Consider a sequence of i.i.d. random Lipschitz functions {Ξ¨nβ}nβ₯0β. Using this sequence we can define a Markov chain via the recursive formula
Rn+1β=Ξ¨n+1β(Rnβ). It is a well known fact that under some mild
moment assumptions this Markov chain has a unique stationary distribution. We
are interested in the tail behaviour of this distribution in the case when
Ξ¨0β(t)βA0βt+B0β. We will show that under subexponential
assumptions on the random variable log+(A0ββ¨B0β) the tail asymptotic in
question can be described using the integrated tail function of log+(A0ββ¨B0β). In particular we will obtain new results for the random difference
equation Rn+1β=An+1βRnβ+Bn+1β.