Continuous Markov chain - expected time












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I have been studying problems from some of my lecture notes and was given this question




A continuous time Markov Chain $X_t$ with state space ${1,2,3}$ has infinitesimal generator
$$A= begin{bmatrix}-6 & 2 & 4\2 &-5 &3 \2 & 3& -5end{bmatrix}$$
If the chain starts in state 2, what is expected total time spent in state 2 before the chain first enters state 3




In the solution the answer is given as $$frac{1}{5}+frac{1}{5}frac{2}{15}+frac{1}{5}(frac{2}{15})^{2}+.......= frac{frac{1}{5}}{1-frac{2}{15}}$$



I know from my lectures notes that
The expected time at state 2 is: $(frac{1}{5})$



The chain jumps to state 1 with probability $(frac{2}{5})$ and state 3 with probability $(frac{3}{5})$



and I know chain jumps from 1 to 2 with probability $(frac{1}{3})$



so the probability that the chain return to state 2 in two steps, starting at 2 is thus $(frac{2}{5})(frac{1}{3})=(frac{2}{15})$



So after finding these probabilities I can understand the right hand side of the equation given in the solution but im confused about how it becomes $$frac{frac{1}{5}}{1-frac{2}{15}}$$



I have been looking through my notes and haven't been able to find anything. Any help would be great






probability statistics markov-chains







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  • 2




    $begingroup$
    $$sum_{n=0}^infty ax^n=frac a{1-x}qquad (a=1/5, x=2/15)$$
    $endgroup$
    – Did
    Dec 13 '18 at 18:15
















0












$begingroup$


I have been studying problems from some of my lecture notes and was given this question




A continuous time Markov Chain $X_t$ with state space ${1,2,3}$ has infinitesimal generator
$$A= begin{bmatrix}-6 & 2 & 4\2 &-5 &3 \2 & 3& -5end{bmatrix}$$
If the chain starts in state 2, what is expected total time spent in state 2 before the chain first enters state 3




In the solution the answer is given as $$frac{1}{5}+frac{1}{5}frac{2}{15}+frac{1}{5}(frac{2}{15})^{2}+.......= frac{frac{1}{5}}{1-frac{2}{15}}$$



I know from my lectures notes that
The expected time at state 2 is: $(frac{1}{5})$



The chain jumps to state 1 with probability $(frac{2}{5})$ and state 3 with probability $(frac{3}{5})$



and I know chain jumps from 1 to 2 with probability $(frac{1}{3})$



so the probability that the chain return to state 2 in two steps, starting at 2 is thus $(frac{2}{5})(frac{1}{3})=(frac{2}{15})$



So after finding these probabilities I can understand the right hand side of the equation given in the solution but im confused about how it becomes $$frac{frac{1}{5}}{1-frac{2}{15}}$$



I have been looking through my notes and haven't been able to find anything. Any help would be great






probability statistics markov-chains







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $$sum_{n=0}^infty ax^n=frac a{1-x}qquad (a=1/5, x=2/15)$$
    $endgroup$
    – Did
    Dec 13 '18 at 18:15














0












0








0


0



$begingroup$


I have been studying problems from some of my lecture notes and was given this question




A continuous time Markov Chain $X_t$ with state space ${1,2,3}$ has infinitesimal generator
$$A= begin{bmatrix}-6 & 2 & 4\2 &-5 &3 \2 & 3& -5end{bmatrix}$$
If the chain starts in state 2, what is expected total time spent in state 2 before the chain first enters state 3




In the solution the answer is given as $$frac{1}{5}+frac{1}{5}frac{2}{15}+frac{1}{5}(frac{2}{15})^{2}+.......= frac{frac{1}{5}}{1-frac{2}{15}}$$



I know from my lectures notes that
The expected time at state 2 is: $(frac{1}{5})$



The chain jumps to state 1 with probability $(frac{2}{5})$ and state 3 with probability $(frac{3}{5})$



and I know chain jumps from 1 to 2 with probability $(frac{1}{3})$



so the probability that the chain return to state 2 in two steps, starting at 2 is thus $(frac{2}{5})(frac{1}{3})=(frac{2}{15})$



So after finding these probabilities I can understand the right hand side of the equation given in the solution but im confused about how it becomes $$frac{frac{1}{5}}{1-frac{2}{15}}$$



I have been looking through my notes and haven't been able to find anything. Any help would be great






probability statistics markov-chains







share|cite|improve this question











$endgroup$




I have been studying problems from some of my lecture notes and was given this question




A continuous time Markov Chain $X_t$ with state space ${1,2,3}$ has infinitesimal generator
$$A= begin{bmatrix}-6 & 2 & 4\2 &-5 &3 \2 & 3& -5end{bmatrix}$$
If the chain starts in state 2, what is expected total time spent in state 2 before the chain first enters state 3




In the solution the answer is given as $$frac{1}{5}+frac{1}{5}frac{2}{15}+frac{1}{5}(frac{2}{15})^{2}+.......= frac{frac{1}{5}}{1-frac{2}{15}}$$



