TITLE: Estimating population size
QUESTION [5 upvotes]: Let's suppose there are $n$ real numbers $a_0 < ... < a_n$ uniformly selected from interval [0, 1). If one knows $k$ numbers on consecutive positions $a_i < ... < a_{i+k-1}$ how good is $(k - 1) / (a_{i+k-1} - a_i)$ an estimator for $n$? What other estimators are possible/better?
NOTE: $n >> k$.

REPLY [4 votes]: As in the book A First Course in Order Statistics (see, in particular, Section 2.5), let us denote by $W_{i,j:n}$ the spacing $W_{i,j:n} = U_{j:n} - U_{i:n}$, $1 \leq i < j \leq n$, where $U_{1:n} < \cdots < U_{n:n}$ are $n$ order statistics from a ${\rm uniform}(0,1)$ distribution. By equation (2.5.21) of that book, the density function of $W_{i,j:n}$ is given by
$$
f_{W_{i,j:n} } (w) = \frac{{n!}}{{(j - i - 1)!(n - j + i)!}}w^{j - i - 1} (1 - w)^{n - j + i} ,\;\; 0 < w < 1,
$$
so that $W_{i,j:n}$ has a ${\rm Beta}(j - i,n - j + i + 1)$ distribution (which depends only on $j-i$ and not on $i$ and $j$ individually). Having the distribution of $W_{i,j:n}$, you can check how good is $\frac{{j - i}}{{U_{j:n}  - U_{i:n} }}$ an estimator for $n$, which is what you asked (in different notation).
EDIT: Using the density function of $W_{i,j:n}$, one can easily calculate the bias (cf. Trevor's answer). My calculation shows that 
$$
{\rm E}\bigg[\frac{{j - i}}{{U_{j:n}  - U_{i:n} }}\bigg] = \frac{{j - i}}{{j - i - 1}}n,
$$
hence the estimator gets better and better as $j-i$ grows. (The expectation is infinite when $j-i=1$; cf. Dinesh's answer.)