# Distribution density for k-th independent variable

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 $$1.$$ ?.

• The keyword for what you are looking for is "order statistics". – 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).