Solve linear system of equations by RREF.












0












$begingroup$



Consider a Markov chain with transition matrix



$$P=begin{bmatrix}1-a&a&0\0&1-b&b\c&0&1-cend{bmatrix},$$



where $0<a,b,c<1.$ Find the stationary distribution.




This is embarassing but somehow I struggle solving a linear system of equations. I am supposed to solve the system $vec{pi} P=vec{pi}$ for $vec{pi}.$



I get the equations



$$begin{align}(1-a)pi_1+cpi_3&=pi_1\
api_1+(1-b)pi_2 &= pi_2\
bpi_2+(1-c)pi_3&=pi_3\
pi_1+pi_2+pi_3&=1end{align}$$



ofcourse adding the last equation since the elements in a probability vector must sum to $1$. Rearranging a bit I get



begin{align}-api_1+cpi_3&=0\api_1-bpi_2&=0\bpi_2-cpi_3&=0\pi_1+pi_2+pi_3&=1,end{align}



adding (2) to (1) and then (3) to (1) leaves me with only the last equation. I'm confused. I also tried using the RREF algorithm, to no success.






linear-algebra







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$endgroup$

















    0












    $begingroup$



    Consider a Markov chain with transition matrix



    $$P=begin{bmatrix}1-a&a&0\0&1-b&b\c&0&1-cend{bmatrix},$$



    where $0<a,b,c<1.$ Find the stationary distribution.




    This is embarassing but somehow I struggle solving a linear system of equations. I am supposed to solve the system $vec{pi} P=vec{pi}$ for $vec{pi}.$



    I get the equations



    $$begin{align}(1-a)pi_1+cpi_3&=pi_1\
    api_1+(1-b)pi_2 &= pi_2\
    bpi_2+(1-c)pi_3&=pi_3\
    pi_1+pi_2+pi_3&=1end{align}$$



    ofcourse adding the last equation since the elements in a probability vector must sum to $1$. Rearranging a bit I get



    begin{align}-api_1+cpi_3&=0\api_1-bpi_2&=0\bpi_2-cpi_3&=0\pi_1+pi_2+pi_3&=1,end{align}



    adding (2) to (1) and then (3) to (1) leaves me with only the last equation. I'm confused. I also tried using the RREF algorithm, to no success.






    linear-algebra







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$



      Consider a Markov chain with transition matrix



      $$P=begin{bmatrix}1-a&a&0\0&1-b&b\c&0&1-cend{bmatrix},$$



      where $0<a,b,c<1.$ Find the stationary distribution.




      This is embarassing but somehow I struggle solving a linear system of equations. I am supposed to solve the system $vec{pi} P=vec{pi}$ for $vec{pi}.$



      I get the equations



      $$begin{align}(1-a)pi_1+cpi_3&=pi_1\
      api_1+(1-b)pi_2 &= pi_2\
      bpi_2+(1-c)pi_3&=pi_3\
      pi_1+pi_2+pi_3&=1end{align}$$



      ofcourse adding the last equation since the elements in a probability vector must sum to $1$. Rearranging a bit I get



      begin{align}-api_1+cpi_3&=0\api_1-bpi_2&=0\bpi_2-cpi_3&=0\pi_1+pi_2+pi_3&=1,end{align}



      adding (2) to (1) and then (3) to (1) leaves me with only the last equation. I'm confused. I also tried using the RREF algorithm, to no success.






      linear-algebra







      share|cite|improve this question









      $endgroup$





      Consider a Markov chain with transition matrix



      $$P=begin{bmatrix}1-a&a&0\0&1-b&b\c&0&1-cend{bmatrix},$$



      where $0<a,b,c<1.$ Find the stationary distribution.




