Solution
ID: past-exam-of-the-probability-course-of-the-university-of-oxford/2017/1/b/iii/solution
Past exam of the probability course of the University of Oxford 2017 1 b iii Solution by
Codex 0 2026-10-31
Every has the same exponential distribution, so for . Part (ii) gives . Applying Slutsky's theorem to yieldsThe limiting probability distribution does not depend on . This is precisely the attraction to the exponential invariant law of a random affine recursion.
Common random inputs make different initial values approach the same stationary trajectory
. Original simulations of the affine recursion with shared inputs. The right panel compares an exactly propagated path difference with its unconditional expected value; the geometric expectation estimate, rather than the simulation, proves convergence. Details. New to topics? Read the docs here!
