Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm asking a general question I think, but I couldn't find the answer myself. I have a system in which the next action depends on some variables, and those variables changes over time. At first I thought of modeling this as a markov chain, but I also need a variable set of probabilities to pass on each state.

Basically, I have 11 states in which I can be, and my probability to translate from state to another depends on the choices of all the other "players".

At the beginning, I start in one state, and at every instant, I look if too many players had choose my state. If so, with some probability I'll switch to another state and so on, until I found my best state.

Is that possible to to with a Markov chain or do I have to look for something else?

In addition, I'd like also to see when my chain gets to an equilibrium, that is, when no players change their state anymore.

share|improve this question

2 Answers 2

up vote 3 down vote accepted

A convenient construction of a (finite) Markov chain $(X_n)$ is to let $X_{n+1}=\xi(X_n,U_n)$, for some deterministic function $\xi$ and a given i.i.d. sequence $(U_n)$ (say, uniformly distributed on $(0,1)$). But your question asks that $X_{n+1}=\xi(X_n,Y_n,U_n)$, for some process $(Y_n)$ describing the positions of the other players.

Since $(Y_n)$ is not i.i.d., a common strategy to save the day is to consider the process $(Z_n)$ defined by $Z_n=(X_n,Y_n)$. Then, everything depends on the structure of the process $(Y_n)$.

If $(Y_n)$ is itself a Markov chain based on a function $\eta$ and a given i.i.d. sequence $(V_n)$ independent from $(U_n)$, that is, if $Y_{n+1}=\eta(Y_n,V_n)$, then $(Z_n)$ is a Markov chain as well. This chain is based on the sequence of innovations $(W_n)$ defined by $W_n=(U_n,V_n)$ since $Z_{n+1}=\zeta(Z_n,W_n)$ for some function $\zeta$ that I will let you write down.

This is the structure of a hidden Markov process: a state process, here $(Y_n)$, which is a Markov chain, and, coupled to the state process, an observation process, here $(X_n)$, which, for a given realization of the state process, is an inhomogenous Markov chain. Then, as we saw above, the couple process $(Z_n)$ defined by $Z_n=(X_n,Y_n)$ is a Markov chain but the observation process $(X_n)$ is not.

With no further hypothesis on the structure of $(Y_n)$, it seems difficult to say anything else. Note that one could simply assume that $(Z_n)$ is a Markov chain, and that this would allow $Y_{n+1}$ to depend on $X_n$ as well. For example, the dynamical structure $X_{n+1}=\xi(X_n,Y_n,U_n)$ and $Y_{n+1}=\eta(Y_n,X_n,V_n)$ could be summarized by $Z_n=\zeta(Z_n,W_n)$ as well.

share|improve this answer
    
Thank you very much! –  lbedogni Feb 1 '12 at 7:08

I do not understand your question completely. But see if this simple analogy helps. Consider a game where players $A,B$ move independently on the states $S = \{{1,2}\}$ with transition probabilities $P= \left( \begin{array}{cc} p_{11} & p_{12} \\ p_{21} & p_{22} \\ \end{array} \right) $ and
$Q= \left( \begin{array}{cc} q_{11} & q_{12} \\ q_{21} & q_{22} \\ \end{array} \right) $ respectively. For simplicity, let us assume $A$ begins on $1$ and $B$ begins on $2$. Suppose our goal is to find the expected time for the game to end. Then one way to approach this is consider the Markov chain on the state space $\hat{S} = \{(1,1), (1,2), (2,1), (2,2)\} $ with the transition probabilities given by $Q= \left( \begin{array}{cccc} 1 & 0& 0 & 0 \\ p_{11}q_{21} & p_{11}q_{22} &p_{12}q_{21} &p_{12}q_{22}\\ p_{21}q_{11} & p_{21}q_{12} &p_{22}q_{11} &p_{22}q_{12}\\ 0 & 0& 0 & 1 \\ \end{array} \right).$ Our goal then reduces to finding the expected time for this Markov chain to hit the set $H=\{(1,1),(2,2)\}$ starting from $(1,2)$.

share|improve this answer
    
Not completely, because A and B change their probability according to the other's moves. –  lbedogni Jan 31 '12 at 20:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.