order statistics question 1

Question

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!

Answer 1

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.