Can I get weighted sample variance from the individual variances? Suppose I have daily aggregates: mean ($m_i$), variance ($v_i$) and number of samples ($n_i$); and I want to calculate weekly aggregates from these. 
Weekly mean, $M$, is simple: its just a weighted mean. $$M = \frac{\sum n_i * m_i} {\sum n_i} $$
How about weekly variance, $V$? Can I just calculate $V$ like this?
$$ V = \frac{\sum n_i * v_i} {\sum n_i} $$
There are formulas on wikipedia that let you get the weighted sample variance if you have access to each sample, but I don't have access to all the samples over the whole week (well, I do have access, but a lot of data if I want to do monthly aggregates, for example). If needed, I can get any other daily aggregate values though.
Thanks!
 A: The answer is not quite as simple as that. You are correct that if the $i$-th
data set consists of  $n_i$ samples with mean $m_i$, then the mean of the
aggregated data set is
$$M = \frac{\sum_i n_i m_i}{\sum_i n_i} = \frac{1}{N}\sum_i n_i m_i\tag{1}$$
where $N$ is the sum of all the $n_i$. 
Variances are a little trickier.  The variance of the $i$-th data set is
$$\begin{align}
v_i &= \frac{1}{n_i-1}\sum_{j=1}^{n_i} \left[ \left(y_j^{(i)}\right)^2 - m_i^2\right]\tag{2}\\
&= \left[\frac{1}{n_i-1} \sum_j  \left(y_j^{(i)}\right)^2\right ]
- \frac{n_im_i^2}{n_i-1}\tag{3}
\end{align}$$
where $y_j^{(i)}$ is the value of the $j$-th sample in the $i$-th data set.
So, from $(2)$,
we can get the value of the sum of the squared sample values as
$$\sum_{j=1}^{n_i} \left(y_j^{(i)}\right)^2 = (n_i-1)v_i + n_im_i^2.\tag{4}$$
Thus, the sum of the squares of all the $N$
samples in the aggregate data set is
$$\sum_i \sum_{j=1}^{n_i} \left(y_j^{(i)}\right)^2
= \sum_i (n_i-1)v_i + n_im_i^2. \tag{5}$$ 
Now, the aggregate data set has $N$ samples 
with mean $M$ as given in $(1)$. The standard formula for
the variance of a data set is a generic version of $(3)$ 
and so we can write the variance of the aggregate
data set as
$$\begin{align}
V &= \frac{1}{N-1}\sum \text{squares of all the sample values} 
- \frac{N}{N-1}\text{square of mean}\\
&= \left [ \frac{1}{N-1}  \sum_i (n_i-1)v_i + n_im_i^2\ \right ]
- \frac{NM^2}{N-1}
\end{align}$$
where we used the result in $(5)$ for the sum of the squares
of all the $N$ sample values.
