Apologies for the confusing title, but I couldn't resist.
Much can and has been said about computing the unbiased variance using a sample of points, weighting by the variances of each point (for example, see this question). But I want to go one step further.
Being a function of random variables, the unbiased weighted sample variance is itself a random variable, and it is natural to study its distribution. According to Wikipedia, for the case where all of the variances are equal, $s^2$ follows a scaled chi-squared distribution:
$$s^2 \sim \frac{\sigma^2}{n-1}\chi^2_{n-1}$$
where $n$ is the number of data points, and $\sigma^2$ is the true variance of the sample. Given this formula, we can compute $\text{Var} (s^2) = \frac{2\sigma^4}{n-1}$l, which tells us the uncertainty of this variance estimate.
Does anyone know how this formula generalises when we consider the weighted variance estimate? That is, if we assume that we have $x_i$ data points, and that each one is normally distributed with variance $\sigma_i^2$, what is the distribution of the weighted variance estimate:
$$\bar V = \frac{1}{1-\sum_k \lambda_k^2} \sum_i \left(\lambda_i\left(x_i -\sum_j \lambda_j x_j\right)^2 \right) $$
where $\lambda_i = \frac{w_i}{\sum_j w_j}$ is the normalised weight of each data point, and each point is weighted based on its precision, or inverse variance: $w_i = 1/\sigma_i^2$.
In particular, I would like to know the variance of the distribution, $\bar V$. That is, what is the variance in this variance-weighted variance estimate?
