Random variates: Why is $Var(\bar{X}_A)+Var(\bar{X}_b) \approx (S_A^2+S_B^2)/N$?
Since I read that $S_A,S_B$ are sample variances which have $/N$ in them as well. So $/N$ would cancel?
However, how are those the variances then?
$\bar{X}_A,\bar{X}_B$ are sample means. $S_a, S_B$ are corresponding sample standard deviations.
Assuming that the random variablesin groups $A$ and $B$ are each IID with respective variances $\sigma_A^2$ and $\sigma_B^2$, you can establish that the variances of sample means are:
$$\mathbb{V}(\bar{X}_A) = \frac{\sigma_A^2}{N} \quad \quad \quad \mathbb{V}(\bar{X}_B) = \frac{\sigma_B^2}{N}.$$
Since the sample variances function as estimators for the true variances, you can use the approximation $S_A^2 \approx \sigma_A^2$ and $S_B^2 \approx \sigma_B^2$. This gives you:
$$\mathbb{V}(\bar{X}_A) + \mathbb{V}(\bar{X}_B) = \frac{\sigma_A^2}{N} + \frac{\sigma_B^2}{N} \approx \frac{S_A^2}{N} + \frac{S_B^2}{N} = \frac{S_A^2 + S_B^2}{N}.$$
Assuming the data in the two groups are IID, and independent of each other, this approximation is an unbiased estimator of the true variance-sum. Using a result in O'Neill (2014) (Result 3, p. 284) it can be shown that the approximation has variance:
$$\mathbb{V} \Bigg( \frac{S_A^2 + S_B^2}{N} \Bigg) = \frac{1}{N^3} \Bigg[ \Big( \kappa_A + \frac{N-3}{N-1} \Big) \sigma_A^4 + \Big( \kappa_B + \frac{N-3}{N-1} \Big) \sigma_B^4 \Bigg],$$
where $\kappa_A$ and $\kappa_B$ are the underlying kurtosis parameters for the two groups.
