# Standard error of sample standard deviation of proportions

I recently started reading Gelman and Hill’s, “Data Analysis Using Regression and Multilevel/Hierarchical Models” and the question is based on that:

The sample contains 6 observations on proportions: $p_{1}, p_{2}, \dots, p_{6}$

Each $p_{i}$ has mean $\pi_{i}$ and variance $\frac{\pi_{i}(1-\pi_{i})}{n_i}$, where $n_{i}$ is the number of observations used to compute proportion $p_{i}$.

The test statistic is $T_{i} =$ sample standard deviation of these proportions.

The book says that Expected value of the sample variance of the six proportions, $p_{1}, p_{2}, \dots, p_{6}$, is $(1/6)\sum_{i=1}^{6} \pi_{i}(1-\pi_{i})/n_{i}$. I understand all this.

What I want to know is the distribution of $T_{i}$ and its variance? Would appreciate if someone could let me know what it is, or guide me to a book or article that contains this information.

Thanks a ton.

The exact distributions for the proportions is $p_i \text{ ~ Bin}(n_i, \pi_i)/n_i$, and the proportions can take on values $p_i = 0, \frac{1}{n_i}, \frac{2}{n_i}, …, \frac{n_i-1}{n_i}, 1$. The resulting distribution of the sample standard deviation $T$ is a complicated discrete distribution. Letting $\boldsymbol{p} \equiv (p_1, p_2, …, p_6)$, it can be written in its most trivial form as:
$$F_T(t) \equiv \mathbb{P}(T \leqslant t) = \sum_{\boldsymbol{p \in \mathcal{P}(t)}} \prod_{i=1}^6 \text{Bin}( n_i p_i|n_i, \pi_i),$$
where $\mathcal{P}(t) \equiv \{ \boldsymbol{p}| T \leqslant t \}$ is the set of all proportion vectors that lead to a sample variance no greater than $t$. There is really no way to simplify this in the general case. Getting an exact probability from this distribution would require you to enumerate the proportion vectors that yield a sample variance in the range of interest, and then sum the binomial products over that enumerated range. It would be an onerous calculation exercise for even moderately large values of $n_1, …, n_6$.