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Let $X$ be a sequence of discrete values from a finite set $V$. Let $A$ be the transition matrix computed by counting instances of $V_i \rightarrow V_j$ from the sequence $X$. Hence, $P(X_{n+1} = V_i | X_n = V_j) = A(i,j)$.

Now, let the events $A = [X_{n+1} = V_i]$ and $B = [X_n = V_j]$. My question is as follows: is $P(A) = P(B)$?

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    Jacob: OK, now we know what we want to prove. But we cannot prove something from nothing, right? Sooo... next step: what is your hypothesis? I see none in your post.2011-06-01

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Reversibility is not required. Let's consider a Markov model of a swimmer. The state WEAR_TRUNKS may be followed by ENTER_WATER (or TAKE_SHOWER). Rarely does one WEAR_TRUNKS after ENTER_WATER.

Look up detailed balance.

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    No, that is not my question. I've updated it to make it clearer.2011-05-31
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If your transition matrix is $\pmatrix{1&1\cr0&0\cr}$, so that $X_{n+1}=V_1$ no matter what $X_n$ is, then the probability that $X_{n+1}=V_1$ is $1$, while the probability that $X_n=V_2$ is zero, provided $n\gt1$, so it would appear that in this case $P(A)\ne P(B)$.

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    If we can assume homogeneity/ergodicity/whatever property is that means that probabilities are independent of $n$, and we assume we have reached an equilibrium state where each state is occupied at time $n$ according to its equilibrium probability (sorry, don't really know the terminology of Markov chains...) then can we make the statement that for $m\neq n$ we have $P(X_n=i)=P(X_m=j)$ if $i=j$, but otherwise they are not in general equal.2011-06-02
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The condition that $P(X_n=V_i)=P(X_{n+1}=V_j)$ for every $n$ and every $(i,j)$ is equivalent to the condition that the distribution of $X_n$ is uniform for every $n$, that is, that $P(X_n=V_i)=1/K$ for every $n$ and $i$, where $K$ denotes the size of $V$.

Now, this happens if and only if (1) the distribution of $X_0$ is uniform and (2) the transition matrix $A$ preserves the uniform distribution. Condition (1) means that, for every $i$, $ P(X_0=V_i)=1/K. $ Condition (2) means that, for every $i$, $ \sum_jA(i,j)=1. $ Note that the dual condition $\displaystyle\sum_iA(i,j)=1$ is always true.