# Distribution for average of multiple binomial proportions

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.

• 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. – kjetil b halvorsen Jan 28 at 18:56