I have a sequence of IID random variables $X_1, X_2, \dots, X_n$. In this particular case, each of the variables is Lévy distributed with PDF
$f(x) = (\lambda / 2 \pi x^3)^{-1/2} \exp(-\lambda/2x)$
for $x > 0$, and $f(x) = 0$ otherwise.
I'm trying to find the probability, given constants $\tau > 0$ and $b < n$, that there exists an interval of length $\tau$ which contains at least $b$ of the $n$ random variables.
My first approach was to use order statistics; for example, if $X_{(1)}, X_{(2)}, \dots, X_{(n)}$ are the order statistics, the probability that the $b$ smallest fall in an interval of length $\tau$ could be found using the joint distribution of $X_{(1)}$ and $X_{(b)}$. From the Wikipedia article,
$f_{X_{(j)},X_{(k)}}(x,y)dx\,dy=n!{[F_X(x)]^{j-1}\over(j-1)!}{[F_X(y)-F_X(x)]^{k-1-j}\over(k-1-j)!}{[1-F_X(y)]^{n-k}\over(n-k)!}f_X(x)f_X(y)\,dx\,dy$
This could then be integrated over the simplex $0 \leq x \leq y \leq x + \tau$, and the result could be repeated for each group of $b$ random variables and summed. However, I feel that this approach leads to double counting (for example, both $X_{(1)} \dots X_{(b)}$ and $X_{(n-b+1)} \dots X_{(n)}$ could fall in disjoint intervals of length $\tau$) and also seems to be difficult to obtain in closed form.
The other approach I considered was to use the joint density of all order statistics:
$f_{X_{(1)},\ldots,X_{(n)}}(x_1,\ldots,x_n)\,dx_1\cdots dx_n=n!f_X(x_1)\cdots f_X(x_n)\,dx_1\cdots dx_n$
However, I can't determine how to express the region of integration in any meaningful way. Any thoughts or pointers would be appreciated!