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Say there are $m+n$ elements split into two groups ($m$ and $n$). The variance of the first group is $\sigma_m^2$ and the variance of the second group is $\sigma^2_n$. The elements themselves are assumed to be unknown but I know the means $\mu_m$ and $\mu_n$.

Is there a way to calculate the combined variance $\sigma^2_{(m+n)}$?

The variance doesn't have to be unbiased so denominator is $(m+n)$ and not $(m+n-1)$.

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  • $\begingroup$ 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 $\mu$ and $\sigma$... $\endgroup$ – Jonathan Christensen Nov 8 '12 at 19:18
  • $\begingroup$ I just used the symbols as a representation. Otherwise, it would have been hard to explain my problem. $\endgroup$ – user1809989 Nov 9 '12 at 0:11
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    $\begingroup$ 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. $\endgroup$ – Jonathan Christensen Nov 9 '12 at 2:31
  • $\begingroup$ No worries, I'll follow that from now on! Cheers $\endgroup$ – user1809989 Nov 9 '12 at 8:42
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    $\begingroup$ @Jonathan Because this is not a question about samples or estimation, one can legitimately take the view that $\mu$ and $\sigma^2$ 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. $\endgroup$ – whuber Nov 9 '12 at 18:36
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The idea is to express quantities as sums rather than fractions.

Given any $n$ data values $x_i,$ use the definitions of the mean

$$\mu_{1:n} = \frac{1}{\Omega_{1;n}}\sum_{i=1}^n \omega_{i} x_i$$

and sample variance

$$\sigma_{1:n}^2 = \frac{1}{\Omega_{1;n}}\sum_{i=1}^n \omega_{i}\left(x_i - \mu_{1:n}\right)^2 = \frac{1}{\Omega_{1;n}}\sum_{i=1}^n \omega_{i}x_i^2 - \mu_{1:n}^2$$

to find the (weighted) sum of squares of the data as

$$\Omega_{1;n}\mu_{1:n} = \sum_{i=1}^n \omega_{i} x_i$$

and

$$\Omega_{1;n} \sigma_{1:n}^2 = \sum_{i=1}^n \omega_{i}\left(x_i - \mu_{1:n}\right)^2 = \sum_{i=1}^n \omega_{i}x_i^2 - \Omega_{1;n}\mu_{1:n}^2.$$

For notational convenience I have written $$\Omega_{j;k}=\sum_{i=j}^k \omega_i$$ for sums of weights. (In applications with equal weights, which are the usual ones, we may take $\omega_i=1$ for all $i,$ whence $\Omega_{1;n}=n.$)

Let's do the (simple) algebra. Order the indexes $i$ so that $i=1,\ldots,n$ designates elements of the first group and $i=n+1,\ldots,n+m$ designates elements of the second group. Break the overall combination of squares by group and re-express the two pieces in terms of the variances and means of the subsets of the data:

$$\eqalign{ \Omega_{1;n+m}(\sigma^2_{1:m+n} + \mu_{1:m+n}^2)&= \sum_{i=1}^{1:n+m} \omega_{i}x_i^2 \\ &= \sum_{i=1}^n \omega_{i} x_i^2 + \sum_{i=n+1}^{n+m} \omega_{i} x_i^2 \\ &= \Omega_{1;n}(\sigma^2_{1:n} + \mu_{1:n}^2) + \Omega_{n+1;n+m}(\sigma^2_{1+n:m+n} + \mu_{1+n:m+n}^2). }$$

Algebraically solving this for $\sigma^2_{m+n}$ in terms of the other (known) quantities yields

$$\sigma^2_{1:m+n} = \frac{\Omega_{1;n}(\sigma^2_{1:n} + \mu_{1:n}^2) + \Omega_{n+1;n+m}(\sigma^2_{1+n:m+n} + \mu_{1+n:m+n}^2)}{\Omega_{1;n+m}} - \mu^2_{1:m+n}.$$

Of course, using the same approach, $\mu_{1:m+n} = (\Omega_{1;n}\mu_{1:n} + \Omega_{n+1;n+m}\mu_{1+n:m+n})/\Omega_{1;n+m}$ can be expressed in terms of the group means, too.


