Mean and Variance of subset of a data set I have a data set of position measurements of an object. However, the data set is split into subsets. The subsets have equal size. I want to find the mean and variance of the whole data set only with access to the subsets, and the not the whole data set at once. How would I go about doing this? Is it correct that the mean of the means of the subsets is equal to the mean of the whole set? What about the variance?
Suppose you have $k$ groups of observations on a variable $x$ (say) where the $i$th group consists of $n_i$ observations, $i=1,\ldots,k$. Let the $j$th observation in the $i$th group be $x_{ij}$ for $i=1,\ldots,k$ and $j=1,\ldots,n_i$.

The $i$th group mean is defined as $$\overline{x_i}=\frac1{n_i}\sum_{j=1}^{n_i} x_{ij}\quad,\,i=1,\ldots,k $$

Then the pooled mean or combined mean is given by $$\overline{\overline x}=\frac{\sum_{i=1}^k n_i\overline x_i }{\sum_{i=1}^k n_i} $$

This is a weighted average with weights being the number of observations in the $i$th group.

The $i$th group variance is defined as

$$s_i^2=\frac1{n_i}\sum_{j=1}^{n_i}\left(x_{ij}-\overline{\overline x}\right)^2\quad,\,i=1,\ldots,k $$

And the pooled variance based on all observations from all groups is given by

$$s^2=\frac{\sum_{i=1}^k\sum_{j=1}^{n_i}\left(x_{ij}-\overline{\overline x}\right)^2}{\sum_{i=1}^k n_i}=\frac{\sum_{i=1}^k n_is_i^2}{\sum_{i=1}^k n_i}+\frac{\sum_{i=1}^k n_i\left(\overline x_i-\overline{\overline x}\right)^2}{\sum_{i=1}^k n_i}$$

You are concerned with the case $n_1=n_2=\cdots=n_k$.
If ${\cal D} = \{ {\cal D}_1, {\cal D}_2, \ldots, {\cal D}_n \}$, each of the same number of points, then the mean of ${\cal D}$ is simply the mean (expectation) of the individual means:

$${\cal E}[{\cal D}] = \frac{1}{n} \sum\limits_{i=1}^n {\cal E}[{\cal D}_i]$$

The variance sum law states is just the sum of the individual variances (for non-zero variances):

$${\rm Var}[{\cal D}] = \sum\limits_{i=1}^n {\rm Var}[{\cal D}_i]$$