Let $U_1,...,U_n \sim Unif(0,1)$ and the order statistics $U_{(1,n)} <...< U_{(n,n)}$, fix $k\leq n$ and prove that:
$$\frac{U_{(i,n)}}{U_{(k+1,n)}} =^d U_{(i,k)}, \forall i=1,2,...,k$$
Well, I need a proof without use the aproximation by Beta distribuction, in my attemps I try to define $T_{(i,k)}=\frac{U_{(i,n)}}{U_{(k+1,n)}} $ and note that $$T_{(1,k)} < ...<T_{(k,k)}$$ but I don't know how to prove with a simple argument that $P[T_{(i,k)} < x]= P[U_{(i,k)} < x]$
Imagine you distributed $n+1$ at random round a circle of circumference $1$. The gaps between points would be identically distributed.
Now split the circle at the final point added and straighten it into a line of length 1. The positions of the $n$ remaining points would each be uniformly distributed between $0$ and $1$.
Alternatively split the circle at the final point added and at the $k+1^{\text{th}}$ point after it, straightening the arc with $k$ points into a line of length $U_{(k+1,n)}$. The positions of the $k$ remaining points would each be uniformly distributed between $0$ and $U_{(k+1,n)}$. Their ordered values $U_{(1,n)}, U_{(2,n)}, \ldots U_{(k,n)}$ would then have essentially the same distributions as if you had performed step 1 with $k$ points, i.e. of $U_{(1,k)}, U_{(2,k)}, \ldots U_{(k,k)}$, apart from a scaling factor of $U_{(k+1,n)}$.