Finding the pdf of an estimator We have a set of unidimensional data, $X_1, \ldots , X_n$ drawn from the positive reals. We define a model for its distribution:
The data are drawn from a uniform distribution on the interval $[0, b]$. This model
has a single positive real parameter, $0 < b$.
now consider the estimator:
$$\hat \mu = \frac{x^{(n)}(n+1)}{2n}$$
where $x^{(n)}$ is the $n^{th}$ order statistic in the distribution of $X_i$. How do I go about finding the pdf of this estimator ?
 A: The probability that the $N$th order statistic is less than $x$ is the joint probability of the union of mutually exclusive events where at least $k \geq N$ of the samples are less than $x$ but the remaining $n-k$ samples are greater than $x$ -- where there are ${n}\choose{k}$ different combinations satisfying this condition. The probaiblity of this union is a sum of probabilities for $k=N$ to $k=n$.
Assuming an iid sample of size $n$, the distribution function for $\hat{\mu}$ is
$$\mathbb{P} (\hat{\mu}\leq x) = \mathbb{P} (X_{(N)}\leq \frac{2nx}{n+1}) = \sum_{k=N}^{n} {{n}\choose{k}}\Big[F\Big(\frac{2nx}{n+1}\Big)\Big]^k\Big[1-F\Big(\frac{2nx}{n+1}\Big)\Big]^{n-k}$$ where F is the uniform distribution function 
$$F(x) = \frac{x}{b} \ (0 \leq x \leq b)$$
Take the derivative with respect to x to find the density function.
$$f_{\hat{\mu}}(x) = n {{n-1}\choose{N-1}}\Big[\frac{2nx}{(n+1)b}\Big]^{N-1}\Big[1-\frac{2nx}{(n+1)b}\Big]^{n-N}\frac{2n}{(n+1)b}$$
You can prove that an order statistic distribution can be expressed as
$$\mathbb{P} (X_{(N)}\leq x)= F_{(N)}(x) = n {{n-1}\choose{N-1}}\int_{0}^{F(x)}t^{N-1}(1-t)^{n-N}dt$$
which makes the derivation of the density more apparent:
$$f_{(N)}(x)= F'_{(N)}(x)= n {{n-1}\choose{N-1}}\Big[F(x)\Big]^{N-1}\Big[1-F(x)\Big]^{n-N}f(x)$$
