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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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  • $\begingroup$ I'm confused by your terminology. What's a "sample estimator"? What are $N_A$ and $N_B$? And "each populations'" doesn't make sense -- did you mean "each population's"? $\endgroup$
    – joriki
    Commented Jun 9, 2018 at 9:38
  • $\begingroup$ There are two populations/random variables $A$ and $B$. I define the sample variance as the variance of the samples I observe. I have $N_A$ samples from population $A$ and $N_B$ samples from population $B$. I used the same random instantiation of both populations for $N_B$ samples, and I have ($N_A$-$N_B$ additional samples of population $A$. I want to estimate the true value of $\frac{\mathrm{Var}(A)}{\mathrm{Var(B)}}$ using the sampled data. I have the choice of using all of my samples or only the identical samples, and it is not clear to me which approach is more accurate. $\endgroup$
    – kilojoules
    Commented Jun 9, 2018 at 18:33
  • $\begingroup$ Please ask more questions if that isn't clear. $\endgroup$
    – kilojoules
    Commented Jun 9, 2018 at 18:33
  • $\begingroup$ I'm calling the $Q$ variables sample estimators. Sorry if that's weird. $\endgroup$
    – kilojoules
    Commented Jun 9, 2018 at 23:32
  • $\begingroup$ I hope I've understood the question now. It's a nice question! It seems that by "sample estimator" you simply mean what is usually called an "estimator". And the apostrophe on "population's" is still in the wrong place. $\endgroup$
    – joriki
    Commented Jun 10, 2018 at 3:10

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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.

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  • $\begingroup$ In what I'm interested in, the correlation is significant, ~0.6-.99. Could you please point me towards the maths that support the second approach being better for highly-correlated variables? It's not even clear to me that the second approach is better with the same random variable. $\endgroup$
    – kilojoules
    Commented Jun 11, 2018 at 3:33
  • $\begingroup$ @kilojoules: If $A$ and $B$ are the same, then the ratio of their variances is $1$, and you'll estimate that precisely if you take the second approach, since you write that your samples are identical, so the numerator and denominator will be equal and the ratio will be exactly $1$. If you use further samples for the numerator, it will only approximately be $1$. $\endgroup$
    – joriki
    Commented Jun 11, 2018 at 5:13
  • $\begingroup$ Can we say anything meaningful about which is the better method for an arbitrary correlation? $\endgroup$
    – kilojoules
    Commented Jun 11, 2018 at 14:44

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