What is the overall mean and variance for partitioned data?

Suppose that:

A normally distributed random variable $x$ is being sampled.
There are k partitions, each with $n_{i}$ samples (where $i=1\ldots k$ ).
For each partition, the mean $\overline{x_{i}}$ and variance $s_{x_{i}}{}^{2}$ are known, but the original observations $x_{i}$ are not available.
The overall mean $\bar{x}$ and variance $s_{x}{}^{2}$ are desired.

Computation of $N$ and $\overline x$ are straightforward:

$N={\displaystyle \sum_{i=1}^{k}}n_{i}$ [1]

$\bar{x}=\frac{{\displaystyle \sum_{i=1}^{k}}\overline{x_{i}}}{N}=\frac{{\displaystyle \sum_{i=1}^{k}}{\displaystyle \sum_{j=1}^{n_{i}}}x_{ij}}{N}$ [2]

However, because the overall mean $\bar{x}$ is not available within the partition where $x_{ij}$ is sampled, the formula for the variance:

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}}{\displaystyle \sum_{j=1}^{n_{i}}}\left(x_{ij}-\bar{x}\right)^{2}}{N-1}$ [3]

must be rewritten as

This formula may be derived as follows.

First introduce $\overline{x_{i}}$ into the formula for $s_{x}{}^{2}$

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}\left((x_{ij}-\overline{x_{i}})+(\overline{x_{i}}-\overline{x})\right)^{2}}}}{(N-1)}$ [5]

Then, simplify to remove $\overline{x}$ from the partition summation $\left({\displaystyle \sum_{j=1}^{n_{i}}}\right)$ :

Apply $(a+b)^{2}=a^{2}+2ab+b^{2}$

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(x_{ij}-\overline{x_{i}})^{2}}}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}\left(2(x_{ij}-\overline{x_{i}})+(\overline{x_{i}}-\overline{x})\right)(\overline{x_{i}}-\overline{x})}}}{(N-1)}$ [6]

Replace $-2\overline{x_{i}}+\overline{x_{i}}$ with $-\overline{x_{i}}$

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(x_{ij}-\overline{x_{i}})^{2}}}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(2x_{ij}-\overline{x_{i}}-\overline{x})(\overline{x_{i}}-\overline{x})}}}{(N-1)}$ [7]

Replace $-\overline{x_{i}}-\overline{x}$ with $-(\overline{x_{i}}+\overline{x})$

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(x_{ij}-\overline{x_{i}})^{2}}}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}\left(2x_{ij}-(\overline{x_{i}}+\overline{x})\right)(\overline{x_{i}}-\overline{x})}}}{(N-1)}$ [8]

Distribute $(\overline{x_{i}}-\overline{x})$

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(x_{ij}-\overline{x_{i}})^{2}}}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}2x_{ij}(\overline{x_{i}}-\overline{x})-(\overline{x_{i}}+\overline{x})(\overline{x_{i}}-\overline{x})}}}{(N-1)}$ [9]

Apply $(a+b)(a-b)=a^{2}-b^{2}$ ;
express summation of sum as sum of summations

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(x_{ij}-\overline{x_{i}})^{2}}}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}\left({\displaystyle \left(\sum_{j=1}^{n_{i}}2x_{ij}(\overline{x_{i}}-\overline{x})\right)-\left(\sum_{j=1}^{n_{i}}\left(\overline{x_{i}}^{2}-\overline{x}^{2}\right)\right)}\right)}}{(N-1)}$ [10]

Because neither $\overline{x_{i}}$ nor $\overline{x}$ depends on j

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(x_{ij}-\overline{x_{i}})^{2}}}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}\left({\displaystyle \left(\sum_{j=1}^{n_{i}}2x_{ij}(\overline{x_{i}}-\overline{x})\right)-\left(\sum_{j=1}^{n_{i}}1\right)\left(\overline{x_{i}}^{2}-\overline{x}^{2}\right)}\right)}}{(N-1)}$ [11]

Apply $\sum_{j=1}^{n_{i}}}1=n_{i$ ;
distribute $\sum_{j=1}^{n_{i}}x_{ij}$ ;
express summation of sum as sum of summations;
simplify

