Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 201 5 d Solution Created 2026-10-03 Updated 2026-10-05
Taking in (c) and using that has no atom at givesThis is the distribution function of the Brownian first-passage time; (b) ensures that it is a proper probability distribution. Comparing these distribution functions gives the Brownian scaling identity for .
For an integer , set and successively hit the levels . The Strong Markov property at these finite stopping times and spatial translation imply thatare independent random variables, each with the probability distribution of . Thus, for independent copies ,ThereforeThe paper's definition is the strictly stable distribution convention, with no centering term, a special case of a stable distribution.
Strictly stable distribution 2026-10-05
A probability distribution is strictly stable with index if for every positive integer , for independent copies. Here no centering term is allowed, unlike in the general definition of a stable distribution.