In the bus ridership example, Sec. 2.12, let A i denote the number of passengers

Question

In the bus ridership example, Sec. 2.12, let Ai denote the number of passengers who alight at stop i. Show by counterexample that A2, A3, A4,... is NOT a Markov chain.

This question is ABOVE!

The Bus example is refers to is below.

Here is the Bus Example:

Please help!

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2.12 Example: Bus Ridership Consider the following analysis of bus ridership. In order to keep things easy, it will be quite oversimplified, but the principles will be clear.) Here is the model At each stop, each passsenger alights from the bus, independently, with probability 0.2 each. Either 0, 1 or 2 new passengers get on the bus, with probabilities 0.5, 0.A and 0.1, respectively. Passengers at successive stops are independent.

Explanation / Answer

For a chain to be a marcov chain it should satisfy the property

P(Xn+1=x | X1=x1, X2=x2, ..., Xn=xn) = P(Xn+1=x | Xn=xn)

ie

probability of next state depends only on the present state and not on the previous states

In this question

lets take a case ( counter example)

B1 = B2 = B3 = 1

so total passengers got on to bus till 3rd stop = 3

and

A3 = 1

now

P(A4 = 1 | A3 =1, A2 = 1 ) = 0.2 since two passengers got down in previous stations

but

P(A4 = 1 | A3 =1, A2 = 2 ) = 0   since all three passengers got down in previous stations

so the Probabilty of A4 doesn't depend only on its previous event (ie A3) as we can see that A3 is same in both cases but the probability distribution of A4 is different in both cases.

so this is not a marcov chain.

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