Assume we have a population $N$ and a proportion $p$ of that population with a characteristic of interest. Both $N$ and $p$ are unknown. Furthermore, assume that we have $k$ random samples $(n_i, x_i)$ of $N$ from which we can estimate the proportion of interest $\hat{p}_i = x_i/n_i$ for each sample, with $i=1,\ldots,k$.

Additionally, $n_i$ and $x_i$ are large enough so that the CLT kicks in and we can say that, with $\mathbb{E}[\hat{p}_i]=p$ and $\mathbb{V}[\hat{p}_i]=\frac{p(1-p)}{n_i}$, then $\hat{p}_i\sim\mathcal{N}\left(p,\frac{p(1-p)}{n_i}\right)$.

I am interested in the behavior (especially towards the right tail) of $\bar{p}=\frac{1}{k}\sum_{i=1}^k \hat{p}_i$. So far, the approach I have thought of is the following: since each $\hat{p}_i$ is normally distributed, then their average $\bar{p}$ also follows a normal distribution since the arithmetic average is just a weighted sum where the weight is constant. Specfit $$\bar{p}=\frac{1}{k}\sum_{i=1}^k \hat{p}_i \sim\mathcal{N}\left(\frac{1}{k}\sum_{i=1}^kp,\frac{1}{k^2}\sum_{i=1}^k\frac{p(1-p)}{n_i}\right) \equiv \mathcal{N}\left(p,\frac{1}{k^2}\sum_{i=1}^k\frac{p(1-p)}{n_i}\right)$$

My questions are:

  1. Is my line of reasoning correct regarding the distributions, both of individual $\hat{p}_i$ and the resulting $\bar{p}$?
  2. Since we do not know the true value $p$, is it reasonable to substitute it with corresponding $\hat{p}_i$ in the calculations? For the last step, this would result to $$ \bar{p} \sim \mathcal{N}\left(\frac{1}{k}\sum_{i=1}^k\hat{p}_i, \frac{1}{k^2}\sum_{i=1}^k \frac{\hat{p}_i (1-\hat{p}_i)}{n_i} \right) \equiv \mathcal{N}\left(\bar{p}, \frac{1}{k^2}\sum_{i=1}^k \frac{\hat{p}_i (1-\hat{p}_i)}{n_i} \right) $$

This approach seemed reasonable to me because we "know" (under some assumptions) the distribution of each component $\hat{p}_i$, and therefore of $\bar{p}$ itself, but if someone has a different proposal on quantifying the (right) tail behavior of $\bar{p}$, please feel free to share it.

  • $\begingroup$ You can rewrite your $\bar{p}$ as a weighted average of the original bernoulli variates that you sum to get the binomials, and then get expectation and variance from there. I guess that would lead to a better normal approximation. $\endgroup$ – kjetil b halvorsen Jan 28 at 18:56

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