How to add and subtract values from an average?

Question

Say I have 100 numbers that are averaged:

number of values = 100
total sum of values = 2000
mean = 2000 / 100 => 20

If I want to add a value and find out the new average:

total sum of values = 2000 + 100
mean = 2100 / 101 => 20.79

If I want to subtract a value and find out the new average:

total sum of values = 2100 - 100
mean = 2000 / 100 => 20

It seems to work, but is the above correct?

Is this the proper way to add/subtract values from a average without having to re-sum all the 100 numbers first?

score 82 · Answer 1 · answered Oct 03 '14 at 23:08

82

I know that's an old thread but I had the same problem. I want to add a value to an existing average without calculate it back to the total sum.

to add an value to an exisitng average we only must know for how much values the average is calculated: $$ average_{new} = average_{old} + \frac{ value_{new} - average_{old}}{size_{new}} $$

answered Oct 03 '14 at 23:08

501 - not implemented

995

26

I may be a smart guy, but the accepted answer above was incomprehensible to me. This answer works perfectly, and I can understand it. Thanks! – Ty H. Nov 13 '14 at 16:37
3

@501 - not implemented thank you! – ThelmaJay Jan 13 '15 at 14:34
1

This answer was exactly what I needed. Elegant! – visc Nov 16 '16 at 23:02
-1 Careful this formula works only for adding 1 new value! Please work out the correct equation using this answer and you'll see! https://math.stackexchange.com/a/1153800/503549 – kouton Nov 16 '17 at 04:25
Is it possible to calculate median in a similar way? – Przemysław Niedziela Mar 09 '20 at 22:14
2

@kouton why would you downvote for an answer that only works for adding one value when the requirement was "...to add A value..." (emphasis added). If I ask you to run a mile and you run a mile, I'm not going to penalize you for not running two miles. – Jamie Apr 05 '22 at 00:14

score 37 · Accepted Answer · edited Jun 05 '20 at 12:41

37

$s=\frac{a_1+...+a_n}{n}$.

If you want the average of $a_1,...,a_n$ and $a_{n+1}$, then $s'=\frac{a_1+...+a_n+a_{n+1}}{n+1}=\frac{ns+a_{n+1}}{n+1} = \frac{(n+1)s+a_{n+1}}{n+1} - \frac{s}{n+1} = s + \frac{a_{n+1}-s}{n+1}$

If you want the average of $a_1,...,a_{n-1}$ then $s''=\frac{a_1+...+a_{n-1}}{n-1}=\frac{ns-a_n}{n-1}= \frac{(n-1)s-a_n}{n-1} + \frac{s}{n-1}=s+\frac{s-a_n}{n-1}$.

edited Jun 05 '20 at 12:41

John St

3

answered Feb 16 '11 at 13:19

Weltschmerz

6,875

3

And $ns$ is the old sum, as you can derive from the first equation. – Ross Millikan Feb 16 '11 at 13:21
2

I'm tempted to downvote as ns is not explained in the answer – Celeritas Aug 22 '14 at 20:32
At first I commented that @501's answer was clearer but realized I didn't fully understand why the question worked. This answer, while denser, actually explains the result. – jds Dec 10 '14 at 23:05
3

@Celeritas $ns = n \cdot s$, $n$ and $s$ are defined quantities in the post. – Axoren Jan 13 '15 at 15:13
This is the working answer. @501's answer doesn't work when increasing n by more than 1. – kouton Nov 16 '17 at 04:26

score 26 · Answer 3 · edited Jan 13 '20 at 05:33

26

To put it programmatically, and since the question was about how to both add and subtract:

Add a value:

average = average + ((value - average) / nValues)

Subtract a value:

average = (average * nValues - value) / (nValues - 1)

edited Jan 13 '20 at 05:33

Dan Dascalescu

300

answered Dec 09 '15 at 10:35

Damien

491

10

Oh thank god there's another programmer on here. Those math equations make my head hurt. – Matt Jan 05 '19 at 06:38
@MattD: see also this answer on StackOverflow for replacing a number in a running average. – Dan Dascalescu Jan 13 '20 at 05:03
2

Shouldn't the first formula end with / nValues + 1) rather than / nValues)? – Magnus Jun 18 '21 at 19:56

score 3 · Answer 4 · edited May 08 '21 at 22:07

3

Another formula can be

$$\text{NewAvg} = \frac{( \text{OldAvg} \cdot \text{OldSize} ) + \text{NewValue} } { \text{NewSize}}$$

edited May 08 '21 at 22:07

Eduardo Magalhães

4,878

answered May 08 '21 at 20:11

Sushrut Thakur

39
1

I was looking for this answer. It's intuitive because newAvg = (sum + newValue) / newSize (by definition of average). And also by definition sum = avg * size. – dekuShrub Mar 03 '22 at 06:57
And also easy for multiple values: newAvg = (sum1 + sum2) / (size1 + size2). Or in the other format newAvg = (avg1 * size1 + avg2 * size2) / (size1 + size2). – dekuShrub Mar 03 '22 at 07:00

How to add and subtract values from an average?

4 Answers4