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Let's say I have an array of $N$ numbers:

$n_1, n_2, .... n_N$

I then calculate a running average of those numbers, by averaging $A$ numbers at a time, which will result in the following array:

$a_1, a_2, .... a_{N-A+1}$ , where

$a_1 = \frac {n_1 + n_2 + ... + n_A}{A}$

$a_2 = \frac {n_2 + n_3 + ... + n_{A+1}}{A}$

...

I then calculate a running average of the running average I obtained above, by averaging B numbers at a time, which will result in the following array:

$b_1, b_2, .... b_{N-A-B+2}$ , where

$b_1 = \frac {a_1 + a_2 + ... + a_B}{B}$

$b_2 = \frac {a_2 + a_3 + ... + a_{B+1}}{B}$

...

My question is, is there some general formula that can get you the second running average (b numbers) directly from original set (n numbers), with arbitrary A and B coefficients?

I'm asking, because I did a few of these by hand, and the results look very much like digital filters (weighted averages). Thanks.

Ivan Neretin
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Eternal21
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    Your examples are unclear. You say "averaging $A$ numbers at a time", but you show averages involving $3$ numbers (e.g., $a_1=\frac{n_1+n_2+n_3}{A}$). – Matthew Conroy Mar 07 '16 at 17:42
  • @MatthewConroy You're right, I'll fix it. – Eternal21 Mar 07 '16 at 17:47
  • If you look at the moving average as a digital filter, this is just passing a signal into two FIRs in cascade. So you can convolve the two impulse responses of the FIRs and get a combined system.

    Remember convolving two box car gives a triangle, that's why you get the increasing and then decreasing and also symmetric coefficients down there in one of the answers.

     – Andrew Au Mar 07 '16 at 17:57

2 Answers2

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For $A=3,B=4$, if we just substitute in we get $b_i=\frac {a_i+a_{i+1}+a_{i+2}+a_{i+3}}4=\frac{n_i+2n_{i+1}+3n_{i+2}+3n_{i+3}+2n_{i+4}+n_{i+5}}{12}$ The pattern should be suggestive. You ramp up the coefficients, stay constant at the peak, then ramp them down. You could write it with summation signs if you want. $$b_i=\frac 1B\sum_{j=i}^{i+B-1}a_j=\frac 1{AB}\sum_{j=i}^{i+B-1}\sum_{k=j}^{j+A-1}n_k$$ and you would like to write this as $$b_i=\frac 1{AB}\sum_{j=i}^{i+A+B-2}c_jn_n$$ The $c_j$ are the coefficients $1,2,3,3,2,1$ above. You can write $c_j=\min(A,B)-$end effects, where the end effects cause the ramp down at the ends. If $j-i \lt \min(A,B), c_j=j-i+1$ and if $i+A+B-2-j \lt \min (A,B), c_j=i+A+B-1-j$

Ross Millikan
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For the sake of illustration, let us average on $3$ numbers, then $4$.

Ignoring the constant denominators for convenience,

$$b_1=a_1+a_2+a_3,b_2=a_2+a_3+a_4,b_3=a_3+a_4+a_5,\\ b_4=a_4+a_5+a_6,b_5=a_5+a_6+a_7,b_6=a_6+a_7+a_8\cdots$$

and

$$c_1=b_1+b_2+b_3+b_4=a_1+2a_2+3a_3+3a_4+2a_5+a_6,\\ c_2=a_2+2a_3+3a_4+3a_5+2a_6+a_7,\\ c_3=a_3+2a_4+3a_5+3a_6+2a_7+a_8\cdots$$

The pattern is a sum of $6$ consecutive elements weighted by $1,2,3,3,2,1$ (to be divided by $3\cdot4$). Such a sliding linear combination is called a convolution operation. (You can express this easily using the Discrete Fourier Transform.)