4
$\begingroup$

Say I have a set S of values, and want to store in a database some summary information about that set, so that later when I acquire a new value v I can make a reasonable estimate of what the summary information would be about the set S ∪ {v} --- although by now I no longer have access to the original members of S. I'd like the summary information to include the mean and variance of these sets, and as minimal additional information as needed. A natural idea for the additional information would be S's cardinality. But I'm willing to save more complicated information about S if needed. My main constraint is to minimize the size of the retained information.

If I only cared about the mean of the sets, then storing the mean plus cardinality of S would obviously be enough. I could update with a new value by just taking a weighted average of the old mean (times the old cardinality) and the new value. But I'd like to be able to keep track of the variance of the sets too. A good estimate is enough; I don't need to be able to reconstruct what the exact mean and variance of S ∪ {v} would be.

I expect that even asking this displays how naive I am about statistics, but I'd appreciate any help. I don't know where to look for answers.

$\endgroup$
2
  • $\begingroup$ Would this give what I'm looking for? $\endgroup$
    – dubiousjim
    Commented Dec 20, 2017 at 20:52
  • $\begingroup$ This problem was discussed, with proof and some alternate methods over on math.stackexchange.com. (Copied from answer by Richmond Newman) $\endgroup$
    – Stephen Rauch
    Commented Dec 20, 2017 at 23:05

2 Answers 2

4
$\begingroup$

This problem was discussed, with proof and some alternate methods over on math.stackexchange.

$\endgroup$
1
  • 1
    $\begingroup$ After further reading, I might also warn that some of the methods there (while correct mathematically) may pose numerical stability issues due to large numbers of multiplications/divisions and accumulated rounding errors. I would probably favor the method given in that discussion by mjqxxxx $\endgroup$ Commented Dec 21, 2017 at 1:15
4
$\begingroup$

Following that link about moving variance in my comment, I came upon this: Welford's online algorithm for calculating variance, which seems to supply what I'm looking for.

Here's the algorithm:

new_count = old_count + 1
d1 = new_value - old_mean
new_mean = old_mean + d1/new_count
d2 = new_value - new_mean
new_sum_squares = old_sum_squares + d1*d2

From a saved (count, mean, sum_squares), the population variance can be computed as sum_squares/count.

Given an initial value v, one can start with:

count = 1
mean = v
sum_squares = 0

If you want a weighted mean and variance, you can modify the algorithm like this:

new_count = old_count + new_weight
d1 = (new_value - old_mean)*new_weight

The other lines stay the same. (Here the mean of values a,b with weights x,y, respectively, is (ax+by)/(x+y); and the weighted sum_squares is x(a-mean)^2 + y(b-mean)^2.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.