variance estimation using order statistics I have four largest samples drawn from a distribution of N i.i.d Gaussian R.V. with standard deviation (Sigma) where sigma is unknown. N is known to be between 50-200. Mean is given to be 0.

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*How do we estimate Sigma (variance) using these 4 largest samples?

*What is the error in the estimate?

 A: Comment: This the germ of an idea, not yet a solution. [I am assuming the mean $\mu$ is also unknown.]
Suppose $n = 100$ observations from a standard normal distribution.
You might find the distance $D$ between order statistics 97 and 100.
Then $1/D$ should estimate $1.$ Several runs give various answers. In R,
set.seed(123)
diff(sort(rnorm(100))[c(97,100)])
[1] 0.4004199
diff(sort(rnorm(100))[c(97,100)])
[1] 1.243827
diff(sort(rnorm(100))[c(97,100)])
[1] 0.337785
diff(sort(rnorm(100))[c(97,100)])
[1] 0.8870224

Find the average of a million runs:
set.seed(2021)
d.4 = replicate(10^6, diff( sort(rnorm(100))[c(97,100)] ) )
mean(d.4)
[1] 0.7060959

sg.est = d.4/.706
mean(sg.est);  sd(sg.est)
[1] 1.000136
[1] 0.5530556

So, on average the distance between the 97th and 100th order statistics
divided by 0.706 estimates $\sigma$ with standard error about 0.553.
Here is a histogram of such estimates of $\sigma$ along with quantiles 0.025 and 0.975
of the estimates.

hist(d.4/.706, prob=T, col="skyblue2")
 abline(v=quantile(sg.est, c(.975,.025)), col="red")

Maybe we can find normal quantiles that give approximately 0.706, without
simulation. One possibility:
diff(qnorm(c(.97,.995)))
[1] 0.6950357

Something like this might generalize for other values of $n$ between 50 and 200.
