The product $z_t$ is not a Markov Chain (MC).
Assume we know that $z_{t_0} = 0$ at
some time $t_0$, meaning that $x_{t_0}=0$ or $y_{t_0}=0$. Then the
past values $\{z_{u};\,u <t_0\}$ of $z_t$ still contain information
about future values $\{z_v;\, v > t_0\}$ in contradiction with
the Markov property. Indeed, let $s_t := [x_t,
\,y_t]$; so that $s_t$ is a MC taking the $4$ values written here as $00$,
$01$, $10$ and $11$. At time $t_0$, only the first $3$
states are possible since the product is $0$. Let $t_0-W$ be the random time of the latest
state change for $z_t$, which was $z: \,1 \rightarrow 0$. If $W$ is small,
then most probably $s_{t_0}$ is $01$ or $10$, but not
$00$: only one change of state occured during the interval
$(t_0-W, \,t_0)$. This in turn tells us that the next transition $0
\rightarrow 1$ will occur more quickly than if a large value $W$ had
been obtained.
Here is a more formal derivation. For a fixed $t$ and a small $h > 0$ we have
$$
\mathrm{Pr}\{z_{t-h} = 1\,\vert \,s_{t} = 00\} = o(h),
\qquad
\mathrm{Pr}\{z_{t-h} = 1\,\,\vert \, s_{t} = 01\} = \lambda h + o(h)
$$
with $\lambda >0$. Indeed, the first probability involves two transitions of $s_t$.
Using Bayes formula
$$
\mathrm{Pr}\{s_{t} = 00 \,\vert\, z_{t-h} = 1, \, z_{t}=0\}
= \frac{\mathrm{Pr}\{ z_{t-h} = 1 \,\vert \,s_{t} = 00\} \,
\mathrm{Pr}\{ s_{t} = 00\, \vert \, z_{t} = 0\}}{ \mathrm{Pr}\{ z_{t-h} = 1 \,\vert \, z_{t} = 0 \} }.
$$
The numerator of the fraction is $o(h)$, while its denominator
is easily found to be $\nu h + o(h)$ for some $\nu >0$, so the probability is $o(h)$.
By contrast
$$
\mathrm{Pr}\{s_{t} = 01 \,\vert\, z_{t-h} = 1, \, z_{t}=0\}
= \rho h + o(h)
$$
for some $\rho > 0$. Thus conditional on $\{z_{t-h} = 1, \, z_{t}=0\}$
the event $\{s_{t} = 01\}$ is much more probable than $\{s_{t} = 00\}$
for small $h$, as claimed.
Here $z_t$ follows a Hidden Markov Model (HMM) with hidden state $s_t$
and simply results from grouping $3$ possible states of $s_t$ as
one $z_t=0$. More generally, grouping states of a MC $s_t$ will
result in a HMM process but not a MC.
The usual definition of limiting distribution is that a Markov chain has a limiting distribution $\pi$ if for every initial distribution $P(0)$,
$
\lim_{n \rightarrow \infty} P(0)P^{(n)}=\pi
$
It's important to get that quantifier right.
A useful example to consider is the Markov chain with
$
P=\left[\begin{array}{cc}
0 & 1 \\
1 & 0
\end{array}
\right].
$
Here, $\pi=[1/2 \;\;1/2]$ is a stationary distribution but not a limiting distribution of the Markov chain. In fact this Markov chain does not have a limiting distribution.
Best Answer
Please read the correct answer from the other post, as mentioned there, this answer incorrectly assumes a rate matrix instead of a transition Matrix.
You can always get a continuous time version of a discrete one by simply "Poissonizing" it. For example, if you have a discrete time Markov chain with transition matrix $T$ you get a continuous time version by considering $$P_t = \sum_{n\geq 0} \frac{t^n}{n!}\exp(-t)T^n $$ Hence the above definition makes sense in the context of continuous time Markov chains.