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I am trying to make a model for a specific stochastic process, however I'm not sure how to tackle this problem.


PROCESS. Discrete time process.

At each time step we have a random number of unit arrivals with known probabilities. The probability that we have $n$ arrivals is given by known $a_n$ and $\sum_{n=0}^\infty a_n = 1$. For example $a_n$ could follow the Poisson distribution.

Right before arrivals we also have possibility of departures. At each time step, each unit has a probability $p$ of departing.

However we have a maximum capacity, say $n_{\text{max}}$, and there is no queueing system, so the exceeding arrivals just disappear.


So my question is the following:

What is the easiest way to model this stochastic process?

Particularly, I'm interested in the number of units at each time step (just after departures and arrivals).

I've been looking at various processes such as Markov Chains, poisson processes, however since I'm not that well-versed in stochastic processes I'm not sure what is the easiest approach.

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  • $\begingroup$ There is no modeling left to do - you've completely specified the model. You have a departure matrix $D$ (indexed from $0$ to $n_{max}$) with row $i$ distributed according to $B(i,1-p)$ (that is $D_{ij}=P(B(i,(1-p))=j)$) and an arrival matrix $A$ s.t. $A_{ij}=a_{j-i}$ for $i\le j<n_{max}$ and $A_{in_{max}}=\sum_{k=n_{max}-i}^{n_{max}} a_k$ where I collapsed all $a_k$ for $k>n_{max}$ into $a_{n_{max}}$. Your single transition step is $P=DA$. There will be a steady state distribution. I doubt you can compactly compute $P^n$ - but a computer can do that easily. $\endgroup$
    – A.S.
    Dec 8, 2015 at 22:50
  • $\begingroup$ @A.S. Ah, thank you. I guess I misspoke, I meant it in the sense of which stochastic process describes this model the best, e.g. poisson process, etc. So I could look up the mathematics for various things that might interested one regarding this model. What is the $B$ in your comment exactly? Is it the binomial distribution or? $\endgroup$
    – Eff
    Dec 8, 2015 at 23:00
  • $\begingroup$ If there was no max capacity (but with $a_i$ collapsed as above), then in steady state $pE(N)=E(A)$, hence $E(N)=E(A)/p$. If this quantity is above $n_{max}$, you'll bunch up close to $n_{max}$ in steady state of the capped model. If the quantity is much less than $n_{max}$, the capped model will be very similar to one without the cap // This process is a Markov chain and choice of distribution of $A$ is important - which only you can make based on problem you are trying to solve. Some choices might lead to easily tractable $P$. $B$ is binomial distribution. $\endgroup$
    – A.S.
    Dec 8, 2015 at 23:08
  • $\begingroup$ @A.S. Thank you for your help, it means a lot. I will see if I can solve my problem tomorrow. I'm interested specifically in the case where $$a_n = \frac{\lambda^n}{n!}e^{-\lambda}. $$ One of the things I'm interested in is the mean number of time steps between where there is maximum capacity. For a certain choice of the parameters, say $p = 0.25$ and $\lambda = 4.5$ and $n_{\text{max}} = 10$. $\endgroup$
    – Eff
    Dec 8, 2015 at 23:40
  • $\begingroup$ Once you reach the max (in less than $5$ expected steps), you'll lose expected 2.5 units upon departure but gain average of $4.5$ upon arrival, so the time between hitting the max will be close to $1$. Do you need better estimates or different ranges of parameters? $\endgroup$
    – A.S.
    Dec 8, 2015 at 23:52

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