Intuitive explanation of why a stochastic process constructed by a uniformly distributed stopping time is not Markov Consider a Markov chain that waits a
time $T^{∗}$ before leaving the current state, where $T^{∗}$ has uniform distribution over the set of times $\{1, 2, 3, 4\}$
If $(W_k)_{k\geq}$ would be a stochastic process constructed so that it stays in each state $i$ for a time distributed to $T^{*}$, what would be an intuitive explanation of why this is not a Markov chain?
 A: Consider such a process with state variable $S$ and probability transition matrix $P$.   At time $t$, we are in state $S(t)$.  Does the probability of being in state $S'$ at time $t+1$, or more generally, time $t+\tau$, depend solely upon the current state $S(t)$?
No, it does not.  Let us assume that you have just made a transition into a new state.  The time to the next transition, label it $T^*$, is now uniformly distributed over the set of times $\{1,2,3,4\}$.  Consequently, as we move forward one "time stamp", we have a 75% probability of not making a transition at all (whenever $T^* > 1$), in which case $S'(t+1) = S(t)$ with probability $1$, and a 25% probability of making a transition in accordance with $P$ (whenever $T^* = 1$).  
On the other hand, let us assume that $T^* = 4$, and three of the intermediate time stamps have passed since the last transition.  We are now certain to make a transition in the next time period, and the probabilities of the next states are determined solely by $P$.  
Consequently, the probability of state $S'(t+1)$ given that we are in $S(t)$ does not depend solely upon $S(t)$, but also upon the time since the last transition.  This breaks the Markov property.
Note, however, that if the time to the next transition has an Exponential distribution, the resulting continuous time Markov chain (CTMC) is Markovian, as the memoryless property of the exponential distribution means that the time to the next transition has the same (exponential) distribution regardless of how much time has passed since the last transition.
Also note, expanding upon @whuber's comment above, that the process can be described by a Markov chain if we expand the state space to include the time since the last transition.  The state space becomes, let us say, $(W,\tau)$, and the transitions to the next state now solely depend upon the current state.  For example, if $(W,\tau)_t = (w,3)$, we know that $(W, \tau)_{t+1} = (w^*, 0)$, where the transition from $w$ to $w^*$ is governed by the transition matrix $P$ and $\tau$ must change from $3$ to $0$ because the maximum time between transitions is four periods.
