omg!
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please excuse any inaccuracies in the following formulation of my problem. here it goes:
1. there are [tex]n[/tex] discrete time steps
2. between every time step, transitions 0 -> 1, 1-> 0 can happen between states denoted by 0 and 1. the two possible transition directions are poisson processes, i.e. number of time steps between transitions of the same kind are exponentially distributed.
3. additionally, there is another unidirectional transition from 1 to 2.
alternatively, you can think of the sequence [tex]X_1, X_2, \ldots, X_n[/tex] of random variables, with [tex]X_i\in\left{0,1,2}\right}[/tex] with the conditions [tex]P[X_{i+1}=1|X_{i}=0]=\text{const.}[/tex], [tex]P[X_{i+1}=0|X_{i}=1]=const.[/tex], [tex]P[X_{i+1}=2|X_{i}=1]=const.[/tex], [tex]P[X_{i+1}=2|X_{i}=0]=0[/tex] and [tex]P[X_{i+1}=0,1|X_{i}=2]=0[/tex].
now take the continuum limit [tex]n\rightarrow\infty[/tex]
the problem is: what is the PDF of time where the state was last 1, i.e. [tex]X(t)=1,X(s)\neq 1,\quad\forall s\geq t[/tex]
1. there are [tex]n[/tex] discrete time steps
2. between every time step, transitions 0 -> 1, 1-> 0 can happen between states denoted by 0 and 1. the two possible transition directions are poisson processes, i.e. number of time steps between transitions of the same kind are exponentially distributed.
3. additionally, there is another unidirectional transition from 1 to 2.
alternatively, you can think of the sequence [tex]X_1, X_2, \ldots, X_n[/tex] of random variables, with [tex]X_i\in\left{0,1,2}\right}[/tex] with the conditions [tex]P[X_{i+1}=1|X_{i}=0]=\text{const.}[/tex], [tex]P[X_{i+1}=0|X_{i}=1]=const.[/tex], [tex]P[X_{i+1}=2|X_{i}=1]=const.[/tex], [tex]P[X_{i+1}=2|X_{i}=0]=0[/tex] and [tex]P[X_{i+1}=0,1|X_{i}=2]=0[/tex].
now take the continuum limit [tex]n\rightarrow\infty[/tex]
the problem is: what is the PDF of time where the state was last 1, i.e. [tex]X(t)=1,X(s)\neq 1,\quad\forall s\geq t[/tex]