Short Answer
Is there annything about the i 's or pi 's or process being described
that leads to ΔS≥0 or not? Would you say that ΔS≥0 is only valid for
natural processes (= physical or spontaneous processes)?
The change in entropy, $\Delta S$ is always greater than or equal to zero, for a Markov process that have symmetric transition rates.
Which is to say, if I have a Markov process such that
$$
\text{Transition Rate} (j \rightarrow i) = \text{Transition Rate} (i \rightarrow j)
$$
Then I can say that in such processes, $\Delta S \ge 0$ with increasing time.
Statement
Now, for Markov processes, which have only one-step memory: The probability at the next instant of time depends only on its previous instant, we can show that microscopic time-reversibility does lead to non-decreasing property of Entropy.
Master Equation: For a Markov process, the time evolution of $P_i(t)$ is governed by the master equation:
$$
\frac{dP_i(t)}{dt} = \sum_{j\ne i} \left( W_{ji} P_j(t) - W_{ij} P_i(t) \right)
$$
where $W_{ij}$ is the probablity transition rate $j \rightarrow i$.
Detailed Balance Condition (Equilibrium): At equilibrium, detailed balance implies:
$$
W_{ij} P_i^\text{eq} = W_{ji} P_j^\text{eq}
$$
where $P_i^\text{eq}$ is the equilibrium probability of state $i$.
Within such equilibrium scenarios, there is a special circumstance when the system has microscopic time-reversibility - meaning the transition rates of $(j\rightarrow i)$ and $(i\rightarrow j)$ are equal. In such cases the transition matrix is symmetric,
$$
W_{ij} = W_{ji}
$$
For Markov processes with underlying microscopic time-reversibility, i.e. $W_{ij} = W_{ji}$, we can explicitly show that $$
\frac{dS}{dt} \ge 0
$$
Proof
Shannon Entropy: For a discrete system with states $i$ and probabilities $P_i(t)$, the Shannon entropy $S(t)$ is given by:
$$
S(t) = -\sum_i P_i(t) \ln P_i(t)
$$
To find the rate of change of Shannon entropy, we differentiate $S(t)$ with respect to time:
$$
\frac{dS(t)}{dt} = -\sum_i \frac{dP_i(t)}{dt} \ln P_i(t) - \sum_i P_i(t) \frac{d}{dt} \ln P_i(t)
$$
Since $\frac{d}{dt} \ln P_i(t) = \frac{1}{P_i(t)} \frac{dP_i(t)}{dt}$, we get:
$$
\frac{dS(t)}{dt} = -\sum_i \frac{dP_i(t)}{dt} \ln P_i(t) - \sum_i \frac{dP_i(t)}{dt}
$$
The second term $\sum_i \frac{dP_i(t)}{dt}$ is zero because the probabilities must sum to 1, so:
$$
\frac{dS(t)}{dt} = -\sum_i \frac{dP_i(t)}{dt} \ln P_i(t)
$$
Using the master equation, we substitute $\frac{dP_i(t)}{dt}$:
$$
\frac{dS(t)}{dt} = -\sum_i \left( \sum_{j\ne i} \left( W_{ji} P_j(t) - W_{ij} P_i(t) \right) \right) \ln P_i(t)
$$
Rearranging terms, we get:
$$
\frac{dS(t)}{dt} = \sum_{i,j \;\;(j \ne i)} \left(\underbrace{ W_{ij} P_i(t)}_{\text{Term-1}} \ln P_i(t) - \underbrace{W_{ji} P_j(t)}_{\text{Term-2}} \ln P_i(t) \right)
$$
Now since the indices $i$ and $j$ are dummy indices, we can exchange them and still get the rate of change of entropy
$$
\frac{dS(t)}{dt} = \sum_{j,i \;\;(j \ne i)} \left(\underbrace{ W_{ji} P_j(t)}_{\text{Term-2}} \ln P_j(t) - \underbrace{W_{ij} P_i(t)}_{\text{Term-1}} \ln P_j(t) \right)
$$
Adding the last two equations, and collecting Term-1 and Term-2, we get,
$$
\frac{dS(t)}{dt} = \frac{1}{2}\sum_{j,i \;\;(j \ne i)} \left[ \ln P_j(t) - \ln P_i(t) \right]\left[W_{ji} P_j(t) - W_{ij} P_i(t) \right]
$$
Under the condition of microscopic time-reversibility, $W_{ij} = W_{ji}$, we can rewrite the above equation as,
$$
\frac{dS(t)}{dt} = \frac{1}{2}\sum_{j,i \;\;(j \ne i)} \left[ \ln P_j(t) - \ln P_i(t) \right]\left[P_j(t) - P_i(t) \right] W_{ij}
$$
It is clear that $W_{ij} > 0$, as it is the transition probability rate. Secondly, if $P_j(t) - P_i(t) > 0$, then it implies that $\left[\ln P_j(t) - \ln P_i(t) \right] > 0$. Therefore, clearly the RHS will always be positive.
Therefore, each term in the sum is non-negative, and thus:
$$
\frac{dS(t)}{dt} \geq 0
$$
This demonstrates that the rate of change of Shannon entropy is always greater than or equal to zero, reflecting the tendency of a Markov process with microscopic reversibility to move towards higher entropy, in line with the second law of thermodynamics.
NOTE
While microscopic reversiblity (and detailed balance) is a sufficient condition for non-negativity of $\Delta S$ with time, it is not a neccessary condition.
