Finding expected order statistics from a normal with known parameters [duplicate]

I'm interested in finding the expected value for the kth ordered observation of a normally distributed variable with known standard deviation, mean and n. Could someone let me know the formula for that?

• Note that "assuming the order holds" is the same as saying "assuming the rank correlation is 1". But with that assumption, the first set of ranks becomes irrelevant to the question; you can drop that and instead deal with the simpler question about expected order statistics from a normal with known parameters. Please edit (since it will make your question clearer); when you do please also remove your name -- you already have a signature next to your gravatar on the right (posts are 'signed' for you already). Mar 11 '15 at 23:07
• thanks for the tip on making the description a bit clearer Glen
– Matt
Mar 11 '15 at 23:22
• Now the first link under "Related" in the sidebar to the right -> is directly relevant. Mar 12 '15 at 0:01
• @Glen_b: I know this is OT, but are Order Statistics used in determining goodness of fit for distributions? I mean, if we suspect our sample of size K comes from, say, a Standard Normal, we compare our ordered sample against the tables of Ordered Statistics from an N(0,1) of sample size K?
– MSIS
Aug 17 '21 at 23:26
• It would be best to post this as a question. If you need this question (or the linked question) for context, link it in the question. The brief answer is "yes, there are tests that make a "comparison" (in a sense) with expected order statistics or approximations to them, like the Shapiro-Francia and the closely related Ryan-Joiner test". Arguably the Shapiro Wilk could also count. Several answers on site discuss one or more of these tests. Aug 17 '21 at 23:36

For a sample of size $n$ from an absolutely continuous distribution, the general formula for the expected value of the $k$-th order statistic is

$$E[X_{k\,:\,n}] = \frac {n!}{(k-1)!(n-k)!}\int_{-\infty}^{\infty}x[F(x)]^{k-1}\cdot [1-F(x)]^{n-k}f(x){\rm d}x$$

where $F$ is the cumulative distribution function and $f$ is the probability density function. This is equivalent to

$$E[X_{k\,:\,n}] = E[F^{-1}(U_{k\,:\,n})]$$

where $U$ is a Uniform $U(0,1)$ random variable, and the "minus one" denotes the inverse, not the reciprocal.

This second expression uses the probability integral transform so one can compute the expected value of interest through simulation:

1) Generate a sample of size $n$ from a Uniform $U(0,1)$.
2) Pick the $k$-th order statistic of the sample.
3) Compute $F^{-1}(U_{k\,:\,n})$ and store, where $F$ is the CDF of interest.
4) Repeat steps 1-3 many times.
5) Take the average of the stored values.