基于齐次泊松过程的带状态转移的A/B类事件计数子过程建模及强度求解问询

Hey there, let's break this down step by step—it's a clever mix of homogeneous Poisson processes and Markovian state transitions, so we'll cover both the nature of $N_A$/$N_B$ and their intensity calculations clearly.

From your problem statement:

• There are two event types, A and B, with transition probabilities $p(A→B)$ (chance next event is B given last was A) and $p(B→A)$ (chance next event is A given last was B).
• The total event process $N(t)$ is a homogeneous Poisson process—meaning event arrival times are independent, exponentially distributed random variables.

1. What Are $N_A$ and $N_B$? #

First, let's formalize the setup to avoid ambiguity:

• Let $\lambda$ denote the intensity of the total Poisson process $N(t)$ (this is the average rate of all events arriving per unit time).
• The event labels follow a binary Markov chain: the label of each new event depends only on the label of the immediately preceding event. We can write the full transition probabilities explicitly:
  • $p(A→A) = 1 - p(A→B)$ (probability next event is A given last was A)
  • $p(B→B) = 1 - p(B→A)$ (probability next event is B given last was B)
  • The transition matrix for the label chain is:

    P = [[p(A→A), p(A→B)],
         [p(B→A), p(B→B)]]
    

With that out of the way, here's how to characterize $N_A(t)$ and $N_B(t)$:

• $N_A(t)$ and $N_B(t)$ are dependent counting processes that track the number of A-labeled and B-labeled events up to time $t$, respectively.
• They're best described as Markov-modulated Poisson processes (MMPPs)—specifically, "thinned" versions of the total Poisson process where the thinning rule (whether we count an event as A or B) updates according to the Markov chain only at each Poisson arrival time.
• Critical note: Unlike independent Poisson processes, $N_A$ and $N_B$ are not independent—their behavior is linked through the shared Markov state (the label of the last event). The only exception is if the transition probabilities are memoryless (i.e., $p(A→B) = p(B→A) = q$, meaning each event's label is an independent Bernoulli trial), in which case $N_A$ and $N_B$ become independent Poisson processes.

2. Intensities of $N_A$ and $N_B$ #

Intensity here has two key interpretations: instantaneous (rate at a specific time) and steady-state (long-term average rate).

2.1 Instantaneous Intensity #

The instantaneous intensity (rate of events of a given type arriving at a specific moment) depends directly on the current state (i.e., the label of the most recent event):

• If the last event was A:
  • $N_A$ has instantaneous intensity $\lambda \cdot p(A→A)$ (each Poisson arrival has a $p(A→A)$ chance to be labeled A)
  • $N_B$ has instantaneous intensity $\lambda \cdot p(A→B)$
• If the last event was B:
  • $N_A$ has instantaneous intensity $\lambda \cdot p(B→A)$
  • $N_B$ has instantaneous intensity $\lambda \cdot p(B→B)$

2.2 Steady-State (Long-Term Average) Intensity #

When the process runs for a sufficiently long time, the Markov chain of event labels will converge to a steady-state distribution $\pi = [\pi_A, \pi_B]$, where:

• $\pi_A$ = long-term probability that the last event was A (or equivalently, the long-term fraction of events that are A)
• $\pi_B = 1 - \pi_A$

To solve for $\pi$, we use the Markov chain steady-state condition: $\pi = \pi P$. Plugging in the transition probabilities and solving gives:
$$\pi_A = \frac{p(B→A)}{p(A→B) + p(B→A)}$$
$$\pi_B = \frac{p(A→B)}{p(A→B) + p(B→A)}$$

The steady-state intensity for each process is the total Poisson intensity $\lambda$ multiplied by the steady-state probability of observing that event type:

• Steady-state intensity of $N_A$: $\lambda \cdot \pi_A = \lambda \cdot \frac{p(B→A)}{p(A→B) + p(B→A)}$
• Steady-state intensity of $N_B$: $\lambda \cdot \pi_B = \lambda \cdot \frac{p(A→B)}{p(A→B) + p(B→A)}$

Quick Example #

Suppose $p(A→B) = 0.2$ and $p(B→A) = 0.3$, with total Poisson intensity $\lambda = 10$ events per hour:

• $\pi_A = 0.3/(0.2+0.3) = 0.6$
• $\pi_B = 0.2/(0.2+0.3) = 0.4$
• $N_A$ has a steady-state intensity of $10*0.6 = 6$ events/hour
• $N_B$ has a steady-state intensity of $10*0.4 = 4$ events/hour

备注：内容来源于stack exchange，提问作者Vladimir Krouglov