- #1

MathematicalPhysicist

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Any hints?

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- Thread starter MathematicalPhysicist
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- #1

MathematicalPhysicist

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Any hints?

- #2

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1,1,2

2,1,1,2

2,2,1,1,2

1,2,1,1,2

...

- #3

MathematicalPhysicist

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I mean for 1,1,2 the probability is: [tex]P_{1,1}\frac{\pi_1}{\pi_1+\pi_2}P_{1,2}[/tex]

2,1,1,2 [tex]\frac{\pi_2}{\pi_1+\pi_2}P_{1,1}P_{2,1}P_{1,2}[/tex]

1,2,1,1,2 [tex]\frac{\pi_1}{\pi_1+\pi_2}P^2_{1,2}P_{1,1}P_{2,1}[/tex]

So my hunch is that if we first get to 1 or 2, we should multiply by pi1/(pi1+pi2) or pi2/(pi1+pi2), and we should always multiply by P1,1, but other than this I don't see a general equation for all cases.

- #4

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I don't have an immediate answer, but you might find the approach in this paper to be helpful:

http://smu.edu/statistics/TechReports/TR211.pdf [Broken]

The authors derive the unconditional distribution of the number of successes (e.g., state 2) in n+1 trials. It seems to me that you need to derive a similar distribution, conditional on having obtained (exactly? or at least?) two consecutive failures (1,1).

http://smu.edu/statistics/TechReports/TR211.pdf [Broken]

The authors derive the unconditional distribution of the number of successes (e.g., state 2) in n+1 trials. It seems to me that you need to derive a similar distribution, conditional on having obtained (exactly? or at least?) two consecutive failures (1,1).

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