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(x_{ij}-\overline{x_{i}})^{2}}}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}\left({\displaystyle 2\overline{x_{i}}\left(\sum_{j=1}^{n_{i}}x_{ij}\right)-2\overline{x}\left(\sum_{j=1}^{n_{i}}x_{ij}\right)-n_{i}\overline{x_{i}}^{2}+n_{i}\overline{x}^{2}}\right)}}{(N-1)}$ [12]

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(x_{ij}-\overline{x_{i}})^{2}}}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}\left({\displaystyle 2\overline{x_{i}}\left(\sum_{j=1}^{n_{i}}x_{ij}\right)-n_{i}\overline{x_{i}}^{2}-2\overline{x}\left(\sum_{j=1}^{n_{i}}x_{ij}\right)+n_{i}\overline{x}^{2}}\right)}}{(N-1)}$ [13]

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(x_{ij}-\overline{x_{i}})^{2}}}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}\left({\displaystyle 2\overline{x_{i}}\left(\sum_{j=1}^{n_{i}}x_{ij}\right)-n_{i}\overline{x_{i}}^{2}}\right)+\sum_{i=1}^{k}\left(n_{i}\overline{x}^{2}-2\overline{x}\left(\sum_{j=1}^{n_{i}}x_{ij}\right)\right)}}{(N-1)}$ [14]

Apply $\left(\sum_{j=1}^{n_{i}}x_{ij}\right)=n_{i}\overline{x_{i}}$ ;
simplify

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(x_{ij}-\overline{x_{i}})^{2}}}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}\left({\displaystyle 2\overline{x_{i}}n_{i}\overline{x_{i}}-n_{i}\overline{x_{i}}^{2}}\right)+\sum_{i=1}^{k}\left(n_{i}\overline{x}^{2}-2\overline{x}n_{i}\overline{x_{i}}\right)}}{(N-1)}$ [15]

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(x_{ij}-\overline{x_{i}})^{2}}}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}\left({\displaystyle n_{i}\overline{x_{i}}^{2}}\right)}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}\left(n_{i}\overline{x}^{2}-2\overline{x}n_{i}\overline{x_{i}}\right)}}{(N-1)}$ [16]

Express summation of sum as sum of summations;
simplify

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(x_{ij}-\overline{x_{i}})^{2}}}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}\left({\displaystyle n_{i}\overline{x_{i}}^{2}}\right)}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}\left(n_{i}\overline{x}^{2}\right)}-{\displaystyle \sum_{i=1}^{k}\left(2\overline{x}n_{i}\overline{x_{i}}\right)}}{(N-1)}$ [17]

Because $\overline{x}$ does not depend on j, factor out $\overline{x}$

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(x_{ij}-\overline{x_{i}})^{2}}}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}\left({\displaystyle n_{i}\overline{x_{i}}^{2}}\right)}}{(N-1)}+\frac{{\displaystyle \overline{x}^{2}\sum_{i=1}^{k}\left(n_{i}\right)}-{\displaystyle 2\overline{x}\sum_{i=1}^{k}\left(n_{i}\overline{x_{i}}\right)}}{(N-1)}$ [18]

Apply ${\displaystyle \sum_{i=1}^{k}}n_{i}=N$ and $\sum_{i=1}^{k}\left(n_{i}\overline{x_{i}}\right)=N\overline{x}$ ;
simplify

$s_{x}{}^{2}=\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle \sum_{j=1}^{n_{i}}(x_{ij}-\overline{x_{i}})^{2}}}}{(N-1)}+\frac{{\displaystyle \sum_{i=1}^{k}{\displaystyle n_{i}\overline{x_{i}}^{2}}}}{(N-1)}+\frac{{\displaystyle \overline{x}^{2}N}-{\displaystyle 2\overline{x}N\overline{x}}}{(N-1)}$ [19]

Reference

This post was largely inspired by
http://stats.stackexchange.com/questions/10441/how-to-calculate-the-variance-of-a-partition-of-variables/10445#10445
which tantalizingly ended with the (under-)statement:

These formulas are easy to derive by writing the desired variance as the scaled sum of $(X_{ji}-\overline{X})^2$ , then introducing $\overline{X_j}$ : $\left[(X_{ji}-\overline{X_{ji}})-(\overline{X_{ji}}-\overline{X})\right]^2$ , using the square of difference formula, and simplifying