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a male and a female go to a $2$-table restaurant on the same day. each day the male sits at one or the other of the $2$ tables, starting at the table $1$, with a Markov chain transition matrix: $$\begin{bmatrix}0.3 & 0.7\\ 0.7 & 0.3\end{bmatrix}$$ similarly the female sits at one or the other of the $2$ tables, starting at the table $2$, with a Markov chain transition matrix: $$\begin{bmatrix}0.4 & 0.6\\ 0.6 & 0.4\end{bmatrix}$$ assume that $2$ chains are independent.

a. model this situation with a three-state Markov chain and transition matrix.

b. find the probability that the male sits at table $1$ and the female sits at table $2$ on day $2,3$ and $4$.

c. if $N$ is the number of days that the male and the female sit the same table, then how can we describe the random variable $N$?

I'm new to markov chain and each time I work out part (a), I get a different answer. Can someone help?

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  • $\begingroup$ Show (at least one of) your answers to part (a). $\endgroup$
    – Did
    May 7, 2013 at 16:40

1 Answer 1

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Hint: The possible states for the markov chain are: {Both sit together, Male sites at Table $1$ and Female at Table $2$, Male sits at Table $2$ and Female at Table $1$}.

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  • $\begingroup$ Why should this yield a Markov chain? $\endgroup$
    – Did
    May 7, 2013 at 16:42
  • $\begingroup$ Because the probability of transition to a state is dependent on only the previous state and not on the entire history of the chain. $\endgroup$
    – response
    May 7, 2013 at 16:45
  • $\begingroup$ This is not true at the state (Both sit together). The transition to (Male at table 1 and Female at table 2) depends on whether Male and Female are both at table 1 or both at table 2. Hence, unless I am missing a miracle somewhere, the lumped process is not Markov. $\endgroup$
    – Did
    May 7, 2013 at 16:59
  • $\begingroup$ Both at table 1 or both at table 2 is included in the state "both sit together". $\endgroup$
    – response
    May 7, 2013 at 17:03
  • $\begingroup$ Sure, and so what? Please read my previous comment. $\endgroup$
    – Did
    May 7, 2013 at 17:04

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