The problem:

  • There are $n$ points uniformly and independently distributed on the segment $\left(0,1\right)$.
  • The points are sorted ascending.
  • Calculate distribution density of $k$-th point's coordinate.

I calculated ( and checked experimentally ) that for $z \in \left(0,1\right)$ probability for the $k$-th point to be less than $z$ is

$\displaystyle P_{k} \equiv \mathrm{P}\left\{x_k < z\right\} = \sum_{i = k}^{n}\binom{n}{i}z^{i}\left(1 - z\right)^{n - i}$

The desired density is

  1. $\displaystyle \frac{dP_k}{dz} = \sum_{i = k}^{n}\binom{n}{i}z^{i - 1}\left(1 - z\right)^{n - i - 1}\left(i - zn\right)$

and ( this is my question ) somehow the latter appears to be equal to

  1. $\displaystyle\frac{dP_k}{dz} = z^{k - 1}k\left(1 - z\right)^{n - k}\binom{n}{k}$

How do you get $2.$ from $1.$ ?.

  • 3
    $\begingroup$ The keyword for what you are looking for is "order statistics". $\endgroup$ – Jean Marie Feb 24 at 18:27

Using induction (base is $k=n$, and then decrement $k$ in induction junction), one can prove that (1) <=> (2).


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