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 October 29th, 2013, 07:09 PM #1 Newbie   Joined: Sep 2013 Posts: 23 Thanks: 0 Steady state vector for the Markov Matrix I have to find the steady state vector for matrix is M= [.5 .8 .5 .2] Would the vector [.8 .5] be in the steady state because [.5 .8 .5 .2] times x = x so the system reduces down to -.5x+.8y=0 ?
October 30th, 2013, 01:20 PM   #2
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Re: Steady state vector for the Markov Matrix

Quote:
 Originally Posted by 84grandmarquis I have to find the steady state vector for matrix is M= [.5 .8 .5 .2] Would the vector [.8 .5] be in the steady state because [.5 .8 .5 .2] times x = x so the system reduces down to -.5x+.8y=0 ?
You are correct. However
Quote:
 so the system reduces down to -.5x+.8y=0
is confusing - what are x and y? You said x = vector (.8,.5).

 October 31st, 2013, 11:48 AM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Steady state vector for the Markov Matrix Don't use "x" to represent a vector and then to represent one of the components of the vector! Yes, if $\begin{bmatrix}x \\ y \end{bmatrix}$ is a "steady state vector" for this matrix, we must have $\begin{bmatrix} .5=&. 8 \\ .5=&.2 \end{bmatrix]\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}.5x+ .8y \\ .5x+ .2y\end{bmatrix}= \begin{bmatrix}x \\ y \end{bmatrix}$ which gives .5x+ .8y= x, .5x+ .2y= y -.5x+ .8y= 0, .5x- .8y= 0. Those are just multiples of each other so as long as $y= \frac{.5}{.8}x= 0.625x$, it is a steady state vector.

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