Random pseudo-walk of Poisson variables

Question

Suppose there is a pool that can contain any non-negative number of objects. At time $t$ it contains $n_t$ objects. Time is discrete.

Before time $t+1$ two things happen, in this order:

1. Unless the pool is empty, one object is removed from it.
2. A number of objects $q \sim \operatorname{Poisson}(\lambda)$ are added to the pool, where $0 < \lambda < 1$. $q$ is drawn from a Poisson distribution with expectation $\lambda$, independently of previous draws.

So $n_{t+1} = n_t + q - 1$, unless $n_t = 0$, in which case $n_{t+1} = n_t + q = q$.

Let $n_0 = 0$. What is the distribution of $n_T$, for any large $T$? Let's call it $n_\infty$. What is $n_\infty$'s expectation, as a function of $\lambda$?

Note that if we allow $\lambda > 1$, $n_\infty$ diverges to $+\infty$. With $\lambda < 1$, $n_\infty$ would diverge to $-\infty$ had we not prevented the pool size from becoming negative.

Answer 1

The situation given above is like a queueing process in discrete time with constant service time and Poissonian arrivals.

Call $D_t$ the number of objects added between times $t+1$. Then $(N_t)_{t \ge 0}$ is a Markov chain governed by the i.i.d. sequence $(D_t)_{t \ge 1}$ via the relation $N_{t+1} = (N_t - 1)_+ + D_{t+1}$. Since the Poisson distribution gives a positive mass to each non-negative integer, this chain is irreducible and aperiodic.

By recursion, $N_t = N_0 + S_t + A_t$ with $$N_t = \sum_{s=1}^t (D_s-1)\ \text{ et }\ A_t = \sum_{s=0}^{t-1} \mathbb{1}_{[X_s=0]}.$$ The law of large numbers yields $S_t/t \to \lambda-1$ a.s.. We recover and precise what you have already noticed.

If $\lambda>1$, then $N_t \to +\infty$ a.s. and $(A_t)_{t \ge 0}$ is stationary: the chain is transient.

If $\lambda<1$, then $A_t \to +\infty$ a.s. and $\liminf A_t/t = 1-\lambda$: the chain is positively recurrent. It has a unique invariant probability measure $\pi$.

If $\lambda=1$, then $\liminf S_t = -\infty$ a.s.., so $\limsup A_t = +\infty$ a.s.., so $A_t \to +\infty$ a.s. : the chain is recurrent, actually null recurrent.

In the last two cases, the distribution of the return tile to $0$ can be computed by applying the optional stopping theorem to the martingale $(z^{S_t+t}/\varphi(z)^t)_{t \ge 0}$. Here $\varphi$ denotes the generating function of $D_1$: for all $z \in \mathbb{C}$, $\varphi(z) = \mathbb{E}[z^{D_1}]$. In the present case $\varphi(z) = e^{\lambda(z-1)}$ for all $z \in \mathbb{C}$.

Let us focus on the case where $\lambda<1$. The invariant probability measure $\pi$ can be determined via its generating function $\psi$: for all $z \in \mathbb{C}$, such that $|z| \le 1$, $$\psi(z) = \sum_{k \ge 0} \pi\{k\}z^k.$$ Indeed, if $N_0$ follows the distribution $\pi$, then $N_1 = (N_0-1)_+ + D_1$ also. Thus, setting $p_n = e^{-\lambda}\lambda^n/n!$, we get
$$\pi\{n\} = p_n \pi\{0\} + \sum_{k=1}^{n+1} p_{n+1-k} \pi\{k\}$$ Thus, for all $z \in \mathbb{C}$ such that $|z|<1$, $$\psi(z) = \Big( \frac{\psi(z)-\pi\{0\}}{z}+ \pi\{0\} \Big) \varphi(z).$$ so $$\psi(z) = \pi\{0\} \varphi(z) \frac{1-z}{\varphi(z)-z}.$$ Letting $z$ go to $1-$ yields $1=\pi\{0\}/(1-\lambda)$, so $\pi\{0\} = (1-\lambda)$. Hence $\psi$ is completely known.