Edit 1

An anonymous contributor points out that when the sample means are equal (so that $\mu_{1:n}=\mu_{1+n:m+n}=\mu_{1:m+n}$), the solution for $\sigma^2_{m+n}$ is a weighted mean of the group sample variances.

Edit 2

I have generalized the formulas to weighted statistics. The motivation for this is a recent federal court case in the US involving a dispute over how to pool weighted variances: a government agency contends the proper method is to weight the two group variances equally. In working on this case I found it difficult to find authoritative references on combining weighted statistics: most textbooks do not deal with this or they assume the generalization is obvious (which it is, but not necessarily to government employees or lawyers!).

BTW, I used entirely different notation in my work on that case. If in the editing process any error has crept into the formulas in this post I apologize in advance and will fix them--but that would not reflect any error in my testimony, which was very carefully checked.

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    $\begingroup$ 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. $\endgroup$ – whuber Nov 9 '12 at 18:33
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    $\begingroup$ @Dario $$\sum(x-\mu)^2+n\mu^2=(\sum x^2 - 2\mu\sum x + n \mu^2)+n\mu^2 = \sum x^2 - 2n\mu^2 + 2n\mu^2 = \sum x^2.$$ $\endgroup$ – whuber Oct 22 '14 at 15:33
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    $\begingroup$ 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. $\endgroup$ – Jonas Lindeløv Dec 7 '14 at 13:45
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    $\begingroup$ @whuber Sorry to bring you back to an old question, but could you shed some light on how you went from $\sigma_{1:n}^2 = \frac{1}{n}\sum_{i=1}^n \left(x_i - \mu_{1:n}\right)^2 = \frac{n-1}{n}\left(\frac{1}{n-1}\sum_{i=1}^n \left(x_i - \mu_{1:n}\right)^2\right)$ to $(m+n)(\sigma^2_{1:m+n} + \mu_{1:m+n}^2) = \sum_{i=1}^{1:n+m} x_i^2$. I don't really understand where the $(m+n)(\sigma^2_{1:m+n} + \mu_{1:m+n}^2)$ came from. Sorry if I'm missing something basic. $\endgroup$ – Chris C Oct 4 '15 at 17:47
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    $\begingroup$ @Chris C It expresses the squares in terms of the variance and the mean by rewriting the usual formula giving a variance in terms of the squares and the mean. Recall that this says the variance of a dataset of $N$ values is $1/N$ times the sum of squares, minus the squared mean. Multiply all of these by $N$ to deduce (easily) that the sum of squares equals $N$ times the variance plus $N$ times the squared mean. That's the entire point of the proceedings: everything is straightforward when expressed as sums instead of averages. $\endgroup$ – whuber Oct 4 '15 at 17:57
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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):

$$\begin{equation} \begin{aligned} s_\text{pooled}^2 &= \frac{1}{n_1+n_2-1} \Bigg[ (n_1-1) s_1^2 + (n_2-1) s_2^2 + \frac{n_1 n_2}{n_1+n_2} (\bar{x}_1 - \bar{x}_2)^2 \Bigg]. \\[10pt] \end{aligned} \end{equation}$$

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.)

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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,

$$\eqalign{ n(\sigma^2 + \mu^2) \space\space \ne \space\space \sum_{i=1}^n x_i^2 }$$

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
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  • $\begingroup$ it's because you forgot the n-1 factor, e.g. try with n*(mean(x)**2+sd(x)**2/(n)*(n-1)) $\endgroup$ – user603 May 13 '13 at 20:11
  • $\begingroup$ user603, what on earth are you talking about? $\endgroup$ – Dave Burton May 15 '14 at 5:09
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    $\begingroup$ 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 "$\sigma$" as the SD of the data and "$\mu$" 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) $\endgroup$ – whuber May 15 '14 at 15:35

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