From discrete-time probability to probability dynamics

Question

This question originates from a conclusion stated in this paper, without any reference, and I am interested in knowing how to prove it.

Assume the following transition probability (respectively, eq (1) and (2) in the reference):

$Pr\{(x_i(t+\Delta t)= 0; x_i(t)=1)\}= \alpha \Delta t$

$Pr\{(x_i(t+\Delta t)= 1; x_i(t)=0)\}= f(s_i(t)) \Delta t$

where $x_i$ are nodes of a network with values in $\{0, 1\}$. This simply states that if a node $x_i$ has value $x_i(t) = 0$, then the probability of it transitioning to $x_i(t+ \Delta t) = 1$ is $Pr = \alpha \Delta (t)$. Note that $\alpha$ is a scalar, while $f(s_i(t))$ is a chosen transition function.

The paper now states that the probability of any node $x_i$ being active (i.e. $x=1$) at any time $t$ evolves as follows (3):

$\dot r_i(t) = -\alpha r_i(t) + f(s_i(t))[1-r_i(t)]$

This looks interesting and looks reasonable (either one of the two addenda is $0$ and the other is $1$ and $r(t)$ evolves as a decaying exponential function).

Would anybody be able to help me understand how (1) and (2) lead to (3)? Thank you!

Answer 1

Let $r_i (t)$ be the probability that the node $x_i$ is active.

Now, $x_i$ can be active at $t+\Delta t$ in 2 cases -

1. $x_i$ was active at $t$, and did not become non-active.
2. $x_i$ was not active at $t$, and became active.

For the $1^{st}$, the probability of not becoming non-active $= 1 -$Probability of becoming non-active $= 1-\alpha\Delta t$ and for the $2^{nd}$, the probability of becoming active is $f(s_i(t))\Delta t$.

Also, probability of $x_i$ not being active at $t = 1-r_i(t)$

Thus, we get - \begin{align} r_i(t+\Delta t) &= (1-\alpha\Delta t) r_i(t) + f(s_i(t))\Delta t (1-r_i(t)) \\ \implies \frac{r_i(t+\Delta t) - r_i(t)}{\Delta t} &= -\alpha r_i(t) + f(s_i(t))\Delta t (1-r_i(t))\\ \end{align}

Thus, by taking the limit $\Delta t \to 0$, we get the desired result.