Adel Makram said:
Homework Statement
Given an initial distribution state vector that represents the probability of the system to be in one of its states. Also given a Markov transition matrix. How to calculate the state vector of the system after n-transition?
Homework Equations
Assuming the initial state vector is a column vector x0, in literature it is considered as a row vector x0T. The state vector after the first transition will be: $$P x_0=x_1$$
where P is the Markov transitional matrix. After n-transitions, $$P^n x_0=x_n$$
The Attempt at a Solution
I tried to decompose P for an easier calculation using singular value decomposition, assuming that it is asymmetric matrix. $$P=UΣV^T$$.
$$P^n=(UΣV^T)^n=(U)^n (Σ)^n (V^T)^n$$.
I know that Σn is a diagonal matrix. Also I know that Un=I. But, I do not know the behavior of (VT)n.
Which convention are you adopting for ##P##? I have the strong impression that most modern treatments take ##p_{ij} = P(X_{t+1}=j|X_t=i)##, so that the
rows of ##P## sum to 1; every book I now own or have ever owned uses that convention. However, the alternative of having the columns summing to 1 is also used, but not as widely.
Anyway, if the rows of ##P## sum to 1 then ##x_n = x_0 P^n##, with row-vectors ##x_0, x_1, x_2, \ldots##. If, instead, the columns of ##P## sum to 1 and you still want row-vector ##x_i##, then you have ##x_n^T = P^n x_0^T##, where ##\{\cdot\}^T## = transpose.
If all you want are numerical results, it is straightforward: ##x_1 = x_0 P##, ##x_2 = x_1 P##, ##\ldots##, ##x_n = x_{n-1} P##. You can compute ##x_1, x_2, \ldots ## sequentially, and you only need to keep track of ##x_k## in order to compute ##x_{k+1}##. On the other hand, if you want an "analytical" expression for ##x_n##, the easiest way is, perhaps, to obtain an analytical expression for ##P^n##, then do a vector-matrix multiplication. Usually we need to use the eigenvalues (and sometimes the eigenvectors) of ##P## to complete the computations, and it can make a difference whether the eigenvalues are all distinct or not. I am reluctant to say more at this point, because it can become lengthy; but I can tell you more if you really want to know.
Note added in edit: I see that you seem to be adopting the convention that columns sum to 1, and where you write ##x^T## I write ##x##.