order statistics question 1 Let $Y1,Y2,...,Yn$ be independent, uniformly distributed random variables on the interval $[0, θ]$, $Y(k)$ the kth-order statistic, where k is an integer between $1$ and $n$.
Find $E(Y(k)- Y(k-1))$, the mean difference between two successive order statistics. Interpret this result.
Since I got $E(Y(k))=Kθ/(n+1)$, how can I find $E(Y(k-1))$? Thanks for help!
 A: We are given that, 
$$f(y) = \frac{1}{\theta}$$. 
First, let's find the density of the $n$th order statistic, so we get
$$f_{n:n}(y) = n \theta ^{-n} y^{n-1}.$$
Now, $$E(Y_{n:n}) = \frac{\theta  n}{n+1}.$$
The quantity $E(Y_{n-1:n})$ is the expectation of the "2nd largest" order statistic.
To compute that quantity, we need the distribution of the 2nd largest order statistic, $f_{n-1,n}(y)$, which turns out to be:
$$f_{n-1:n}(y) = \frac{(n-1) n (\theta -y) \left(\frac{y}{\theta }\right)^n}{y^2}.$$
Then, by routine calculations, it's easy to show that
$$E(Y_{n-1:n}) = \frac{\theta  (n-1)}{n+1}.$$
By the linearity of expectation, we assume that $E(Y_{n:n} - Y_{n-1:n})$ = $E(Y_{n:n}) - E(Y_{n-1:n})$. Given this, we get
$$E(Y_{n:n}) - E(Y_{n-1:n}) = \frac{\theta  n}{n+1} - \frac{\theta  (n-1)}{n+1}$$
which turns out to be:
$$\frac{\theta }{n+1}.$$
The "interpretation" is that for uniform order statistics, all of the expected order statistics are equally spaced on the interval from $[0,1]$, which makes sense since the distribution is "uniformly distributed" over that interval. For other distributions, we cannot expect such an easy physical interpretation.
