Isn’t the variance 0? Since mean(x)/mean(y) is a number and not a distribution?
No, I don’t think that’s correct. I take it that with “mean(x)” and “mean(y)” you mean the sample averages of x and y. In this case, these means will have variances equal to Var(x)/N and Var(y)/N. Consequently, the ratio of mean(x) and mean(y) will also have a variance. See here and here.
But now I’m still confused: why is this “not quite what we’re after” and why can’t we use it to express the uncertainty around the cost/effectiveness ratio?
But doesn’t this tend to 0 if we consider enough samples? (N very large)
Hey!
No, I don’t think that’s correct. I take it that with “mean(x)” and “mean(y)” you mean the sample averages of x and y. In this case, these means will have variances equal to Var(x)/N and Var(y)/N. Consequently, the ratio of mean(x) and mean(y) will also have a variance. See here and here.
Thanks for commenting, this is interesting!
(For other people that had never heard of this and are curious about the derivation)
So the variance would be Var(¯X¯Y)=Var(∑X∑Y) which can be estimated running a simulation or with these fancy forumulas https://en.wikipedia.org/wiki/Ratio_estimator#Variance_estimates
But now I’m still confused: why is this “not quite what we’re after” and why can’t we use it to express the uncertainty around the cost/effectiveness ratio?But doesn’t this tend to 0 if we consider enough samples? (N very large)