# “Standard error” of a sample's 90th percentile for a normally distributed population

When sampling from a normally distributed population, I understand that the expected deviation between the sample mean and the population mean can be calculated using the standard error

$$\text{standard error} = \frac{\sigma_{\text{population}}}{\sqrt{n}}$$

Is there a way to calculate the expected deviation between a sample's 90th percentile and the population's true 90th percentile?

Edit: Here's my attempt to formalize this idea:

$\sigma= \frac{\sum_i^n{((\pi_{90}^*-\pi_{90})^2})}{n}$ where $\pi_{90}$ is the truly such that/$\pi_{90}^*$ is the sample value such that $P(f(X) < \pi_{90}) = 0.9$

My question is: "Can $\sigma$ be expressed in terms of $\sigma = g(f(X))$," where g is some mapping from f's formulation to a description of how $\sigma$ scales with X? I realize that there may be different answers for different types of PDFs - I'm curious if this can be solved for any specific PDF (uniform, Gaussian, or whatever else lends itself well to the mathematics).

• I think this is quite sensitive to the specifics of the distribution that you have. In particular, what you basically want is the standard deviation of the $0.9n$th order statistic (rounding $0.9n$ to the nearest integer if necessary). But this standard deviation depends in a complicated way on the CDF of the underlying distribution. Are you interested in just the case of the normal distribution? In this case the standard deviation can in principle be calculated using formulas given at www2.bc.edu/~baglivo/MT427/notebook04.pdf (see page 4), but I doubt they simplify nicely. – Ian Apr 20 '16 at 19:48
• Page 4 tells you what the PDF of the $0.9n$th order statistic is in terms of the underlying normal CDF. If this is $f_n$ then the standard deviation (which is the number you want) is $\sqrt{\int_{-\infty}^\infty x^2 f_n(x) dx - \left ( \int_{-\infty}^\infty x f_n(x) dx \right )^2}$. – Ian Apr 20 '16 at 20:12
• Actually, I don't think that's what the OP wants. He/she said "expected deviation", not "root-mean-squared deviation", and in addition the expected value of the sample percentile is not the same as the percentile of the distribution. – Robert Israel Apr 20 '16 at 20:18
• @RobertIsrael I was working by analogy to the case given for the mean, which is indeed RMS deviation. Your second remark is interesting, I wasn't aware of that. – Ian Apr 20 '16 at 23:49
• @RobertIsrael Indeed it seems like it's not even all that close for small $n$: the average of the 9th element in a sample of size 10 from the standard normal is about 1 (my numerical values are between 0.98 and 1.02) while the 90th percentile of the normal distribution is about 1.28. That's weird... – Ian Apr 21 '16 at 0:00

Using the form from the websites mentioned in the comments, we have sample size N=10n+9 for any positive integer n, and k=9n+9 for the same positive integer n. Also $f(x)$ is the normal PDF and $F(x)$ is the normal CDF.

The 90th sample percentile has PDF for a sample size $10n+9$ of:

$$\binom{10n+9}{9n+8,1,n}(F(x))^{9n+9}f(x)(1-F(x))^n=\frac{(10n+9)!}{(9n+8)!n!}(F(x))^{9n+9}f(x)(1-F(x))^n$$

If we take the binomial approximation for the factors which are polynomials in $F(x)$, we get:

$$\approx \frac{(10n+9)!}{(9n+8)!n!}(1-(9n+9)(1-F(x)))f(x)(1-nF(x))=\frac{(10n+9)!}{(9n+8)!n!}((9n+9)F(x)-9n-8)f(x)(1-nF(x))=\frac{(10n+9)!}{(9n+8)!n!}[-(9n+9)F(x)^2+(9n^2+17n+9)F(x)-(9n+8)]f(x)$$

We approximate further by throwing out all terms which are not $\Theta(n^2)$ (ultimately we will be taking the large n limit, so this approximation seems at least plausible):

$$\approx \frac{(10n+9)!}{(9n+8)!n!}(9n^2F(x))f(x)$$

Approximating even further, we now make use of Stirling's Formula:

$$\approx \frac{\sqrt{2\pi(10n+9)}(\frac{10n+9}{e})^{10n+9}}{2\pi\sqrt{(9n+8)n}(\frac{9n+8}{e})^{9n+8}(\frac{n}{e})^n}(9n^2F(x))f(x)=\frac{\sqrt{2\pi(10n+9)}(10n+9)^{10n+9}}{e\sqrt{(9n+8)n}(9n+8)^{9n+8}(n)^n}(9n^2F(x))f(x)$$

Neglecting all parts of factors which are not at least $\Omega(n)$ and attempting to cancel terms of similar orders (this basically has no analytical justification whatsoever, but none of these approximations actually simplify to anything ):

$$\approx \frac{\sqrt{2\pi}}{e}\frac{10}{9}F(x)f(x)$$

which isn't even a probability density function.

Hence, as the comments above point out, while such an expectation for the deviation from the 90th sample percentile does theoretically exist, even using incredibly aggressive approximations, it is very difficult to identify even an approximate simple closed form for it that holds for arbitrarily large sample sizes (i.e. in the large n limit, and thus which is independent of the actual sample size).

• Could this be considered an unsolved problem in maths? – kilojoules Aug 14 '16 at 17:36
• @kilojoules Perhaps? One would have to do a literature review first -- just because I wasn't able to solve it doesn't mean that someone else hasn't. ) – Chill2Macht Aug 14 '16 at 17:39

This site describes how a $$100(1-\alpha)\%$$ confidence interval for any percentile of any arbitrary continuous distribution can be estimated using a sample from the distribution. This is not exactly what you've asked for, but perhaps close enough to be useful to some reading this question.

The approach is simple and uses a binomial distribution or the normal approximation thereof (for samples larger than ~20). You can't exactly compute any arbitrary $$100(1-\alpha)\%$$ confidence interval because you need to pick sampled data points as proposed endpoints for the confidence interval, but with some trial-and-error and enough samples you should be able to get pretty close to any particular $$100(1-\alpha)\%$$ interval you're looking for.