Let's suppose we want to estimate $p$ from $m$ independant realisations of $X\sim Bin(n,p)$: $x_1,x_2,\dots,x_m$, with respective size $n_i$ $i$ in $\{1,\dots,m\}$. Let $p_i$ be $p_i:=x_i/n_i$. To calculate the estimate $p$, inverse variance weighting can be used, as it is the estimator with minimal variance.

My question: Is this fact (inverse variance weight -> estimator with minimal variance) directly implied by the Gauss-Markov theorem?

I'm a bit lost because, according to me, to fulfil Gauss-Markov conditions, we need a homoscedastic variable, which in the case of binomial variable, is done by a Freeman-Tuckey (arcsine transformation), while here we only do an inverse variance weighting. Does it mean that inverse variance weighting does not give the best linear unbiased estimator (BLUE)?

  • $\begingroup$ Your description is confusing. Are you trying to say that each $x_i$ is itself the sum of $n_i$ iid Binomial$(p)$ variables and that you are trying to estimate $p$ from observing the $x_i$?? $\endgroup$ – whuber Mar 20 at 12:11
  • $\begingroup$ It was confusing, I changed it. $\endgroup$ – Anthony Hauser Mar 20 at 13:17
  • $\begingroup$ Thank you. I am still confused about something, though: exactly where will the variances come from that you use for inverse variance weighting? Surely not from the observations, because whenever $x_i$ is $0$ or $n_i$ you will estimate the variance as $0$! And why do you propose this method when you have available the simple, straightforward procedure of computing the mean of all $n_1+\cdots+n_m$ observations? It is of course equal to $\hat p = \sum_i x_i / \sum_i n_i.$ $\endgroup$ – whuber Mar 20 at 16:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.