I was following the question asked here where the distribution has bounds 0 and 1. I am trying to derive the same thing, but from $X_i \sim U(0,\theta)$.
This gives $$ E(X_{(k)}) =\frac{n!}{(k-1)!(n-k)!}\int_0^\theta x^{k}[1-x]^{n-k}dx $$
And if we use transformation and define $Z = \frac{X}{\theta}$ we can replace in the formula we get $$E(X_{(k)}) = \theta \frac{k}{n+1}$$
I am lost in this, I presume it does not seem right. If anyone could help me with that I would be grateful. If I am wrong, I want to understand why