TITLE: If parametric quantile esimation estimates $p$ by computing the MLE, then how to get non-parametric $p$?
QUESTION [2 upvotes]: For non-parametric or parametric quantile estimation.
If parametric quantile esimation estimates $p$ by computing the MLE, then how to get non-parametric $p$?
Related:
https://mathoverflow.net/questions/48223/parametric-vs-non-parametric-estimation-of-quantiles
But I don't understand how to do non-parametric quantile estimation, nor how is $p$ estimated in that case. Does one still have to use the inverse CDF?

REPLY [1 votes]: Suppose you have a sample of size $n = 1000$ from an unknown normal distribution. We want to estimate the 65th percentile of the unknown distribution. I just
happen to have one in my computer now, in a vector $\mathbf{x}.$
As you say, there are two possible approaches. One is to take the 65th percentile
of the sample. The result is 105.45, which I get using R statistical software.
quantile(x, .65)
##      65% 
## 105.4531 

The second approach is to estimate the mean $\mu$ and standard deviation $\sigma$ of the normal population, by using the sample mean and sample standard
deviation. I get $\hat \mu = \bar X = 100.40$ and $\hat \sigma = S = 14.74.$
mean(x);  sd(x)
## 100.4008
## 14.74482

But the 65th percentile of $\mathsf{Norm}(100.50, 14.745)$ is 105.98.
qnorm(.65, 100.40, 14.475)
## 105.9775

The question now arises: which is closer to the right answer? The first (nonparametric) estimate 105.45
or the second (parametric) estimate 105.98.  In a real life situation we
would never know for sure, but might expect the parametric estimate based
on MLEs $\hat \mu$ and $\hat \sigma$ would be better.
But in this case, we can know for sure because I simulated my sample of 1000
from $\mathsf{Norm}(100, 15),$ which has 65th percentile 105.78. So the parametric estimate is a little closer.
 qnorm(.65, 100, 15)
 ## 105.7798

The data were simulated by the following R code. Because I set a seed, you
can replicate the experiment precisely in R.
set.seed(2017); x = round(rnorm(1000, 100, 15), 3)

Addendum per question in Comment:  Here is a brief demo of quantile
in the current context.
quantile(x, .65)
##      65% 
## 105.4531 
sx = sort(x);  sx[650];  sx[651]
## 105.423
## 105.509

Note: The superiority of the parametric estimator in the example above
is not a one-time accidental result. In a simulation of 100,000 samples of
size $n=200,$ the mean of the parametric estimators was 105.77 with
a mean squared error of 1.22; the mean of the nonparametric estimators
was 105.74 with a MSE of 1.85.
{"set_name": "stack_exchange", "score": 2, "question_id": 2507913}