normal distributionquantilesskewness
If it is of any one's interest, the data is taken from the duration of men "last in bed" as studied here.
I want to calculate the percentiles (say, 95th) of a non-normal distribution without raw data. What are the minimum constraints needed? I have the following and I think some are not necessary.
One possible pragmatic response is inspired by the motivation behind the logit transformation used in logistic regression. In that situation, the original response is constrained between 0 and 1 and hence treating it as though it has a normal distribution causes all sorts of problems. Part of the response is to transform by log(y/(1-y)), where y is the modelled response.
Your problem is comparable, just that the limits are 1 and 5 rather than 0 and 1. If you want to analyse your data using methods motivated by Normal-distribution assumptions (into which I would categorise use of squared errors), you could consider transforming it on to a scale with no upper or lower bounds as below (in R, but the code should make sense if you're not familiar with it; the key line is the one that creates the "z" variable from the original "y"):
# generate some skewed data in the [1,5] space:
y <- rnorm(1000,4,1)
y[y<1] <- runif(sum(y<1),1,3)
y[y>5] <- runif(sum(y>5),3.5,5)
# transform it similar to logit transform
z <- log((y-1)/4 / (1-(y-1)/4))
# plot the results
par(mfrow=c(1,2))
hist(y); hist(z)
The new z variable is much more suited for OLS or whatever other similar techniques you might want to try.
If the uncontaminated data in your sample is drawn from an asymmetric distribution and the measure of scale you use to determine the width of the rejection region assumes that the good part of your data is symmetric, then, these rejection regions will be larger than they need to be. For illustration, if the distribution of the data is really right skewed. This would lead you to
Overall, the combined effect would be that your (inappropriately) cleaned dataset will look more symmetric than it really is.
The alternative here is to use an outlier detection rule that treats the left and right tails of your sample separately. Of course, compared to the mad and median, this will also halve the breakdown point of your procedure (this is inevitable because the contamination rate of an half sample can be potentially twice as high as the contamination rate the full sample).
In my opinion, the best procedure for this problem is to use the rejection regions from the adjusted boxplots. In my experience (drawn from numerical simulation), they can be expected to reliably detect asymmetric contaminations even when the data contains as much as 10-15% outliers concentrated in one tail. Adjusted boxplots are widely implemented and their connection with the classical boxplots makes them easy to understand and use. This answer explains and illustrates the use of adjusted boxplots in a context quiet like yours.
Best Answer
This is not possible. The percentile depends on the exact shape of the distribution. So unless you make a parametric assumption about the distribution or an assumption directly about the 95th percentile you cannot estimate the percentile well without the original data.