      This is embarassing but somehow I struggle solving a linear system of equations. I am supposed to solve the system $vec{pi} P=vec{pi}$ for $vec{pi}.$



      I get the equations



      $$begin{align}(1-a)pi_1+cpi_3&=pi_1\
      api_1+(1-b)pi_2 &= pi_2\
      bpi_2+(1-c)pi_3&=pi_3\
      pi_1+pi_2+pi_3&=1end{align}$$



      ofcourse adding the last equation since the elements in a probability vector must sum to $1$. Rearranging a bit I get



      begin{align}-api_1+cpi_3&=0\api_1-bpi_2&=0\bpi_2-cpi_3&=0\pi_1+pi_2+pi_3&=1,end{align}



      adding (2) to (1) and then (3) to (1) leaves me with only the last equation. I'm confused. I also tried using the RREF algorithm, to no success.






      linear-algebra




      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 13 at 22:05









      ParsevalParseval

      2,9171719




      2,9171719






















          2 Answers
          2






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          0












          $begingroup$

          Notice that $vec{pi}P = vec{pi} iff vec{pi}left(P - Iright)= vec{0}$ and evaluate $vec{pi}P$:



          $
          begin{bmatrix}
          pi_1 & pi_2 & pi_3
          end{bmatrix}
          begin{bmatrix}
          -a & a & 0\
          0 & -b & b\
          c & 0 & -c
          end{bmatrix}
          =
          begin{bmatrix}
          -api_1 + cpi_3 & api_1 - bpi_2 & bpi_2 - cpi_3
          end{bmatrix}
          $          .



          Now use that $$vec{pi}left(P - Iright)= vec{0} iff
          begin{bmatrix}
          -api_1 + cpi_3\ api_1 - bpi_2\ bpi_2 - cpi_3
          end{bmatrix} =
          begin{bmatrix}
          0\ 0\ 0
          end{bmatrix}.$$



          Adding the first equation to the second we get the third and we can thus eliminate it:



          $$vec{pi}left(P - Iright)= vec{0} iff
          begin{bmatrix}
          -api_1 + cpi_3\ api_1 - bpi_2
          end{bmatrix} =
          begin{bmatrix}
          0\ 0
          end{bmatrix}.$$



          Next let's replace $pi_3$, our free variable, with the more neutral variable $x$:



          $$begin{bmatrix}
          -api_1 + cx\ api_1 - bpi_2
          end{bmatrix} =
          begin{bmatrix}
          0\ 0
          end{bmatrix}.$$



          Lastly, solve: $pi_1 = frac{c}{a}x$ and $pi_2 = frac{a}{b}pi_1 = frac{c}{b}x$. Thus, your general solution is
          $$vec{pi} = xbegin{bmatrix}c/a & c/b & 1end{bmatrix}.$$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Stationary distributions are left eigenvectors of $1$. Thus, they are elements of the null space of $$P^T-I = begin{bmatrix}-a&0&c \ a&-b&0 \ 0&b&-c end{bmatrix}.$$ Row-reducing this matrix results in $$begin{bmatrix}1&0&-frac ca\0&1&-frac cb \ 0&0&0end{bmatrix}$$ from which we can read that the null space is spanned by $(c/a,c/b,1)$. To get the stationary distribution $vecpi$, normalize this vector by dividing by the sum of its elements.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              active

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              2 Answers
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              0












              $begingroup$

              Notice that $vec{pi}P = vec{pi} iff vec{pi}left(P - Iright)= vec{0}$ and evaluate $vec{pi}P$:



              $
              begin{bmatrix}
              pi_1 & pi_2 & pi_3
              end{bmatrix}
              begin{bmatrix}
              -a & a & 0\
              0 & -b & b\
              c & 0 & -c
              end{bmatrix}
              =
              begin{bmatrix}
              -api_1 + cpi_3 & api_1 - bpi_2 & bpi_2 - cpi_3
              end{bmatrix}
              $              .



              Now use that $$vec{pi}left(P - Iright)= vec{0} iff
              begin{bmatrix}
              -api_1 + cpi_3\ api_1 - bpi_2\ bpi_2 - cpi_3
              end{bmatrix} =
              begin{bmatrix}
              0\ 0\ 0
              end{bmatrix}.$$



              Adding the first equation to the second we get the third and we can thus eliminate it:



              $$vec{pi}left(P - Iright)= vec{0} iff
              begin{bmatrix}
              -api_1 + cpi_3\ api_1 - bpi_2
              end{bmatrix} =
              begin{bmatrix}
              0\ 0
              end{bmatrix}.$$



