# Absorbing state for a collection of random walks

Further to this question; having learned some stuff since I posed it.

Consider a collection of random walks $X_i$ which take finite integer values. These evolve as time-inhomogeneous Markov Chains.

Consider $Y=\sum X_i$; this is clearly a (more complex) random walk: the "Super Walk"!

If we set any states of $Y$ where $Y<y$ as absorbing states, I feel that this could be modeled by creating an absorbing state in each $X_i$ and transitioning to this state based on the status of the Super Walk.

How would the transition matrices work for this?

To give a concrete (and simple) example:

Let $X_1=X_2$ have three states 0, 1 and 2 with a $transition matrix of: $$P_X=\begin{bmatrix} 0.5 &0.5&0\\ 0.25&0.5&0.25\\ 0&0.5&0.5\\ \end{bmatrix}$$$Y$then has the states 0,$\dots$,9 and even for this simple example its transition matrix, viz. is rather complex, viz: $$P_Y=\begin{bmatrix} 0.25 &0.25 &0 &0.25&0.25&0&0&0&0\\ 0.125&0.25&0.125&0.125&0.25&0.125&0&0&0\\ 0&0.25&0.25&0&0.25&0.25&0&0&0\\ 0.125&0.125&0&0.25 &0.25&0&0.125&0.125&0\\ 0.065&0.125&0.065&0.125&0.25 &0.125&0.065&0.125&0.065\\ 0&0.125&0.125&0&0.25&0.25 &0&.125&0.125\\ 0&0&0&0.25&0.25&0&0.25 &0.25&0\\ 0&0&0&0.125&0.25&0.125&0.125&0.25 &0.125\\ 0&0&0&0&0.25&0.25&0 &0.25&0.25\\ \end{bmatrix}$$ If we now define,$Y<2$as absorbing states, the transition matrix becomes: $$P_{Y'}=\begin{bmatrix} 1 &0 &0 &0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0&0\\ 0&0.25&0.25&0&0.25&0.25&0&0&0\\ 0&0&0&1&0&0&0&0&0\\ 0.065&0.125&0.065&0.125&0.25 &0.125&0.065&0.125&0.065\\ 0&0.125&0.125&0&0.25&0.25 &0&.125&0.125\\ 0&0&0&0.25&0.25&0&0.25 &0.25&0\\ 0&0&0&0.125&0.25&0.125&0.125&0.25 &0.125\\ 0&0&0&0&0.25&0.25&0 &0.25&0.25\\ \end{bmatrix}$$ How can I determine the transition matrix for X (with the first state being the absorbing one)? $$P_{X'}=\begin{bmatrix} 1&0&0&0\\ ?&?&?&?\\ ?&?&?&?\\ ?&?&?&?\\ \end{bmatrix}$$ Thanks to @Did, whose answer has made me realize that while$Y$is a Markov Chain, these revised$X_i$are not Markov Chains because their transition does not depend on their own state only. I now realize that I will have to consider the "Super Walk" as the fundamental entity. - ## 1 Answer Consider$Y=\sum X_i$; this is clearly a (more complex) random walk Well, no... In general, and in particular in the example you suggest, the fact that the processes$X_1=(X_1(t))_t$and$X_2=(X_2(t))_t$are independent real valued Markov chains does not imply that the process$Y=(Y(t))_t$defined by$Y(t)=X_1(t)+X_2(t)$, is a Markov chain. In your case, consider the event$[Y(t)=2]$. Depending on whether$[X_1(t)=X_2(t)=1]$or$[X_1(t)=2,X_2(t)=0]$, the possible states for$Y(t+1)$are$\{0,1,2,3,4\}$or$\{1,2,3\}$. Hence,$Y=(Y(t))_t$is not a Markov chain. - I can see that the way I have constructed the matrix above is wrong and will edit, but is not$Y$a Markova chain considering the 9 states labeled with the tupple$(X_1,X_2)$? That is$Y=2$can occur from 2 distinct states? – Dale M Jul 2 '13 at 20:27 Once you received an answer, it is considered bad form to modify the question. New problem$\implies\$ new question. – Did Jul 2 '13 at 20:32
Same question - error in example – Dale M Jul 2 '13 at 23:38