Stochastic MatricesDate: 03/17/98 at 09:46:14 From: Wayland Subject: eigenvectors Hi: In an earlier answer you responded to a student's question about eigenvectors. In stochastic hydrology there is an equation that is used: p(t+1) = p(t)T where p(t+1) and p(t) are nx1 probability row vectors at two different time periods t and t+1, and T is a called a transitional probability matrix (nxn) whose rows sum to one. As t approaches infinity, p(t+1) approaches p(t). According to Eigenvector theory, the eigenvalue of this equation should then be one and solving for T should yield the identity matrix. It is true that the equation is satisfied for the identity matrix, but for other T matrices it converges to a unique vector that depends on the matrix, but not on p(0). I have tried to solve for this vector but when I put in the conditions that the probabilities sum to one, the problem is overspecified. Please help me with this dilemma. Thanks Wayland Date: 03/17/98 at 15:36:07 From: Doctor Anthony Subject: Re: eigenvectors I have worked through an example and hope it answers your queries on stochastic matrices. Below is an example of the distribution of trade between depots. The service specialist of Metrobug Heating and Air Conditioning Company make their calls with a fleet of radio-dispatched service trucks. Established procedure is that when a call requesting service is received by the office, the dispatcher sends the next available truck to respond to the call. One of the service specialists keeps records of his service calls, and his data are summarize in the table below. Formulate this situation. Find the transition diagram and the one-step transition matrix. --------------------------------------------------------------------- District of District of Percent of Current Call Next Call Calls --------------------------------------------------------------------- East East 50 Central 40 West 10 --------------------------------------------------------------------- Central East 10 Central 60 West 30 --------------------------------------------------------------------- West East 30 Central 60 West 10 --------------------------------------------------------------------- It is more usual to write the transition matrix with columns adding to 1 rather than rows in the way you have shown it. This allows you to operate on a vector representing the present state as a product M.x where x is the vector of present position. Example below. East Central West East [.5 .1 .3][1] [.5] Central |.4 .6 .6||0| = |.4| West [.1 .3 .1][0] [.1] If we take the stationary vector, that is we find the vector that does not change any more, we have: M[a] [a] |b| = |b| [c] [c] we get the vector [ 9/38] |21/38] [ 8/38] This vector is found by solving the three equations below for a, b, c. a(.5-1) + .1b + .3c = 0 .4a + b(.6-1) + .6c = 0 .1a + .3b + c(.1-1) = 0 we get a = 9/38, b = 21/38, c = 8/38 This is in fact a quick way to find M^(infinity) as you will see below. This will always work for a stochastic matrix but is obviously not a general method for powers of matrices. So, over the long term, East gets 9/38 of the trade, Central gets 21/38 of the trade and West gets 8/38 of the trade. If you raised M to an infinite power you will find that it settles down to the matrix shown below, and ANY starting vector will end up with the vector shown on the right hand side. [9/38 9/38 9/38][1/2] [ 9/38] |21/38 21/38 21/38||1/4| = |21/38] [8/38 8/38 8/38][1/4] [ 8/38] This shows that after an infinite application of the matrix every starting vector will end up with the same distribution. The columns of M^(infinity) are all the same and are given by the vector: [a] |b| [c] -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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