# Expected value of kth order statistic drawn from ANY uniform distribution

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

• Where did $500$ come from? Are you assuming that $\theta = 500$? Nov 25 at 10:27
• oh, my bad. I had two screens open. probably wrote 500 to the wrong place. It is $\theta$ indeed Nov 25 at 10:30
• Cool. In that case, $\frac{\theta k}{n+1}$ is correct. Intuitively your sample is just $\theta$ times a random sample from $U(0,1)$ so the $k$-th order statistic's mean is $\theta$ times that of the $k$-th order statistic of $U(0,1)$. Nov 25 at 10:34