expected value of order stats for uniform distribution

Question.

Let $\displaystyle Y_1, ..., Y_n$ be a random sample from uniform distribution $\displaystyle U[0,\theta]$, where $\displaystyle \theta>0$ is an unknown parameter. Let $\displaystyle Y_{(k)}$ be the k-th smallest value among $\displaystyle Y_1, ..., Y_n$ for $\displaystyle 1\leqk\leqn$.

(a) find density function of $\displaystyle Y_{(k)}$

(b) Find $\displaystyle E[Y_{(k)}]$.

You may use the identity $\displaystyle \int_0^1u^i(1-u)^jdu=\frac{i!j!}{(i+j+1)!}$ for integers i,j>0.

Answer.

(a) can be found in a textbook so I'll skip it here.

(b)

$\displaystyle E[Y_{(k)}]=\int_0^{\theta}yf_{Y_{(k)}}(y)dy=\int_0^{\theta}y \frac{n!}{(k-1)!(n-k)!}\frac{1}{\theta}(\frac{y}{\theta})^{k-1}(1-\frac{y}{\theta})^{n-k}dy$

$\displaystyle =\int_0^{\theta}y\frac{n!}{(k-1)!(n-k)!}\frac{1}{\theta}(\frac{y}{\theta})^k(\frac{y}{ \theta})^{-1}(1-\frac{y}{\theta})^{n-k}dy=\int_0^{\theta}\frac{y{\theta}}{{\theta}y}\fr ac{n!}{(k-1)!(n-k)!}(\frac{y}{\theta})^k(1-\frac{y}{\theta})^{n-k}dy$

I will now integrate not y but $\displaystyle \frac{y}{\theta}$ which will (1) change the upper bound of integration to 1 and (2) $\displaystyle d(\frac{y}{\theta})=\frac{1}{\theta}dy$ so I need to add one more theta into the integrand. I then take all non-y items outside the integral, and apply the 'identity' given above to what's left under the integral sign

$\displaystyle \theta\frac{n!}{(k-1)!(n-k)!}\int_0^1(\frac{y}{\theta})^k(1-\frac{y}{\theta})^{n-k}d(\frac{y}{\theta})=\theta\frac{n!}{(k-1)!(n-k)!}\frac{k!(n-k)!}{(k+n-k+1)!}=\theta\frac{n!(k-1)!k}{(k-1)!n!(n+1)}=\theta\frac{k}{n+1}$

Now, does it make sense as an expected value of k-th order statistic? It is between 0 and theta (since k/(n+1) is between 0 and 1). But I cannot say anything else, I cannot visualise a distribution where any value of y between 0 and theta has equal probability to happen.