Как рассчитать объединенную дисперсию двух или более групп с учетом известных групповых дисперсий, средних значений и размеров выборки?

32

Скажем, есть m+n элементов, разбитых на две группы ( m и n ). Дисперсия первой группы σm2 и дисперсия второй группы σn2 . Предполагается, что сами элементы неизвестны, но я знаю, что означает μm и μn .

Есть ли способ расчета комбинированной дисперсии σ(m+n)2 ?

The variance doesn't have to be unbiased so denominator is (m+n) and not (m+n1).

user1809989
источник
When you say you know the means and variances of these groups, are they parameters or sample values? If they are sample means/variances you should not use μ and σ...
Jonathan Christensen
I just used the symbols as a representation. Otherwise, it would have been hard to explain my problem.
user1809989
1
For sample values, we usually use Latin letters (e.g. m and s). Greek letters are usually reserved for parameters. Using the "correct" (expected) symbols will help you communicate more clearly.
Jonathan Christensen
No worries, I'll follow that from now on! Cheers
user1809989
1
@Jonathan Because this is not a question about samples or estimation, one can legitimately take the view that μ and σ2 are the true mean and variance of the empirical distribution of a batch of data, thereby justifying the conventional use of greek letters rather than latin letters to refer to them.
whuber

Ответы:

36

Use the definitions of mean

μ1:n=1ni=1nxi

and sample variance

σ1:n2=1ni=1n(xiμ1:n)2=n1n(1n1i=1n(xiμ1:n)2)

(the last term in parentheses is the unbiased variance estimator often computed by default in statistical software) to find the sum of squares of all the data xi. Let's order the indexes i so that i=1,,n designates elements of the first group and i=n+1,,n+m designates elements of the second group. Break that sum of squares by group and re-express the two pieces in terms of the variances and means of the subsets of the data:

(m+n)(σ1:m+n2+μ1:m+n2)=i=11:n+mxi2=i=1nxi2+i=n+1n+mxi2=n(σ1:n2+μ1:n2)+m(σ1+n:m+n2+μ1+n:m+n2).

Algebraically solving this for σm+n2 in terms of the other (known) quantities yields

σ1:m+n2=n(σ1:n2+μ1:n2)+m(σ1+n:m+n2+μ1+n:m+n2)m+nμ1:m+n2.

Of course, using the same approach, μ1:m+n=(nμ1:n+mμ1+n:m+n)/(m+n) can be expressed in terms of the group means, too.


An anonymous contributor points out that when the sample means are equal (so that μ1:n=μ1+n:m+n=μ1:m+n), the solution for σm+n2 is a weighted mean of the group sample variances.

whuber
источник
4
The "homework" tag doesn't mean the question is elementary or stupid: it's used for self-study questions that can even include research-level queries. It distinguishes routine, more or less context-free questions (of the sort that might ordinarily grace the math forum) from specific applied questions.
whuber
I cannot understand your first passage: n(σ2+μ2)=(xμ)2+nμ2=?x2 In particular I get [(xμ)2+μ2]=[x22xμ] which requires μ=0 Am I missing something? Could you please explain this?
DarioP
2
@Dario
(xμ)2+nμ2=(x22μx+nμ2)+nμ2=x22nμ2+2nμ2=x2.
whuber
Oh yes, I did a stupid sign mistake in my derivation, now is clear, thanks!!
DarioP
4
I guess this can be extended to an arbitrary number of samples as long as you have the mean and variance for each. Calculating pooled (biased) standard deviation in R is simply sqrt(weighted.mean(u^2 + rho^2, n) - weighted.mean(u, n)^2) where n, u and rho are equal-length vectors. E.g. n=c(10, 14, 9) for three samples.
Jonas Lindeløv
3

I'm going to use standard notation for sample means and sample variances in this answer, rather than the notation used in the question. Using standard notation, another formula for the pooled sample variance of two groups can be found in O'Neill (2014) (Result 1):

spooled2=1n1+n21[(n11)s12+(n21)s22+n1n2n1+n2(x¯1x¯2)2].

This formula works directly with the underlying sample means and sample variances of the two subgroups, and does not require intermediate calculation of the pooled sample mean. (Proof of result in linked paper.)

Reinstate Monica
источник
-3

Yes, given the mean, sample count, and variance or standard deviation of each of two or more groups of samples, you can exactly calculate the variance or standard deviation of the combined group.

This web page describes how to do it, and why it works; it also includes source code in Perl: http://www.burtonsys.com/climate/composite_standard_deviations.html


BTW, contrary to the answer given above,

n(σ2+μ2)    i=1nxi2

See for yourself, e.g., in R:

> x = rnorm(10,5,2)
> x
 [1] 6.515139 8.273285 2.879483 3.624233 6.199610 3.683164 4.921028 8.084591
 [9] 2.974520 6.049962
> mean(x)
[1] 5.320502
> sd(x)
[1] 2.007519
> sum(x**2)
[1] 319.3486
> 10 * (mean(x)**2 + sd(x)**2)
[1] 323.3787
Dave Burton
источник
it's because you forgot the n-1 factor, e.g. try with n*(mean(x)**2+sd(x)**2/(n)*(n-1))
user603
user603, what on earth are you talking about?
Dave Burton
3
Dave, mathematics is a more reliable teacher than software. In this case R computes the unbiased estimate of the standard deviation rather than the standard deviation of the set of numbers. For instance, sd(c(-1,1)) returns 1.414214 rather than 1. Your example needs to use sqrt(9/10)*sd(x) in place of sd(x). Interpreting "σ" as the SD of the data and "μ" as the mean of the data, your BTW remark is wrong. A program demonstrating this is n <- 10; x <- rnorm(n,5,2); m <- mean(x); s <- sd(x) * sqrt((n-1)/n); m2 <- sum(x^2); c(lhs=n * (m^2 + s^2), rhs=m2)
whuber