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Eigen vector problem

  • 25-08-2009 9:26pm
    #1
    [Deleted User]
    Posts: 5,589 ✭✭✭


    Right, this is driving me up the walls...

    0.981 , 0.019
    0.019 , 0.981

    is a transition matrix, I want to get to the eigenvalues and vectors.
    Now I'm getting that 1 and 0.962 are the eigenvalues and that
    -0.70711 , 0.70711
    0.70711 , 0.70711
    are the eigenvalues, but those don't make sense economically (at least I don't think they do)
    as

    (p(a)) = c1 (-0.70711).(1) + c2 (0.70711).962
    (p(b)) = ....(.70711).........+.....(0.70711)

    leads to a p(a) = p(b) = .5, which is not correct


Comments

  • #2
    [Deleted User]
    Posts: 5,589 ✭✭✭ [Deleted User]


    Its grand.
    Transition matrix was wrong. Stupid decimal places... :(


  • #3
    alan4cult
    Registered Users, Registered Users 2 Posts: 1,583 ✭✭✭alan4cult


    If you don't mind me asking what were you using this matrix for? What are p(a) and p(b)? Is it economics?


  • #4
    [Deleted User]
    Posts: 5,589 ✭✭✭ [Deleted User]


    Yeah, its to calculate the expected probabilitiy of a future state.


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