Past exam of the probability course of the University of Oxford / 2017 / 1 / b / iii

Table of contents
- Solution iii

Solution

Every

X_{n}

has the same exponential distribution, so

X_{n} d X

for

X \sim Exp (1)

. Part (ii) gives

Z_{n} - X_{n} p 0

. Applying Slutsky's theorem to

Z_{n} = X_{n} + (Z_{n} - X_{n})

yields

Z_{n} d Exp (1) .

The limiting probability distribution does not depend on

c

. This is precisely the attraction to the exponential invariant law of a random affine recursion.

Figure 1.
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.

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