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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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    What do you mean by $P(X_{n+1} = V_i)$? This is either zero or one. Ditto for the conditional probabilities.2011-05-30
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    It's the probability that the current event is $V_i$ --- it's not binary since it depends on the $X$2011-05-30
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    Jacob: for any events A and B of positive probability, what is the *definition* of P(A|B)? And of P(B|A)? Ergo?2011-05-31
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    @Didier: I'm sorry, I don't understand what you're getting at. I've updated my question to make it clearer.2011-05-31
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    Forget the update. Quote: *My question is as follows: is P(A|B)=P(B|A) since [meaning: if one assumes that] P(A)=P(B)*. The answer follows from the very definition of P(A|B) and P(B|A), so I am trying to understand what prevents you from seeing it. Once again: if I give you two events A and B, how do you compute P(A|B)?2011-05-31
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    @Jacob, I don't see why the probability of $X_{n+1}=V_i$ should equal the probability of $X_n=V_j$.2011-05-31
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    @Didier: $P(A|B) = A(i,j)$ where $A$ is determined by observing $X$ as mentioned in the question. I'm sorry if the problem wasn't clear before, but I'm actually asking whether $P(A) = P(B)$.2011-05-31
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    You are asking whether P(A)=P(B)??? Well, I can tell you THIS question is NOT in your post... OK, what is your hypothesis? Is it that P(A|B)=P(B|A)?? (I cannot believe one could write *is property Q since property R?* to ask for a proof that *if Q then R*...)2011-05-31
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    Jacob: what is your question?2011-06-01
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    @Didier: The question is whether P(A) = P(B) but I don't think that's the case2011-06-01
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    Sorry, your question does not make sense at all to me. The event A is 'state=i at time=n+1' . The event B is 'state=j at time=n'. How could one say anything about P(A) P(B), knowing only that there is a (general) transition matrix, I cannot imagine.2011-06-01
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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.