Suppose I have 2 Markov processes with transition kernels Q\_1(y|x) and Q\_2(y|x). Suppose i also have Lyapunov functions V\_1, V\_2 for these processes w.r.t. a common set, i.e. there exists a compact set S such that for x outside S, the drift of the respective Lyapunov function is bounded above by a negative number for both the processes. 


Now suppose I construct a new Markov processes by using Q\_1 kernel at odd times and Q\_2 kernel at even times.
Can I guarantee stability of this new process in the set S i.e. can I construct a Lyapunov function for this third process on set S such that the drift of the Lyapunov function is negative outside the set S?



Is there any foster-Lyapunov criterion for time varying Markov processes?
The answer is negative- the combined process obtained by alternating the kernels need not be stable. 


On the state space 
$\Lambda=\{(x,y) \in {\bf Z}^2 : x,y\ge 0\}$ (The non-negative quadrant in the square lattice) consider the following two kernels. Along the two axes both kernels will send the particle toward the origin. Elsewhere, $Q\_1$ will have a strong drift right and a weak drift down, while $Q\_2$ will have a strong drift up and a weak drift left. Each of these kernels will send any particle to the axes, and then to zero, but alternating them will yield a drift up and right.


Formally, for $x,y>0$ let
$Q\_1((x',y')|(x,y))=1/2$ iff $x'=x+1$ and $(y'=y \, $ or $\, y'=y-1)$.


Also for $x,y>0$ let $Q\_2((x',y')|(x,y))=1/2$ iff $y'=y+1$ and 
$(x'=x\, $ or $\, x'=x-1)$.


If $x>0$ then $Q\_i((x-1,0)|(x,0))=1$. If $y>0$ then $Q\_i((0,y-1)|(0,y)=1$. Finally,
 $S=\{(0,0)\}$ is absorbing for both kernels: $Q\_i((0,0)|(0,0))=1$.


Then $L\_1(x,y)=x+4y$ is a Lyapunov function for $Q\_1$ with drift at most $-1$ off $S$. Similarly, $L\_2(x,y)=4x+y$ is a Lyapunov function for $Q\_2$ off $S$. However, alternating $Q\_1$ and $Q\_2$ yields a process that tends to infinity from any initial lattice point $(x,y)$ with $x,y>1$.


