the $W$-matrix are positive, any non negative initial distribution remains non negative.

The Chapman-Kolmogorov condition and time reversal symmetry. {#ChapmanKRelations}
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One may wonder if one realizes the same process but instead to look at the system at times $t$ and $t+1$, one looks at time $t$ and $t+2$ or more generally at $t$ and $t+\tau$. Let us call $\hat W^{(\tau)}$ the resulting probability transfer matrix after $\tau=\tau_1 + \tau_2$ iterations, therefore it is easy to show that this matrix should verify the Chapman-Kolmogorov or compatibility condition [@nicolis], $$\begin{aligned}
 \hat W^{(\tau)}=  \hat W^{(\tau_1)} \cdot  \hat W^{(\tau_2)} . \label{ChapmanK} \end{aligned}$$

In particular, for $\tau_1=\tau_2=1$ one should satisfy $$\hat W^{(2)}=  \hat W\cdot  \hat W=  \hat W ^2 ,$$ which is true for conditional probabilities, because of the relation $P(M_{t+2} \in I_k | M_t \in I_i)= \sum_j P(M_{t+2} \in I_k | M_{t +1}\in I_j) P(M_{t+1} \in I_j| M_t \in I_i),$ which is equivalent to the right hand side.

Other compatibility conditions are $$\begin{aligned}
 \hat W^{(3)}& = &  \hat W^{(2)}\cdot  \hat W, \nonumber \\
\hat W^{(3)} & = & \hat W \cdot  \hat W^{(2)}, \nonumber \\
\hat W^{(4)} & = &\hat W^{(2)} \cdot  \hat W^{(2)}, \nonumber \\
\hat W^{(4)} & = & \hat W 