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 Incoming links: Exponential invariant law of a random affine recursion

Past exam of the probability course of the University of Oxford / 2017 / 1 / a / iii / Solution 2026-10-31

Put

W = X + Y

. The independent pair

(U, W)

has the same joint distribution as

(V, W)

from part (ii), because both first coordinates are uniform and independent of the same sum law. Hence

U (X + Y) = d VW = X .

Consequently

U (X + Y) \sim Exp (1) .

The rate is one. This multiplication identity is the mechanism behind the exponential invariant law of a random affine recursion.

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Past exam of the probability course of the University of Oxford / 2017 / 1 / b / iii / Solution 2026-10-31

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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