I know from my lectures notes that
The expected time at state 2 is: $(frac{1}{5})$



The chain jumps to state 1 with probability $(frac{2}{5})$ and state 3 with probability $(frac{3}{5})$



and I know chain jumps from 1 to 2 with probability $(frac{1}{3})$



so the probability that the chain return to state 2 in two steps, starting at 2 is thus $(frac{2}{5})(frac{1}{3})=(frac{2}{15})$



So after finding these probabilities I can understand the right hand side of the equation given in the solution but im confused about how it becomes $$frac{frac{1}{5}}{1-frac{2}{15}}$$



I have been looking through my notes and haven't been able to find anything. Any help would be great






probability statistics markov-chains




probability statistics markov-chains






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edited Dec 13 '18 at 22:32









Tianlalu

3,08421138




3,08421138










asked Dec 13 '18 at 18:12









Rito LoweRito Lowe

516




516








  • 2




    $begingroup$
    $$sum_{n=0}^infty ax^n=frac a{1-x}qquad (a=1/5, x=2/15)$$
    $endgroup$
    – Did
    Dec 13 '18 at 18:15














  • 2




    $begingroup$
    $$sum_{n=0}^infty ax^n=frac a{1-x}qquad (a=1/5, x=2/15)$$
    $endgroup$
    – Did
    Dec 13 '18 at 18:15








2




2




$begingroup$
$$sum_{n=0}^infty ax^n=frac a{1-x}qquad (a=1/5, x=2/15)$$
$endgroup$
– Did
Dec 13 '18 at 18:15




$begingroup$
$$sum_{n=0}^infty ax^n=frac a{1-x}qquad (a=1/5, x=2/15)$$
$endgroup$
– Did
Dec 13 '18 at 18:15










1 Answer
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$begingroup$

It is a geometric series. In this particular case you could say:



$$S = frac{1}{5}+frac{1}{5}frac{2}{15}+frac{1}{5}left(frac{2}{15}right)^{2}+cdots$$



multiplying by $frac2{15}$ gives
$$frac2{15}S = frac{1}{5}frac{2}{15}+frac{1}{5}left(frac{2}{15}right)^{2}+frac{1}{5}left(frac{2}{15}right)^{3}+cdots$$



and subtracting gives
$$left(1-frac2{15}right)S = frac{1}{5}$$



so
$$S = frac{frac{1}{5}}{1-frac2{15}}$$






share|cite|improve this answer









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    $begingroup$

    It is a geometric series. In this particular case you could say:



    $$S = frac{1}{5}+frac{1}{5}frac{2}{15}+frac{1}{5}left(frac{2}{15}right)^{2}+cdots$$



    multiplying by $frac2{15}$ gives
    $$frac2{15}S = frac{1}{5}frac{2}{15}+frac{1}{5}left(frac{2}{15}right)^{2}+frac{1}{5}left(frac{2}{15}right)^{3}+cdots$$



    and subtracting gives
    $$left(1-frac2{15}right)S = frac{1}{5}$$



    so
    $$S = frac{frac{1}{5}}{1-frac2{15}}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      It is a geometric series. In this particular case you could say:



      $$S = frac{1}{5}+frac{1}{5}frac{2}{15}+frac{1}{5}left(frac{2}{15}right)^{2}+cdots$$



      multiplying by $frac2{15}$ gives
      $$frac2{15}S = frac{1}{5}frac{2}{15}+frac{1}{5}left(frac{2}{15}right)^{2}+frac{1}{5}left(frac{2}{15}right)^{3}+cdots$$



      and subtracting gives
      $$left(1-frac2{15}right)S = frac{1}{5}$$



      so
      $$S = frac{frac{1}{5}}{1-frac2{15}}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        It is a geometric series. In this particular case you could say:



        $$S = frac{1}{5}+frac{1}{5}frac{2}{15}+frac{1}{5}left(frac{2}{15}right)^{2}+cdots$$



        multiplying by $frac2{15}$ gives
        $$frac2{15}S = frac{1}{5}frac{2}{15}+frac{1}{5}left(frac{2}{15}right)^{2}+frac{1}{5}left(frac{2}{15}right)^{3}+cdots$$



        and subtracting gives
        $$left(1-frac2{15}right)S = frac{1}{5}$$



        so
        $$S = frac{frac{1}{5}}{1-frac2{15}}$$






        share|cite|improve this answer









        $endgroup$



        It is a geometric series. In this particular case you could say:



        $$S = frac{1}{5}+frac{1}{5}frac{2}{15}+frac{1}{5}left(frac{2}{15}right)^{2}+cdots$$



        multiplying by $frac2{15}$ gives
        $$frac2{15}S = frac{1}{5}frac{2}{15}+frac{1}{5}left(frac{2}{15}right)^{2}+frac{1}{5}left(frac{2}{15}right)^{3}+cdots$$



        and subtracting gives
        $$left(1-frac2{15}right)S = frac{1}{5}$$



        so
        $$S = frac{frac{1}{5}}{1-frac2{15}}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 9:21









        HenryHenry

        100k481168




        100k481168






























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