How to calculate the variance of a partition of variables

Question

I'm running an experiment where I'm gathering (independent) samples in parallel, I compute the variance of each group of samples and now I want to combine then all to find the total variance of all the samples.

I'm having a hard time finding a derivation for this as I'm not sure of terminology. I think of it as a partition of one RV.

So I want to find $Var(X)$ from $Var(X_1)$, $Var(X_2)$, ..., and $Var(X_n)$, where $X$ = $[X_1, X_2, \dots, X_n]$.

EDIT: The partitions are not the same size/cardinality, but the sum of the partition sizes equal the number of samples in the overall sample set.

EDIT 2: There is a formula for a parallel computation here, but it only covers the case of a partition into two sets, not $n$ sets.

Answer 1

The formula is fairly straightforward if all the sub-sample have the same sample size. If you had $g$ sub-samples of size $k$ (for a total of $gk$ samples), then the variance of the combined sample depends on the mean $E_j$ and variance $V_j$ of each sub-sample: $$ Var(X_1,\ldots,X_{gk}) = \frac{k-1}{gk-1}(\sum_{j=1}^g V_j + \frac{k(g-1)}{k-1} Var(E_j)),$$ where by $Var(E_j)$ means the variance of the sample means.

A demonstration in R:

> x <- rnorm(100)
> g <- gl(10,10)
> mns <- tapply(x, g, mean)
> vs <- tapply(x, g, var)
> 9/99*(sum(vs) + 10*var(mns))
[1] 1.033749
> var(x)
[1] 1.033749

If the sample sizes are not equal, the formula is not so nice.

EDIT: formula for unequal sample sizes

If there are $g$ sub-samples, each with $k_j, j=1,\ldots,g$ elements for a total of $n=\sum{k_j}$ values, then $$ Var(X_1,\ldots,X_{n}) = \frac{1}{n-1}\left(\sum_{j=1}^g (k_j-1) V_j + \sum_{j=1}^g k_j (\bar{X}_j - \bar{X})^2\right), $$ where $\bar{X} = (\sum_{j=1}^gk_j\bar{X}_j)/n$ is the weighted average of all the means (and equals to the mean of all values).

Again, a demonstration:

> k <- rpois(10, lambda=10)
> n <- sum(k)
> g <- factor(rep(1:10, k))
> x <- rnorm(n)
> mns <- tapply(x, g, mean)
> vs <- tapply(x, g, var)
> 1/(n-1)*(sum((k-1)*vs) + sum(k*(mns-weighted.mean(mns,k))^2))
[1] 1.108966
> var(x)
[1] 1.108966

By the way, these formulas are easy to derive by writing the desired variance as the scaled sum of $(X_{ji}-\bar{X})^2$, then introducing $\bar{X}_j$: $[(X_{ji}-\bar{X}_j)-(\bar{X}_j-\bar{X})]^2$, using the square of difference formula, and simplifying.

Answer 2

This is simply an add-on to the answer of aniko with a rough sketch of the derivation and some python code, so all credits go to aniko.

derivation

Let $X_j \in X = \{X_1, X_2, \ldots, X_g\}$ be one of $g$ parts of the data where the number of elements in each part is $k_j = |X_j|$. We define the mean and the variance of each part to be $$\begin{align*} E_j & = \mathrm{E}\left[X_j\right] = \frac{1}{k_j} \sum_{i=1}^{k_j} X_{ji}\\ V_j & = \mathrm{Var}\left[X_j\right] = \frac{1}{k_j-1} \sum_{i=1}^{k_j} (X_{ji} - E_j)^2 \end{align*}$$ respectively. If we set $n = \sum_{j=1}^g k_j$, the variance of the total dataset is given by: $$\begin{align*} \mathrm{Var}\left[X\right] & = \frac{1}{n-1} \sum_{j=1}^{g} \sum_{i=1}^{k_j} (X_{ji} - \mathrm{E}\left[X\right])^2 \\ & = \frac{1}{n-1} \sum_{j=1}^{g} \sum_{i=1}^{k_j} \big((X_{ji} - E_j) - (\mathrm{E}\left[X\right] - E_j)\big)^2 \\ & = \frac{1}{n-1} \sum_{j=1}^{g} \sum_{i=1}^{k_j} (X_{ji} - E_j)^2 - 2(X_{ji} - E_j)(\mathrm{E}\left[X\right] - E_j) + (\mathrm{E}\left[X\right] - E_j)^2 \\ & = \frac{1}{n-1} \sum_{j=1}^{g} (k_j - 1) V_j + k_j (\mathrm{E}\left[X\right] - E_j)^2. \end{align*}$$ If we have the same size $k$ for each part, i.e. $\forall j: k_j = k$, above formula simplifies to $$\begin{align*} \mathrm{Var}\left[X\right] & = \frac{1}{n-1} \sum_{j=1}^g (k-1) V_j + k(g-1) \mathrm{Var}\left[E_j\right] \\ & = \frac{k-1}{n-1} \sum_{j=1}^g V_j + \frac{k(g-1)}{k-1} \mathrm{Var}\left[E_j\right] \end{align*}$$

python code

The following python function works for arrays that have been splitted along the first dimension and implements the "more complex" formula for differently sized parts.

import numpy as np

def combine(averages, variances, counts, size=None):
    """
    Combine averages and variances to one single average and variance.

    # Arguments
        averages: List of averages for each part.
        variances: List of variances for each part.
        counts: List of number of elements in each part.
        size: Total number of elements in all of the parts.
    # Returns
        average: Average over all parts.
        variance: Variance over all parts.
    """
    average = np.average(averages, weights=counts)

    # necessary for correct variance in case of multidimensional arrays
    if size is not None:
        counts = counts * size // np.sum(counts, dtype='int')

    squares = (counts - 1) * variances + counts * (averages - average)**2
    return average, np.sum(squares) / (size - 1)

It can be used as follows:

# sizes k_j and n
ks = np.random.poisson(10, 10)
n = np.sum(ks)

# create data
x = np.random.randn(n, 20)
parts = np.split(x, np.cumsum(ks[:-1]))

# compute statistics on parts
ms = [np.mean(p) for p in parts]
vs = [np.var(p, ddof=1) for p in parts]

# combine and compare
combined = combine(ms, vs, ks, x.size)
numpied = np.mean(x), np.var(x, ddof=1)
distance = np.abs(np.array(combined) - np.array(numpied))
print('combined --- mean:{: .9f} - var:{: .9f}'.format(*combined))
print('numpied  --- mean:{: .9f} - var:{: .9f}'.format(*numpied))
print('distance --- mean:{: .5e} - var:{: .5e}'.format(*distance))