estimating variance using only data at the tails without resorting to Gibbs sampling Suppose we know that the population size is $n=1,000$ but for whatever reason, we only have the bottom $n_1=100$ observations and the top $n_2 = 200$ observations. Furthermore, suppose we know the data $X_i \overset{\text{iid}}{\sim} N(0,\sigma^2)$. Basically, we are missing the middle $700$ observations but would like to estimate the variance $\sigma^2$. 
Normally, I would sample the missing observations from a truncated Normal as part of a simple Normal-Inverse Gamma model. But suppose we can't do that. It seems to me that because we know the bounds of the observations we should have an analytical solution or at least some frequentist solution. Ideas? 
 A: Here is an analytical solution:
If the data $X_1,\ldots,X_n\sim \mathcal{N}(0,\sigma^2)$, then we expect that
$$
\hat{F}_n(x)\approx F(x),
$$
where $\hat{F}_n$ is the empirical CDF and $F$ is the true normal CDF,
$$
F(x)=\Phi\left(\frac{x}{\sigma}\right),
$$
where $\Phi$ is the standard normal CDF. Let $x_{(1)},\ldots,x_{(n)}$ be the order statistics. We know the first 100 and last 200 of them. By definition of the empirical CDF,
$$
\hat{F}_n(x_{(k)})=\frac{k}{n}.
$$
Therefore,
$$
\Phi\left(\frac{x_{(k)}}{\sigma}\right)\approx\frac{k}{n}.
$$
There is a little technical problem with this approximation for $k=n$ (the argument of $\Phi$ becomes infinite). The standard remedy is to replace $\frac{k}{n}$ with $\frac{k}{n+1}$:
$$
\Phi\left(\frac{x_{(k)}}{\sigma}\right)\approx\frac{k}{n+1},
$$
which for large $n$ (as in our case) works well. The above approximation gives an estimate for $\sigma$:
$$
\sigma\approx\frac{x_{(k)}}{\Phi^{-1}\left(\frac{k}{n+1}\right)}.
$$
Since we know $m=n_1+n_2=300$ order statistics, we have $m$ estimates for $\sigma$. To cook up one estimate, we can take the average of them:
$$
\sigma\approx \frac{1}{n_1+n_2}\left(\sum_{k=1}^{n_1}\frac{x_{(k)}}{\Phi^{-1}\left(\frac{k}{n+1}\right)}+\sum_{k=n-n_2+1}^n\frac{x_{(k)}}{\Phi^{-1}\left(\frac{k}{n+1}\right)}\right).
$$
