Markov chain - stationary distribution -

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SUMMARY

The discussion centers on finding the stationary distribution of a Markov chain representing a taxi's trips between the airport and the city. The transition matrix provided is [[0, 1], [1/4, 3/4]]. The stationary distribution exists and can be determined by solving the linear equations derived from the transition probabilities. The solution yields a stationary distribution of [1/5, 4/5], indicating the long-term probabilities of the taxi being at the airport and in the city, respectively.

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Carolyn
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Homework Statement



-A taxi is located either at the airport or in the city. From the city, the next trip is to the airport with 1/4 probability, or to somewhere else in the city with 3/4. From the airport, the next trip is always to the city.

(a) find the stationary distribution
(b) starting from the airport, what is the expected number of trips until its next visit to the airport?

Homework Equations





The Attempt at a Solution



The matrix I got is

0 1
1/4 3/4

but I am having trouble solving for the stationary distribution and expected number of trips. Somehow, I don't think the stationary distribution even exists. But it should. What am I missing? Thanks.
 
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hello carolyn

your transition matrix is same as expected.And the stationary distribution exists. you can find the stationary distribution by solving it as two linear equations.
let 'Pij' be the element, i=1,2 and j=1,2 (for the given transition matrix).

P0=P0*P00+P1*P01; P1=P0*P10+P1*P11

here P0 and P1 are unknowns , the elements of stationary distribution.
Pij(P00,P01,P10,P11) are the transition matrix elements.

By substituting you will get P0=P0*0+P1*(1/4); P1=P0*1+P1*(3/4).

Also the condition to find P0 and P1 is P0+P1=1(the raw sum =1);

then u will get a stationary distribution of
[1/5 4/5]
 

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