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I have just started learning markov chains and I need help on the following question:

Alice and Bob vote in each parliamentary election. If, in a certain election, Alice and Bob vote for the same party, they vote for it again in the next election. If they vote for different parties, then in next election each of them switches their opinion independently with probability 1/4.

How do I draw the transition graph for this one?

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    $\begingroup$ There is more than one way to answer this, but they all begin by enumerating the possible states in your model. Could you tell us what you have decided about that? $\endgroup$
    – whuber ♦
    Commented Oct 5, 2020 at 20:46
  • $\begingroup$ I think there would be 2 states , one is "voting for the same party" and the other one would be "voted for different parties"? $\endgroup$
    – bluelagoon
    Commented Oct 5, 2020 at 20:48
  • $\begingroup$ If that is going to work, then you should be able to determine the two transition probabilities between those states. The next step, then, is to try to compute those probabilities. If you can, you're done; if you can't, your effort ought to reveal what additional detail is needed in describing the states. $\endgroup$
    – whuber ♦
    Commented Oct 5, 2020 at 21:01
  • $\begingroup$ Please add new info in comments as an edit to the post! Not everybody reads comments ... $\endgroup$
    – kjetil b halvorsen ♦
    Commented Oct 8, 2020 at 1:58

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The 2x2 transition matrix consists of 4 probabilities: $p(s|s)$, $p(d|s)$, $p(s|d)$ and $p(d|d)$, where $s$ and $d$ stand for "same votes" and "different votes" respectively.

In case when Alice and Bob had different (previous) votes, then if both switch or both do not switch their choices, they will have different votes again. Otherwise, if only one of them switches his (her) choice, they will have same vote. Thus,

$p(s|d)=\frac{1}{4}*\frac{3}{4}+\frac{3}{4}*\frac{1}{4}=\frac{3}{8}$

$p(d|d)=\frac{1}{4}*\frac{1}{4}+\frac{3}{4}*\frac{3}{4}=\frac{5}{8}$

In case when Alice and Bob had same (previous) vote, obviously, $p(s|s)=1$ and $p(d|s)=0$.

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