I'm studying a simplification of a biological neuron model with $n$ neurons. We are describing the evolution of the membrane potential of each neuron. Let $(X_t)_{t\geq 0}$ be a Markov Jump Process in $\mathbb{N}_0^n$ with the following generator $Q$:

\begin{equation} \begin{cases} (i)\ \ \ \ q(x,(x)_i^c) = b(x_i) \\ \\ (ii)\ \ \ q(x,x+e_i)=\lambda x_i\ \ \ \ i=1,...,n \\ \\ (iii)\ \ q(x,x-e_i) =\mu x_i \end{cases} \end{equation}

Where $(x)_i^c: = x+c(1,...,1)-(x_i+c)e_i\ \ \ (c\in \mathbb{N})$; $\ \ \mu > \lambda\ $; and $b:\mathbb{R}_{\geqslant 0}\to \mathbb{R}_{\geqslant 0}$ is a non-negative, non-decreasing function with $b(0)=0$.

Looking carefully at $(x)_i^c$, you can see that it represents the discharge of neuron $i$. It goes to zero, whereas the other neurons recieve a fixed potential $c>0$.

The goal is to prove that this process has almost certain extinction. That is, we need to prove that

$$ \mathbb{P}(T_0<+\infty) =1 $$

Where $T_0:=\inf\{t>0:X_t=(0,...,0)\}$

Here are some observations and an idea of what may be a path towards a proof:

For a start, see that if we ignore condition $(i)$, then $(ii)$ and $(iii)$ represent in each coordinate an independent Birth-and-Death process in $\mathbb{N}_0$ with almost certain extinction, so if we ignored $(i)$ then we would have almost certain extinction for $(X_t)$.

I expect that the perturbation introduced by $(i)$ does not affect so much: What $(i)$ says is that when a coordinate is too "high", then it increasingly tends to "discharge". The total potential $||X_t||_1$ is increased by $c(n-1)$ and decreased by $x_i(t)$, so it only increases if the potential $x_i(t)$ was below the threshold $c(n-1)$. A discharge of a neuron with such a potential occurs with rate at most $b(c(n-1))$.

The idea I have for a proof is to define some neighborhood around $(0,...,0)$ (say $\mathcal{C}:=\{ ||X||_1 \leqslant k\}$ for some $k>0$) where I can have positive probability of extinction $p$, independently of the starting point. And then I'd like to prove that if you are out of $\mathcal{C}$, then you come back with probability 1.

Finally I would define an associated random variable with Geometric distribution $p$ , thinking as the experiment " whether or not $(X_t)$ starting inside $\mathcal{C}$ reaches extinction".

Are my ideas correct?

What about the idea of a proof?

If correct, are there some details I need to be careful with?

EDIT 1: I've already proved that

$$ \mathbb{P}_x(T_{\mathcal{C}}<+\infty) = 1 \ \ \ \forall x\notin \mathcal{C}$$

Where $\mathcal{C}=\{x\in\mathbb{N}_0^n:||x||_1\leqslant K \}$ for some $K>0$ (which is a finite set).

Now the issue would be to write properly the idea that I've exposed above.


Did you mean $\lambda < \mu$? (As you stated the question, the birth/death process ignoring discharges grows to infinity.) Also, is $n = N$, is $c$ an integer, and is $X$ constrained to be nonnegative coordinate-wise?

I'm going to assume the answer to all the above questions is "yes". In that case, your approach looks exactly right. Let $\mathcal{C}$ be the set where neuron discharge can increase $\|X\|_1$. If you're outside $\mathcal{C}$, then $\|X\|_1$ is dominated by a birth/death process, which means that in finite expected time you'll re-enter $\mathcal{C}$. Since $\mathcal{C}$ is finite, this means $X$ is positive recurrent, i.e. gets absorbed at 0.

If $c$ is rational, the argument stays essentially the same (just multiply everything through by the denominator of $c$). If $c$ is irrational, as stated it's impossible to reach the 0 state after a neuron discharge happens, since coordinates can be changed only by multiples of 1 and $c$. You still have the result that you'll return to the compact neighborhood $\mathcal{C}$ of the origin infinitely often and in finite expected time, but you'll need to do something to make sure 0 is reachable from all of $\mathcal{C}$.

  • $\begingroup$ Thank you for the answer. I've already edited my question to correct what you pointed out. And I'll write down the proof properly with this idea and your comments. $\endgroup$ – Max Sep 14 '17 at 17:11
  • $\begingroup$ I've got a question regarding your answer. You said $\mathcal{C}$ is finite, but as you defined $\mathcal{C}$ ( the set where neuron discharge can increase $|| X ||_1$ i.e., where there exists $i$ such that $x_i < c(n-1)$) it is not finite $\endgroup$ – Max Sep 14 '17 at 18:02
  • $\begingroup$ How do you think this could be fixed? $\endgroup$ – Max Sep 14 '17 at 18:08
  • $\begingroup$ Oh, good point. On the other hand, the rate at which $\|X\|_1$ goes up (by at most $cn$) through this mechanism is bounded above by $b(n)$, which will become negligible relative to to the birth/death dynamics for $\lambda < \mu$ when $\|X\|_1$ is large enough. (That is, your BDP will -- eventually -- be dominated by a BDP with a slightly larger value of $\lambda$ and slightly odd offspring.) $\endgroup$ – Elena Yudovina Sep 15 '17 at 14:15
  • $\begingroup$ Could you explain to me why you say that the rate at which $||X||_1$ goes up through this mechanism is bounded above by $b(n)$? The closer true statement I can think of is that it is bounded by $b(c(n-1))$ inside a modified $\hat{\mathcal{C}}$ the finite set where $x_i\leqslant c(n-1)\ \forall i$. Maybe you are assuming something I'm not seeing. $\endgroup$ – Max Sep 18 '17 at 20:01

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