Initial state of a Markov Process Intuitively (without using matrices) , why in a Markov process the probability vector , after a huge number of iterations, is indipendent from the initial probability vector ?
(Im considering the particular case in which this happens) 
Thanks
 A: Intuitively: If a Markov process has a limiting distribution (which is the  "probability vector after a huge number of iterations [that is] independent from the initial probability vector that you mention), that means the process will reach a kind of equilibrium over time. For example, consider a marathon runner that reaches a steady marathon pace as the race goes on, regardless of the speed of her initial sprint. Recall that a Markov process has the property that the future, given the present, is independent of the past. So very loosely speaking, if the present time is "far enough along" (i.e. "after a huge number of iterations" as you say), then the initial state, which is part of the distant past, "washes out."
Warning: Not every Markov process has a limiting distribution, let alone a unique one. The limiting distribution But of course, we need a rigorous characterization of "huge number of iterations," so we use limits.
The technical details: Suppose we have a Markov chain $\{X_n\}$ with states $\{1, 2, \dots, m\}$. Then the limiting distribution is a probability vector $\boldsymbol{\pi}$ such that 
$$\boldsymbol{\pi} = (\pi_1, \pi_2, \dots, \pi_m) = \Big(\lim_{n\rightarrow\infty} \mathbb P(X_n = 1), \lim_{n\rightarrow\infty}\mathbb P(X_n = 2), \dots, \lim_{n\rightarrow\infty}\mathbb P(X_n = m)\Big).$$
An irreducible and aperiodic Markov chain on a finite state space $S$ has a unique limiting distribution $\boldsymbol{\pi}$ that can be obtained by solving the system of linear equations $\sum_{i \in S} \pi_i = 1$ and $\boldsymbol{\pi} = \boldsymbol{\pi} \boldsymbol{P}$, where $\boldsymbol{P}$ is the probability transition matrix of the Markov chain.
But if a Markov chain is periodic, limiting distributions are not guaranteed (i.e. the theorem above does not hold if the aperiodicity assumption is relaxed). It is easy to check that a Markov chain on $\{0,1\}$ with probability transition matrix given below cannot have a limiting distribution.
$$\boldsymbol{P} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
