Distances to the center of points uniformly distributed in a disk We choose $n$ points at random  from the surface of disk of  radius $1$ (the points are chosen with equal probability). If we omit the point furthest from the center (from $n$ points), what is the expected value of the distance of the point furthest from the center (from the $n-1$ points). 
In his solution Stephen Siklos writes:
If we reject the outermost point, the probability that the radius of the circle is between
$z$ (the expected value)
and
$z+\mathrm dz$ is $n\cdot (n-1)\cdot z^{2n-4}\cdot (1-z^2)\cdot 2z\cdot \mathrm dz$
the factors being: the number of ways the outermost point and the second outermost point can be
chosen; the probability that $(
n
−
2)$ lie within a distance
$z$
from the centre; the probability that
the
$n$th point is further than
$z$
from the centre; the probability that the $(
n
−
1)$th point is between
distances
$z$
and
$z+\mathrm dz$
from the centre. At this stage, we should integrate this expression from $0$ to
$1$ to check that it really is a probability density function.

We could also find the probability that $(
n
−
1)$ points are within a distance
  $z$
  from the centre,
  which is
  $nz^{2n-2}(1-z^2)+z^{2n}$
  (the second term arising because $(n−1)$ points will certainly be within
  within a distance
  $z$ from the centre if all
  $n$
  are), and differentiate.

Could someone explain me more clearly why do we have to include the factor $z^{2n}$ in the second method because it's not really clear to me from Siklos's explanation?
 A: The probability of being within distance $z$ from the center is $z^2$ and the probability of being more than $z$ from the center is $1-z^2$. For $n-1$ points to be within distance $z$ from the center, it can happen in 2 ways:


*

*Some point $P_0$ is more than $z$ from the center (probability $1-z^2$) and the remaining $n-1$ points within $z$ from the center (probability $(z^2)^{n-1}$). But there are $n$ ways to pick $P_0$, so the total probability of having exactly $n-1$ points within distance $z$ from the center is 
$$
n(z^2)^{n-1}(1-z^2).
$$

*All $n$ points are within $z$ from the center. The probability of this of course is 
$$
(z^2)^n.
$$


The two events above are exclusive so it remains to add the two expressions.
A: An alternate method to compute $E[X]$ (where $X$ is the non-negative random  variable "distance of second-furthest point from the origin") is to note that 
$$E[X]=\int_0^\infty P(X>r)\,\mathrm dr =\int_0^1 P(X>r)\,\mathrm dr =\int_0^1 \bigl(1-P(X\le r)\bigr)\,\mathrm dr $$
If $p_r$ is the probability that a single random point is at distance $\le r$ from the origin, then we have from the theory of Bernoulli experiments that
$$ P(X\le r)=p_r^n+n(1-p_r)p_r^{n-1}=np_r^{n-1}-(n-1)p_r^n$$
(namely, either all $n$ or exactly $n-1$ points must be within radius $r$). 
Since $p_r$ is proportional to the area of the circle of radius $r$, we have $p_r\sim r^2$ and from $p_1=1$ we get $p_r=r^2$ for $0\le r\le 1$. Thus
$$ \begin{align}E[X]&=\int_0^1\bigl(1-nr^{2n-2}+(n-1)r^{2n}\bigr)\,\mathrm dr\\&=\left.r-\frac{n}{2n-1}r^{2n-1}+\frac {n-1}{2n+1}r^{2n+1}\right|_0^1\\&=\frac{4n(n-1)}{4n^2-1}.\end{align}$$
