Random affine recursion ( $X_{n} = A_{n} X_{n - 1} + B_{n}$ )

A random affine recursion updates a real random variable by

X_{n} = A_{n} X_{n - 1} + B_{n}

. Assume the coefficient pairs

(A_{n}, B_{n})

are independent and identically distributed random variables, independent of the initial value. The two coordinates within a pair may be dependent. Fresh coefficients make the recursion a time-homogeneous Markov process: its next conditional distribution depends on the current value alone.

The competition between the random multiplier and the fresh input determines whether initial conditions are forgotten. A synchronous coupling makes this particularly visible by using the same coefficient sequence for two initial values.

Table of contents
- Exponential invariant law of a random affine recursion Random affine recursion
- Synchronous contraction of a random affine recursion Random affine recursion

Exponential invariant law of a random affine recursion ( $X_{n} = U_{n} (X_{n - 1} + E_{n}), X_{\infty} \sim Exp (λ)$ )

Let

U_{n}

have the uniform distribution on

[0, 1]

and let

E_{n}

have an exponential distribution of rate

λ > 0

, with all these inputs independent. The random affine recursion

X_{n} = U_{n} (X_{n - 1} + E_{n})

has an invariant exponential law of the same rate. Indeed, for independent exponential

X, E

, beta-gamma independence says that

V = X / (X + E)

is uniform and independent of

W = X + E

. Thus an independent uniform

U

gives

U W

the same law as

VW = X

.

For any deterministic real initial value

c

, couple the recursion to a stationary copy. Synchronous contraction of a random affine recursion gives an expected difference

2^{- n} E ∣ c - X_{0} ∣ \leq 2^{- n} (∣ c ∣ + λ^{- 1})

. Hence the one-time laws converge to

Exp (λ)

, independently of

c

. An invariant initialization keeps every one-time law unchanged while the chain continues to fluctuate.

Figure 1.
Shared random inputs contract the affine recursion toward its invariant exponential law
. Original calculations and simulations. The accompanying proof establishes the limiting result; finitely many simulated paths illustrate it. Details.

Synchronous contraction of a random affine recursion ( $E ∣ X_{n} - Y_{n} ∣ = ρ^{n} E ∣ X_{0} - Y_{0} ∣$ )

Drive two copies of a random affine recursion with the same coefficient pairs, independent of their initial coupling. Then

X_{n} - Y_{n} = A_{n} (X_{n - 1} - Y_{n - 1})

. If

ρ = E ∣ A_{1} ∣ < 1

and

E ∣ X_{0} - Y_{0} ∣ < \infty

, independence of random variables gives

E ∣ X_{n} - Y_{n} ∣ = ρ^{n} E ∣ X_{0} - Y_{0} ∣ ⟶ 0.

This proves convergence in L1 and hence convergence in probability of the difference. If one copy starts in a stationary distribution, Slutsky's theorem shows that the other copy converges to that distribution. The argument compares paths before taking their one-time probability distributions; it does not require the paths ever to become exactly equal.

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