Variance of event count in a two-state Markov Modulated Poisson Process (2-MMPP) during a given time period

Question

I have a two-state Markov Modulated Poisson Process (2-MMPP). It's a two-state continuous time Markov process, with $\mu_{12}$ the switching rate from state 1 to state 2 ; $\mu_{21}$ the switching rate from state 2 to state 1. In state 1, events happen as a Poisson process of rate $\lambda_1$. In state 2, events happen as a Poisson process of rate $\lambda_2$. Here's a representation of the 2-MMPP.

I'm observing the process during a duration $T$. I'm trying to find out the variance of the number of events $N_T$ observed during this duration.

Here's what I have so far:

• The steady-state vector is $\Pi = \begin{pmatrix} \pi_1 & \pi_2 \end{pmatrix} = \begin{pmatrix} \frac{\mu_{21}}{\mu_{12} + \mu_{21}} & \frac{\mu_{12}}{\mu_{12} + \mu_{21}}\end{pmatrix}$. So, on average, the system is in state 1 for a duration of $\pi_1 \times T$ during $T$.

• "the number of points in a region of finite size is a random variable with a Poisson distribution" (Wikipedia page for Poisson point processes).

  • This means that the mean number of events in each state is the event rate. Therefore, the mean number of events can be calculated as: $\mathbb{E}[N_T] = \pi_1 \lambda_1 + \pi_2 \lambda_2$
  • This also means that the variance of the numbers of events in each state is the event rate. If we're in state 1 during the entire duration of $T$, the variance is $Var(N_T | \text{{In state 1 during $T$}}) = \lambda_1$

What I can't figure out now is how do I calculate the variance of the time spent in each state while I observe the process?

Once I find that, I think I'll manage to get my final result $Var(N_T)$ using conditional variance.

Thanks in advance!

Answer 1

I found the answer in this paper: https://doi.org/10.1016/0305-0548(92)90014-V

With the following notations, as used in the paper:

• $n(t)$ the number of events during time interval $(0,t]$
• $s_i$ the random variable of the time spent in state $i$ before there is a switch
• $Z_{it}$ the random variable of the time spent in state $i$ during $(0,t]$

In this paper, the variance of the time spent in each state $\mathbb{V}(Z_{it})$ is presented in the 4th equation. To get there, they assume that $s_1$ and $s_2$ have finite means and variances. They also use properties of renewal processes, described by Cox (D. R. Cox, Renewal Theory. Methuen, London (1962) - p.101), but I could not get an access to this book to dig deeper there. They find:

$$ \mathbb{V}[Z_{it}] = \frac{\mathbb{E}[s_1]^2 \mathbb{V}[s_2] + \mathbb{E}[s_2]^2 \mathbb{V}[s_1]}{\left( \mathbb{E}[s_1] + \mathbb{E}[s_2] \right)^3} \times t + o(1) $$

The 6th equation gives the variance of the number of events happening during $T$ while in state $i$, which they note $Y_{it}$.

$$ \mathbb{V}[Y_{it}] = \mathbb{E} \left[ \mathbb{V} \left[ \frac{n(t)}{Z_{it}} \right] \right] + \mathbb{V} \left[ \mathbb{E} \left[ \frac{n(t)}{Z_{it}} \right] \right] = \mathbb{E} [Z_{it}] \lambda_i + \mathbb{V}[Z_{it}] \lambda_i^2 $$

To get to what I wanted in the end, which is the the variance of the total number of events happening during $T$, since $Y_{1t}$ and $Y_{2t}$ are independant, I had $\mathbb{V}[Y_{1t} + Y_{2t}] = \mathbb{V}[Y_{1t}] + \mathbb{V}[Y_{2t}]$ .