_i$$

Then you can derive

$$n \mu_n = \sum_{i=1}^nx_i = x_n + \sum_{i=1}^{n-1} x_i = x_n + (n-1)\mu_{n-1}$$

and hence, dividing by $n$,

$$\mu_n = \frac{(n-1) \mu_{n-1} + x_n}{n}$$

i.e. to calculate the new average after then $n$th number, you multiply the old average by $n-1$, add the new number, and divide the total by $n$.

In your example, you have the old average of 2.5 and the third number is 10. So you multiply 2.5 by 2 (to get 5), add 10 (to get 15) and divide by 3 (to get 5, which is the correct average).

Note that this is functionally equivalent to keeping a running sum of all the numbers you've seen so far, and dividing by $n$ to get the average whenever you want it (although, from an implementation point of view, it may be better to compute the average as you go using the formula I gave above. For example, if the running sum ever gets larger than $10^{308}$ish then it may be too large to represent as a standard floating point number, even though the average can be represented).

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Thanks alot man. You saved me alot of time :) – bksi Dec 8 '12 at 12:51
Thank you for this, found it with a search and was exactly what I needed. – Ashigore Feb 21 at 19:21
As you're computing the previous sum $(n-1)\mu_{n-1}$ as part of the formula, I think this wouldn't help is the sum gets too large. – danijar Jun 23 at 1:12
@danijar Completely true - a better approach is to keep running sums of the $x_n$ using a method that is robust to rounding error (e.g. Kahan summation) and a separate running count of $n$, and divide whenever you need the mean. That way, your error is bounded by the accuracy of your floating point type. – Chris Taylor Jun 23 at 7:44
If you are likely to end up with numbers larger than $10^{308}$ then you either need to scale down your inputs, or use a more capacious floating point type. – Chris Taylor Jun 23 at 7:46

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