I'm interested in the relation between the variance of the mean of set of real numbers, $A$, and sum of the variances of the means of any arbitrary partition of $A$ into a smaller number of sets.
Let's call the variance of the mean of the original set
$$ V_A = \frac{ \frac{1}{n} \sum^n_{i=1} (A_i - \bar{A})^2 }{n} $$
and the weighted sum of the variances of the means of some arbitrary partitioning of the set
$$ V_P = \left( \frac{n}{n_x}\right)^2 \frac{ \frac{1}{n_x} \sum^{n_x}_{i=1} (A_i - \bar{A_x})^2 } {n_x} + \left( \frac{n}{n_y}\right)^2 \frac{ \frac{1}{n_y} \sum^{n_y}_{i=1} (A_i - \bar{A_y})^2 } {n_y} + ... + \left( \frac{n}{n_k}\right)^2 \frac{ \frac{1}{n_k} \sum^{n_k}_{i=1} (A_i - \bar{A_k})^2 } {n_k} $$
where $n = n_x + n_y + ... + n_k$. (In case this notation isn't clear, I provide a short demonstration below in R.)
Obviously, $V_P$ is minimized when $k$ is set to $n$ such that $V_P = 0$. Thus, $V_P$ can be arbitrarily smaller than $V_A$. By contrast, I haven't thought of many cases where $V_A < V_P$, and I'm wondering if it's possible to find a bound for the difference between $V_A$ and $V_P$ when $V_A < V_P$.
Here's a simple example.
t <- 1
A <- rep(c(0, 20), t)
B <- rep(c(11, 9), t)
C <- rep(c(11, 9), t)
n <- length(A)
# Calculate variances (not sample variances)
var_A <- var(A) * (n / (n-1))
var_B <- var(B) * (n / (n-1))
var_C <- var(C) * (n / (n-1))
var_ABC <- var(c(A, B, C)) * (3 * n / (3 * n-1))
var_ABC / (3 * n)
(1/n^3) * (var_A / n) + (1/n^3) * (var_B /n ) + (1/n^3) * (var_C /n)
In this case, for $t = 1$, $V_A < V_P$ but for $t > 1$, $V_P < V_A$.
I suppose there is not a novel question, but I haven't found an answer so would be grateful if someone can point me in the right direction. If there's a bound, how do we establish it; if not, perhaps an example to show that there is no bound.