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.

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 got different answer. any can help? thanks

share|improve this question
    
Show (at least one of) your answers to part (a). –  Did May 7 '13 at 16:40

1 Answer 1

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}.

share|improve this answer
    
Why should this yield a Markov chain? –  Did May 7 '13 at 16:42
    
Because the probability of transition to a state is dependent on only the previous state and not on the entire history of the chain. –  response May 7 '13 at 16:45
    
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. –  Did May 7 '13 at 16:59
    
Both at table 1 or both at table 2 is included in the state "both sit together". –  response May 7 '13 at 17:03
    
Sure, and so what? Please read my previous comment. –  Did May 7 '13 at 17:04

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.