Suppose I have three i.i.d. random variables $X_1, X_2, X_3$, and I do "max-min normalization" on them.

$$X_i \mapsto \frac{X_i - \min\limits_i X_i}{\max\limits_i X_i - \min\limits_i X_i} $$

Let $Y$ be the location of the "middle" point after normalization. (It lies in $[0,1]$.) What is the distribution of $Y$?

How can I extend to the case of $n$ i.i.d. variables $X_1, \dotsc, X_n$, asking about the distribution on the $n-2$ middle points?

We could give $X_i$ specific distributions such as $\mathcal{N}(0,1)$ for concreteness.

  • $\begingroup$ This isn't a real answer, but we could consider trying just "max-normalization". Suppose $X$ is a positive RV with continuous CDF. Let $X_1,\dots,X_n$ be iid random samples. Let $X_i \mapsto X_i / \max_j X_j$. With probability $1/n$, $X_i$ is the maximum element, in which case its distribution is the "n"th max distribution. With probability $n-1/n$, $X_i$ is not the maximum element, in which case the denominator is the "n-1"th max distribution (up to truncating the support and rescaling depending on $X_i$) (its ugly but you can write down the new PDF in terms of the old PDF!). $\endgroup$ – alw Sep 13 '18 at 13:49
  • $\begingroup$ @alw Can you explain just a little more? I'm having trouble understanding "nth max distribution", "n-1th max distribution", and how this allows us to write down the new PDF. It may be that I'm familiar with these concepts, but just not in this terminology. $\endgroup$ – Eric Auld Sep 13 '18 at 18:24
  • 2
    $\begingroup$ First of all the joint pdf of the ordered statistics are well known and easily derived - here I assume all are continuous random variables. E.g. The joint pdf of $(X_{(1)}, X_{(i)}, X{(n)})$ is $\displaystyle \frac {n!} {0!1!(i-2)!1!(n-i-1)!1!0!} f(x_1)[F(x_i)-F(x_1)]^{i-2}f(x_i)[F(x_n)-F(x_i)]^{n-i-1}f(x_n)$, $x_1<x_i<x_n$ where $f$ and $F$ are the common pdf and CDF. With this in hand, the CDF of $\displaystyle \frac {X_{(i)}} {X_{(n)}-X_{(1)}}$ is just an integral, theoretically. Practically this integral is intractable for many distributions as many CDFs are not simple enough. $\endgroup$ – BGM Sep 14 '18 at 7:27

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