Asymmetric Markov process

Question

The limiting distribution $\xi(x)$ of a Markov process

$$x_0=1\text{ and }x_{i+1}=x_i+\Delta x_i,\tag1$$

where $\Delta x_i=-ax_i$ and $\Delta x_i=a$ occur with equal probability for every $i$, and $a\in(0,1)$ is a fixed parameter,

is a good approximation, for small values of $a$, of the solution of

$$p(x)=\frac{1}{2(1-a)}p\left(\frac{x}{1-a}\right)+\frac{1}{2a}\int_0^x p\left(\frac{x-x'}{a}\right)p(x')\,dx',\tag2$$

where $p(x)$ is treated as a probability distribution with the unit mean, $\int_0^\infty p(x)\,dx=1$ and $\int_0^\infty p(x)x\,dx=1$.

I can solve (2) numerically using Laplace transform and the solution matches well the limiting distribution I get by running the Markov process on a computer. In the limit $a\to0$, both $p(x)$ and $\xi(x)$ tend to a delta function at $x=1$. I am interested in the difference between $p(x)$ and $\xi(x)$ when $a$ is small, say, $a=0.1$.

Question 1: The process (1) is asymmetric, additive in the positive direction and multiplicative in the negative direction. Have such processes been studied by anyone?References to publications will be much appreciated.

Question 2: What is the limiting distribution of (1)? Is it possible to get it analytically? If not, perhaps you could say something about the asymptotics, especially for small values of $a$.

Question 3: What is a good way to describe the difference between $p(x)$ and $\xi(x)$ when $a$ is small?

Answer 1

(Partial answer) For every $a$, the drift is $\leqslant-a$ on $(3,+\infty)$ hence there exists a unique stationary distribution on $(0,+\infty)$. If $X$ follows this distribution, its Laplace transform $L$, defined by $L(s)=E[\mathrm e^{-sX}]$ for every $s\geqslant0$, is such that $$ L(s)=\tfrac12E[\mathrm e^{-sX}(\mathrm e^{-sa}+\mathrm e^{asX})]=\tfrac12(\mathrm e^{-sa}L(s)+L(s(1-a)), $$ hence $$ L(s)=\frac{L((1-a)s)}{2-\mathrm e^{-sa}}. $$ Iterating this yields $$ L(s)=\prod_{n\geqslant0}(2-\mathrm e^{-sa(1-a)^n})^{-1}. $$ Thus, one can represent $X$ as the sum of the random series $$ X=\sum_{n\geqslant0}a(1-a)^nY_n, $$ where $(Y_n)_n$ is i.i.d. and geometrically distributed with parameter $\frac12$. Since $E[Y]=1$ and $\mathrm{var}(Y)=2$, this representation yields $$ E[X]=1,\qquad\mathrm{var}(X)=\frac{2a}{2-a}. $$ This suggests that, when $a\to0$, one should look for the asymptotics of $$Z_a=\frac{X-1}{\sqrt{a}}. $$ An exact computation is, for every real number $t$, $$E[\mathrm e^{\mathrm itZ_a}]=\prod_{n\geqslant0}(2\mathrm e^{\mathrm it\sqrt{a}(1-a)^n}-\mathrm e^{\mathrm it\sqrt{a}(1-a)^n})^{-1}, $$ and nonrigorous expansions suggest that, when $a\to0$, $$ E[\mathrm e^{\mathrm itZ_a}]\to\mathrm e^{-t^2/2}, $$ which would indicate (but this should be checked more seriously) that $Z_a$ is asymptotically standard normal. One could start by rewriting $Z_a$ as $$ Z_a=\sum_{n\geqslant0}z_n^aU_n,\qquad U_n=\frac{Y_n-1}{\sqrt2},\qquad z_n^a=\sqrt{2a}(1-a)^n, $$ where $(U_n)$ is i.i.d. centered with unit variance and $z_n^a\to0$ for every $n$ and $\sum\limits_n(z_n^a)^2\to1$ when $a\to0$, and to apply a Lyapunov type central limit theorem.