\(T_i \)= \(A_i\) + \(Z_i \)*\( B_i\),\(i\)=1,2...,6.

Suppose for all \(i\)(i=1,2,...,6) data generation of \(A_i\) and \(B_i\) are exactly same for two methods. The two methods differ only in generating data of \(Z_i\). For method 1, \(Z_i\) is generated from Bernoulli distribution with probability 0.5 for all \(i\), that is, Z <- rbinom(6, 1, 0.5). But for method 2, \(Z\) is a vector of size 6 and generated from a probability vector whose length is also 6, that is, Z <- rbinom(6, 1, c(.5,.3,.2,.2,.3,.5)).

Consequently, probability distributions of \(T\) for two methods are different.

Now if the population probability by the method 1 is 0.425 and the population probability by the method 2 is 0.358 and I estimate the probability for 1000 Monte Carlo data sets for both methods and calculate the relative efficiency (relative efficiency=sample variance of method 1 divided by sample variance of method 2), will it be logical?

I have doubt that I can't calculate the relative efficiency, since the two method yielding two population probabilities. But relative efficiency is unitless. Doesn't this feature allow to calculate the relative efficiency?