              Next let's replace $pi_3$, our free variable, with the more neutral variable $x$:



              $$begin{bmatrix}
              -api_1 + cx\ api_1 - bpi_2
              end{bmatrix} =
              begin{bmatrix}
              0\ 0
              end{bmatrix}.$$



              Lastly, solve: $pi_1 = frac{c}{a}x$ and $pi_2 = frac{a}{b}pi_1 = frac{c}{b}x$. Thus, your general solution is
              $$vec{pi} = xbegin{bmatrix}c/a & c/b & 1end{bmatrix}.$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Notice that $vec{pi}P = vec{pi} iff vec{pi}left(P - Iright)= vec{0}$ and evaluate $vec{pi}P$:



                $
                begin{bmatrix}
                pi_1 & pi_2 & pi_3
                end{bmatrix}
                begin{bmatrix}
                -a & a & 0\
                0 & -b & b\
                c & 0 & -c
                end{bmatrix}
                =
                begin{bmatrix}
                -api_1 + cpi_3 & api_1 - bpi_2 & bpi_2 - cpi_3
                end{bmatrix}
                $                .



                Now use that $$vec{pi}left(P - Iright)= vec{0} iff
                begin{bmatrix}
                -api_1 + cpi_3\ api_1 - bpi_2\ bpi_2 - cpi_3
                end{bmatrix} =
                begin{bmatrix}
                0\ 0\ 0
                end{bmatrix}.$$



                Adding the first equation to the second we get the third and we can thus eliminate it:



                $$vec{pi}left(P - Iright)= vec{0} iff
                begin{bmatrix}
                -api_1 + cpi_3\ api_1 - bpi_2
                end{bmatrix} =
                begin{bmatrix}
                0\ 0
                end{bmatrix}.$$



                Next let's replace $pi_3$, our free variable, with the more neutral variable $x$:



                $$begin{bmatrix}
                -api_1 + cx\ api_1 - bpi_2
                end{bmatrix} =
                begin{bmatrix}
                0\ 0
                end{bmatrix}.$$



                Lastly, solve: $pi_1 = frac{c}{a}x$ and $pi_2 = frac{a}{b}pi_1 = frac{c}{b}x$. Thus, your general solution is
                $$vec{pi} = xbegin{bmatrix}c/a & c/b & 1end{bmatrix}.$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Notice that $vec{pi}P = vec{pi} iff vec{pi}left(P - Iright)= vec{0}$ and evaluate $vec{pi}P$:



                  $
                  begin{bmatrix}
                  pi_1 & pi_2 & pi_3
                  end{bmatrix}
                  begin{bmatrix}
                  -a & a & 0\
                  0 & -b & b\
                  c & 0 & -c
                  end{bmatrix}
                  =
                  begin{bmatrix}
                  -api_1 + cpi_3 & api_1 - bpi_2 & bpi_2 - cpi_3
                  end{bmatrix}
                  $                  .



                  Now use that $$vec{pi}left(P - Iright)= vec{0} iff
                  begin{bmatrix}
                  -api_1 + cpi_3\ api_1 - bpi_2\ bpi_2 - cpi_3
                  end{bmatrix} =
                  begin{bmatrix}
                  0\ 0\ 0
                  end{bmatrix}.$$



                  Adding the first equation to the second we get the third and we can thus eliminate it:



                  $$vec{pi}left(P - Iright)= vec{0} iff
                  begin{bmatrix}
                  -api_1 + cpi_3\ api_1 - bpi_2
                  end{bmatrix} =
                  begin{bmatrix}
                  0\ 0
                  end{bmatrix}.$$



                  Next let's replace $pi_3$, our free variable, with the more neutral variable $x$:



                  $$begin{bmatrix}
                  -api_1 + cx\ api_1 - bpi_2
                  end{bmatrix} =
                  begin{bmatrix}
                  0\ 0
                  end{bmatrix}.$$



                  Lastly, solve: $pi_1 = frac{c}{a}x$ and $pi_2 = frac{a}{b}pi_1 = frac{c}{b}x$. Thus, your general solution is
                  $$vec{pi} = xbegin{bmatrix}c/a & c/b & 1end{bmatrix}.$$






                  share|cite|improve this answer









                  $endgroup$



                  Notice that $vec{pi}P = vec{pi} iff vec{pi}left(P - Iright)= vec{0}$ and evaluate $vec{pi}P$:



                  $
                  begin{bmatrix}
                  pi_1 & pi_2 & pi_3
                  end{bmatrix}
                  begin{bmatrix}
                  -a & a & 0\
                  0 & -b & b\
                  c & 0 & -c
                  end{bmatrix}
                  =
                  begin{bmatrix}
                  -api_1 + cpi_3 & api_1 - bpi_2 & bpi_2 - cpi_3
                  end{bmatrix}
                  $                  .



