Variance of combined samples?

Factual Questions
1

Say you’ve got two sample groups of single real-valued measurements. Call them S and T, with |S| = m and |T| = n. Call the combined group of measurements X.

Since the boards don’t offer great notation, I’ll use ave() to denote the sample mean, and var() to denote the (unbiased) sample variance. It’s easy to derive that ave(X) = m/(m + n) * ave(S) + n/(m + n) * ave(T). What is var(X) in terms of var(S) and var(T)?

If I was at home, I could do this–I know all the formulas but the algebra is involved.

2

Well, var(S) = avg(S[sup]2[/sup]) - avg(S)[sup]2[/sup], so avg(S[sup]2[/sup]) = var(S) - avg(S)[sup]2[/sup]
so var(X) = m/(m+n) * (var(S) - avg(S)[sup]2[/sup]) + n/(m+n) * (var(T) - avg(T)[sup]2[/sup]) - avg(X)[sup]2[/sup]

3

D’oh.

avg(S[sup]2[/sup]) = var(S) + avg(S)[sup]2[/sup]

so make that

var(X) = m/(m+n) * (var(S) + avg(S)[sup]2[/sup]) + n/(m+n) * (var(T) + avg(T)[sup]2[/sup]) - avg(X)[sup]2[/sup]

math is hard

4

I hadn’t considered that approach, but it looks like it’ll work nicely, and it generalizes to an arbitrary number of samples as well. Thanks.