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I've tried some recursive moving average formulae (to reuse a previous output instead of summing the whole n-long set for every i) I've managed to find but none of them produces the same results as a bare moving mean does. Is there a reliable recursive formula which would produce exactly (or almost exactly) the same output as a bare moving mean?

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It seems that you are using some kind of software. If this is true, please tell us, it is easier to give an answer then. – mpiktas Mar 16 '12 at 4:23
I program in Scala. – Ivan Mar 16 '12 at 4:26

2 Answers

up vote 6 down vote accepted

Just try to remove the last value of the window and add the new one.

If

$$MA(t)=\frac{1}{w}\sum\limits_{i=t-w+1}^t{y_i}$$

then

$$MA(t+1)=MA(t)+\frac{y(t+1)-y(t-w+1)}{w}.$$

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1  
Yes, this formula looks reasonable but somehow the result I get is always below the non-recursive moving mean. And the difference looks constant. Perhaps I have a bug in my code, the difference would hardly be constant otherwise. – Ivan Mar 16 '12 at 4:37
The possible reason might be the following. If the values of y(t) are relatively small and w is big, divisions for sum(y(t))/w and (y(t)+y(t-w))/w have different precision orders. (for example if y(t) is approximately 0.001, and n is 10000, division 0.001/10000 is less accurate, compared to 10/10000). Then if you add real numbers with different precisions, the smaller one is truncated. – chab Mar 16 '12 at 4:54
This formula is not correct because it subtracts the new value rather than adds it! – whuber♦ Mar 16 '12 at 16:15
1  
Yes, you was right. Just corrected it. – chab Mar 17 '12 at 13:05
Thank you! (+1). I have taken the liberty of also correcting a slight indexing error. – whuber♦ Mar 17 '12 at 14:34
up vote 1 down vote 
double mean(const double F, const double C, unsigned int *n)
{
  return (F*(*n)+C)/(++*n);
}

F is the old average number, C is a new addition to the avarage. *n is the number of values in F. This does not need a buffer.

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This is a way of updating a cumulative mean but not a moving mean: that's why no buffer is needed. – whuber♦ Feb 4 at 13:53

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