All Order Statistics are statistics but not estimators 

Any order statistic is a statistic but not an estimator. Discuss along with an example. 


Can anyone help me out with this question, please?    
 A: Must be some misunderstanding here. Order statistics can be estimators--in some cases even 'best' estimators. Several examples:
(a) Maximum of uniform: In sampling from $\mathsf{Unif}(0, \theta),$ the maximum observation is MLE for $\theta,$ even if biased. 
For example if $n = 10$ and $\theta = 5,$ then $W = X_{(10)} = \max(X_1,\dots,X_{10})$ is MLE and $W^\prime=\frac{n+1}{n}W$
has $E(W^\prime) = \theta = 5.$  Simulation illustrates this:
mx.unb =replicate(10^6, 1.1*max(runif(10, 0, 5)))
mean(mx.unb)
[1] 4.999921  $ aprx E(W') = 5

(b) Sample minimum to estimate exponential mean: In sampling from an exponential distribution, the minimum observation can be used to estimate the mean, although there are better estimators. 
For example, if we take a sample of 10 observations from
$\mathsf{Exp}(\mu = 1),$ then $V = X_{(1)}= \min(X_1,\dots,X_{10}),$
then $nV$ is an unbiased estimator of $\mu.$
mn = replicate(10^6, 10*min(rexp(10,1)))
mean(mn)
[1] 1.00015    # aprx population mean = 1

(c) Median of Laplace: In many situations the median observation can be used to estimate the population median; for Laplace data the median is MLE and unbiased.
Example: If we take 9 observations from a Laplace distribution
with mean (and median) $\eta = 7,$ then the sample median $H =X_{(5)}$ is MLE and $E(H) = 7.$
med = replicate(10^6, median( rexp(9)-rexp(9)+7 ))
mean(med)
[1] 7.000467  # aprs E(H) = 7

Note on simulation: The difference of two independent exponential random variables with the same rate is Laplace with median $0.$
(d) Various percentiles: Quite generally the $p$th percentile (except possibly for the min or max) of a sample is a good
estimator of the $p$th percentile of the population from which the sample was taken. The variability of the estimate is smallest in regions where the density is largest. 
(For small samples, these estimates may be
slightly biased, depending on the convention for finding sample percentiles. About a dozen slightly different conventions are in common use; for nine of the most common, see R documentation for quantile.)
Here is an illustration of using 40th percentiles of samples of size ten from a normal population to estimate the the 40th percentile of the population.
q.40 =replicate(10^6, quantile(rnorm(10, 100, 15),.4))
mean(q.40);  qnorm(.4, 100, 15)
[1] 96.64472   # aprx mean of sample 40th percentile
[1] 96.19979   # exact 40th population percentile

