Does it matter if the division between two sample variances use identical samples when estimating division between two population variances?

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Say that I have two uncertain parameters, A and B. I estimate each population's mean and variance using estimators of the mean, $Q_A:=\frac{1}{N_A}\sum_{i=1}^{N_A}a_i, a_i \sim A$ and $Q_B:=\frac{1}{N_B}\sum_{i=1}^{N_B}b_i, b_i \sim B$, and variance, $\hat{s}_A:=\sum_{i=1}^{N_A}\frac{(a_i-Q_A)^2}{N_A}$ and $\hat{s}_B:=\sum_{i=1}^{N_B}\frac{(b_i-Q_B)^2}{N_B}$.

Say that I want to estimate $\frac{{\mathrm{Var}}(A)}{\mathrm{Var}(B)}$. If I'm doing Monte Carlo simulation and $N_a$>$N_b$, then I can either approximate it using

$\frac{{\mathrm{Var}}(A)}{\mathrm{Var}(B)} \approx \frac{\hat{s}_A}{\hat{s}_B}$,

or I can approximate it using the identical samples,

$\frac{{\mathrm{Var}}(A)}{\mathrm{Var}(B)} \approx \sum_{i=1}^{N_B} \frac{(a_i-Q_A)^2}{(b_i-Q_B)^2}$.

Is there any advantages or disadvantages to one of these approaches to the other? Is one expected to be more accurate?

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You can't tell, just from the information you've given us, which approach will yield a better estimate. It depends on the correlation between $A$ and $B$.

In the extreme case that $A$ and $B$ are the same quantity, the second approach will clearly be better, whereas if there's no correlation between $A$ and $B$ (so the event of $B$ deviating a certain amount from its mean for a certain sample is independent of $A$ deviating by a certain amount from its mean for the same sample), then the first approach would be better.