# Is the pooled variance equal to the variance of the centered populations?

Suppose two data arrays of length $$n$$ with variance $$\sigma^2$$ and mean $$\mu$$.

Is the pooled variance $$\sigma_P^2 = \frac{(n_1-1)\sigma_1^2 + (n_2-1)\sigma_2^2}{(n_1-1) + (n_2-1)}$$ equal to the variance of the centered concatenated data $$\textrm{Var}(x_{1,1}-\mu_1, ..., x_{1,n_1}-\mu_1,\,x_{2,1}-\mu_2, ..., x_{2,n_2}-\mu_2)$$?

I've tested this in python as follows:

import math
from random import gauss
import numpy as np

var1 = 3
var2 = 10

mean1 = 3
mean2 = 50

n1 = 500
n2 = 1000

x1 = [gauss(mean1, math.sqrt(var1)) for i in range(n1)]
x2 = [gauss(mean2, math.sqrt(var2)) for i in range(n2)]

pooled = ((len(x1)-1)*np.var(x1, ddof=1) + (len(x2)-1)*np.var(x2, ddof=1)) / ((len(x1)-1) + (len(x2)-1)) # 7.007545276099887

concd = np.var(np.concatenate((x1-np.mean(x1), x2-np.mean(x2)))) # 6.998201882398422


For both approaches the variance is approximately 7.

Only if you estimate the concatenated variance with degrees of freedom equals to $$2$$:
np.var(np.concatenate((x1-np.mean(x1), x2-np.mean(x2))), ddof=2)

Because, $$(n_i-1)S_i^2=\sum_{n=1}^{n_i} (x_{i,n}-\hat\mu_i)^2$$ and when summed over $$i$$, this is going to be equal to the unnormalized variance of the concatenated array. And you need to normalize it with $$n_1+n_2-2$$.