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This question is a twist on a question I asked here Random pseudo-walk of Poisson variables, but with randomly 'disappearing' objects. I do not know how to generalize the (satisfactory) answer given there.

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

Before time 𝑑+1 three things happen, in this order:

  1. Unless the pool is empty, one object is removed from it.
  2. Each of the remaining $n_t-1$ objects independently disappears from the pool, with probability $\delta$. So we remain with $m_t \sim Bin(n_t-1, 1 - \delta)$ objects.
  3. A number of objects π‘žβˆΌPoisson(πœ†) are added to the pool, where 0<πœ†<1. π‘ž is drawn from a Poisson distribution with expectation πœ†, independently of previous draws. So $𝑛_{𝑑+1}=m_𝑑+π‘žβˆ’1$, unless $𝑛_𝑑=0$, in which case $𝑛_{𝑑+1}=π‘ž$.

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 πœ†?

With $\lambda>1$, does $𝑛_\infty$ necessarily diverge to +∞, as it did without disappearances?

This is my motivation: Objects are bidders in a recurring auction, with values i.i.d. drawn from a distribution $F(x)$, and whose values exceed some $y > 0$. The removed object is an auction winner, who leaves. In this question, the bidders discount their values each round, so with some probability fall below $y$.

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Here's a partial answer, that I hope someone can complete.

Assuming there is a stationary state, let $X$ be a random variable that is the number of objects. Let $p_n := \Pr[X = n]$.

Let $G(z) := \sum_{n=0}^\infty p_n z^n$ be the probability generating function for $X$. After removing one object, its PGF is $$ p_0 + p_1 + \sum_{n=2}^\infty p_n z^{n-1}, $$ and after each remaining element disappears with probability $\delta$, its PGF is $$ p_0 + p_1 + \sum_{n=2}^\infty p_n [\delta + (1-\delta)z]^{n-1}. $$ The Poisson arrivals have PGF $\phi(z) = e^{\lambda(z-1)}$. Their distribution is independent of the previous distribution, so by the law for independent distributions, the resultant PGF is their PGFs multiplied $$ \Big\{p_0 + p_1 + \sum_{n=2}^\infty p_n [\delta + (1-\delta)z]^{n-1}\Big\}\phi(z). $$ Finally, since this probability is stationary, this equals the original probability, i.e., $$ G(z) = \Big\{p_0 + p_1 + \sum_{n=2}^\infty p_n [\delta + (1-\delta)z]^{n-1}\Big\}\phi(z). $$ This is an implicit equation for $G(z)$. If it can be solved, analytically or numerically, we are done. This eludes me, so far.

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