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Consider the discrete time random process $X_n,n\in \mathbb N$, with $$X_{n+1}=(1-K)\cdot X_n+K\cdot\frac{G_n}{c}\cdot X_n$$ where $G_n$ is a random variable with expectation $\mathbb E[G_n\mid X_n]=\lambda\cdot (1-F(X_n))$ for some strictly increasing and continuous distribution function $F$ and $K,c$ are constants with $K\in (0,1)$ (think of $K=0.1$) and $c\in(0,\lambda)$.

If I substitute the expectation of $G_n$ in the formula above, I get that $$X_{n+1}=X_n\left(1+K\cdot\frac{\lambda\cdot (1-F(X_n))-c}{c}\right)$$

So, it is obvious to me that $X^*=F^{-1}(1-c/\lambda)$, is some kind of fixed point of this random process. Moreover, this is a stable fixed point: if $X_n>X^*$, then $X_{n+1}<X_n$ and if $X_n<X^*$, then $X_{n+1}>X_n$, i.e., $X^*$ is attracting. However, this argument is non-rigorous. So, my question is

  • Can I prove rigorously that $X_n$ converges to $X^*$.
  • In what sense? Is this convergence in expectation? Can I show something stronger, e.g., convergence in probability?

Footnote: for completeness, $G_n$ is determined as follows: there are Poisson arrivals with rate $\lambda$. Each arrival has a value drawn randomly from a distribution $F$. $G_n$ is the random sum of all arrivals that have a value above $X_n$.

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  • $\begingroup$ Is it correct that you only have $\mathbb{E}[G_n | X_n] = \ldots$ or do you have $\mathbb{E}[G_n | \mathcal{F}_n] = \ldots$ (same expression on the right side) with $\mathcal{F}_n = \sigma(X_0,G_0,X_1,G_1,\ldots,G_{n-1},X_n)$? $\endgroup$
    – Dieter Kadelka
    Commented Nov 4, 2020 at 11:18
  • $\begingroup$ @DieterKadelka Yes, you also know $\mathcal F_n$. But since you only use $X_n$, does this make a difference? Sorry for asking, I am just not too familiar with these notions. $\endgroup$
    – Jimmy R.
    Commented Nov 4, 2020 at 11:53
  • $\begingroup$ The difference is that with $\mathcal{F}_n$ you get a Markovian model, I think. This simplifies investigations of convergence. $\endgroup$
    – Dieter Kadelka
    Commented Nov 4, 2020 at 13:25
  • $\begingroup$ @DieterKadelka Yes, the model is Markovian. It only depends on the previous state. Plese use the notation that feels correct. Thank you. $\endgroup$
    – Jimmy R.
    Commented Nov 4, 2020 at 13:57
  • $\begingroup$ You need at least additional assumptions. You get simple counterexamples if $G_n = \lambda \cdot (1-F(X_n)) + U_n$ with $(U_n)_{n \in \mathbb{N}}$ a sequence of independent (not iid) random variables with $\mathbb{E}U_n = 0$, but $|U_n|$ "large". $\endgroup$
    – Dieter Kadelka
    Commented Nov 4, 2020 at 15:43

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