                  Now use that $$vec{pi}left(P - Iright)= vec{0} iff
                  begin{bmatrix}
                  -api_1 + cpi_3\ api_1 - bpi_2\ bpi_2 - cpi_3
                  end{bmatrix} =
                  begin{bmatrix}
                  0\ 0\ 0
                  end{bmatrix}.$$



                  Adding the first equation to the second we get the third and we can thus eliminate it:



                  $$vec{pi}left(P - Iright)= vec{0} iff
                  begin{bmatrix}
                  -api_1 + cpi_3\ api_1 - bpi_2
                  end{bmatrix} =
                  begin{bmatrix}
                  0\ 0
                  end{bmatrix}.$$



                  Next let's replace $pi_3$, our free variable, with the more neutral variable $x$:



                  $$begin{bmatrix}
                  -api_1 + cx\ api_1 - bpi_2
                  end{bmatrix} =
                  begin{bmatrix}
                  0\ 0
                  end{bmatrix}.$$



                  Lastly, solve: $pi_1 = frac{c}{a}x$ and $pi_2 = frac{a}{b}pi_1 = frac{c}{b}x$. Thus, your general solution is
                  $$vec{pi} = xbegin{bmatrix}c/a & c/b & 1end{bmatrix}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 13 at 23:49









                  PiKindOfGuyPiKindOfGuy

                  18611




                  18611























                      0












                      $begingroup$

                      Stationary distributions are left eigenvectors of $1$. Thus, they are elements of the null space of $$P^T-I = begin{bmatrix}-a&0&c \ a&-b&0 \ 0&b&-c end{bmatrix}.$$ Row-reducing this matrix results in $$begin{bmatrix}1&0&-frac ca\0&1&-frac cb \ 0&0&0end{bmatrix}$$ from which we can read that the null space is spanned by $(c/a,c/b,1)$. To get the stationary distribution $vecpi$, normalize this vector by dividing by the sum of its elements.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Stationary distributions are left eigenvectors of $1$. Thus, they are elements of the null space of $$P^T-I = begin{bmatrix}-a&0&c \ a&-b&0 \ 0&b&-c end{bmatrix}.$$ Row-reducing this matrix results in $$begin{bmatrix}1&0&-frac ca\0&1&-frac cb \ 0&0&0end{bmatrix}$$ from which we can read that the null space is spanned by $(c/a,c/b,1)$. To get the stationary distribution $vecpi$, normalize this vector by dividing by the sum of its elements.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Stationary distributions are left eigenvectors of $1$. Thus, they are elements of the null space of $$P^T-I = begin{bmatrix}-a&0&c \ a&-b&0 \ 0&b&-c end{bmatrix}.$$ Row-reducing this matrix results in $$begin{bmatrix}1&0&-frac ca\0&1&-frac cb \ 0&0&0end{bmatrix}$$ from which we can read that the null space is spanned by $(c/a,c/b,1)$. To get the stationary distribution $vecpi$, normalize this vector by dividing by the sum of its elements.






                          share|cite|improve this answer









                          $endgroup$



                          Stationary distributions are left eigenvectors of $1$. Thus, they are elements of the null space of $$P^T-I = begin{bmatrix}-a&0&c \ a&-b&0 \ 0&b&-c end{bmatrix}.$$ Row-reducing this matrix results in $$begin{bmatrix}1&0&-frac ca\0&1&-frac cb \ 0&0&0end{bmatrix}$$ from which we can read that the null space is spanned by $(c/a,c/b,1)$. To get the stationary distribution $vecpi$, normalize this vector by dividing by the sum of its elements.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 14 at 1:03









                          amdamd

                          30.2k21050




                          30.2k21050






























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