One solution that does not require matrix multiplication:
Remember that the steady state vector is in the nullspace of I - T.
Therefore:
require(MASS)
n1 <- Null(t(diag(nrow(T)) - T))
n1 / sum(n1)
On Wed, Dec 12, 2012 at 2:19 AM, annek <an...@ifm.liu.se> wrote:
> Hi,
> I have a transition matrix T for which I want to find the steady state
> matrix for. This could be approximated by taking T^n , for large n.
>
> T= [ 0.8797 0.0382 0.0527 0.0008
> 0.0212 0.8002 0.0041 0.0143
> 0.0981 0.0273 0.8802 0.0527
> 0.0010 0.1343 0.0630 0.9322]
>
> According to a text book I have T^200 should have reached the steady state
> L
>
> L =[0.17458813 0.17458813 0.17458813 0.17458813
> 0.05731902 0.05731902 0.05731902 0.05731902
> 0.35028624 0.35028624 0.35028624 0.35028624
> 0.44160126 0.44160126 0.44160126 0.44160126]
>
> I am addressing the problem using a for loop doing matrix multiplication
> (guess there might be better ways, please suggest) and find a steady state
> matrix after n=30. But if I run the code with n=100 or more I get "Inf" for
> all entities in L. Does anyone know why is that?
>
> The code I use look like this
> #------------------------------------
> rep<-20
>
> T <- Ttest
> for(i in 1:rep){
> print(i)
> T<-T%*%Ttest
> Ttest<-T
> }
> L<-T
> print(L)
> #----------------------------------
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