# Thread: number of visits in a non homogeneous Markov chain

1. ## number of visits in a non homogeneous Markov chain

Hi
I have a stochastic process $\{X_t, t \geq 1\}$ representing a non homogeneous Markov chain. The transition matrix between $X_1$ and $X_2$ is given by
$\begin{pmatrix}A &B& C\cr D&E&F\cr 0&0&1
\end{pmatrix}$

From the time $t=2$, the transition matrix between $X_t$ and $X_{t+1}$ becomes time independent and is given by
$\begin{pmatrix}G&H&K\cr
L&M&N\cr 0&0&1\end{pmatrix}$

For homgeneous Markov chains, it is well know that
$n_{ij}=(I-Q)^{-1},$
where
$n_{ij}$ is the number of times the process visits the transient state $j$ before it eventually enters a recurrent state, having initially started from the transient state $i$.
$I$ is the identity matrix.
$Q$ is the transition matrix between transient states.
How can I obtain the number of visits for my case?
Can I transform the non homogeneous model to a homogeneous one?

2. i assume you mean the expected number of visits.

If the process is in any given state at time 2, you can use your formula to get a conditional expectation. So just find the probability that the process is in each state at time 2, multiply by the conditional expectation, and sum up over all possible states.

3. Hi
Exactly, I mean the expected number of visits. Below the proof of
the formula for the homogeneous case. I wish it could help. Assume
that we have 3 states, state 3 is an absorbing one (to be close to
my case). let the transition probability matrix be
$
P^{(1)}=\begin{pmatrix} p_{11}&p_{12}&p_{13}\\
p_{21}&p_{22}&p_{23}\\
0&0&1
\end{pmatrix}$

Let $N_{ij}$ be a random variable denoting the number of times
the process visits the transient state j before it eventually
enters the absorbing state 3, having initially started from
state i.
Let $r_{ij}=E(N_{ij})$, we have
Theorem: For transient states i and j,
$\|r_{ij}\|=M=(I-Q)^{-1})$.
Proof:
If in one step, it enters a recurrent state, the number of visits
to j is zero unless j=i. If $\delta_{ij}$ is kronecker function
such that $\delta_{ij}=1$ if i=j and 0 otherwise. Then
$N_{ij}=\delta_{ij}$ with probability $p_{i3}$. On the other hand,
suppose that the Markov chain moves to a transient state at the
first step (with probability $p_{ir}$, from that position onward
the number of visits to j is $N_{rj}$. Then,
$
N_{ij} = \left\{ \begin{array}{cc}
\delta_{ij} & \mbox{with probability p_{i3}}\\
\delta{ij}+N_{rj} & \mbox{with probability p_{ir}, r a transient state.}
\end{array}
\right.
$

It follows that
$E(N_{ij})=\delta_{ij}p_{i3}+\sum_{r={1,2}}
(\delta_{ij}+E(N_{rj}))*p_{ir}.$

$E(N_{ij})=\delta_{ij}+\sum_{r={1,2}}p_{ir}E(N_{rj} )$.
Let the matrix A be
$\begin{pmatrix} \sum_{r={1,2}} p_{1r}
E(N_{r_1})&\sum_{r={1,2}} p_{1r} E(N_{r_2})\\
\sum_{r={1,2}} p_{2r} E(N_{r_1})& \sum_{r={1,2}} p_{2r}
E(N_{r_2}).\end{pmatrix}$

The transition matrix between the transient states is
given by $Q=\begin{pmatrix} p_{11}&p_{12}\\
p_{21}&p_{22}\end{pmatrix}.$

Let
$\|r_{ij}\|=\begin{pmatrix}E(N_{11})&E(N_{12})\\E(N _{21}&
E(N_{22})\end{pmatrix}$
, then $\|r_{ij}\|=I+A$ where $I$ is the
identity matrix. By using the fact that $A=Q*\|r_{ij}\|$ it follow
that $\|r_{ij}\|=(I-Q)^{-1}$.
I wish yhis could help.
If the Markov chain that I described in the beginning was a homogeneous one, then the expected value of the random variable that I needed for my problem would be
$\mathbf{i}(I-Q)^{-1}\mathbf{1}$
with $\mathbf{i}$ is the vector of the probabilies of being in states 1 and 2 at time t=1 and $\mathbf{1}$ is a unit vector.

4. The proceedure i already gave you should work then.

If the process is in any given state at time 2, you can use your formula to get a conditional expectation. So just find the probability that the process is in each state at time 2, multiply by the conditional expectation, and sum up over all possible states.

5. Hi I don't think I can apply the solution you give. Because, if at the first step, the process moves to state 3 the number of visits to j will be 0. However, if it moves to a transient state the formula may be applied.
I'm thinking of using two random variables Z and Y, with Z: the expected number of visits before absorption between the time t=1 and t=2.
Y: The expected number of visits from time 2 to absorption.
After, I think I need the random variable B=X+Y with B=Z if at the first step it enters state 3, and B=Z+Y if at the first step it enters a transient state.
This is not yet mathematically very clear for me. I don't know if my idea is correct or not.

6. i still think my procedure will work so i'll explain it once more and then leave the thread to more experienced contributors:

Consider the homogenous markov chain defined by the transition matrix
$A=\begin{pmatrix}G&H&K\cr
L&M&N\cr 0&0&1\end{pmatrix}$

I will call this "process A".

Given that you start in state $i$, define $N_{ij3}$ as the expected number of visits to state j before reaching state 3. You already know how to calculate $N_{ij3}$ for any values of i and j.

Now, consider the heterogenous markov chain defined on t=1,2,3,.... with transsition matrix
$B = \begin{pmatrix}A &B& C\cr D&E&F\cr 0&0&1
\end{pmatrix}$
for the transition between t=1 -> t=2 ; and the matrix A thereafter.

if you start in state 1, there are only 3 possible outcomes:
1. You remain in state 1 for 1 period, then begin process A

You can easily work out the probability of each of these happening by looking at matrix B. I will define the chance of being in state j at time 2 as $P_{1j}$

The expected number of visits to state 2 for the whole combined process is then:

$P_{1,2} \times \left(N_{2,2,3}+1 \right)$
$+P_{1,1} \times N_{1,2,3}$
$+P_{1,3} \times 0$

Where $P_{ij}$ is calculated based on matrix B and $N_{i,j,k}$ is calculated from matrix A.

The only assumption i have made in this answer is that the process starts in state 1 at time 1, and you can easily modify it if that is not the case.