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- Apr 14, 2013

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We consider the equation \begin{equation*}u_{k+1}=\begin{pmatrix}a & b \\ 1-a & 1-b\end{pmatrix}u_k \ \text{ with } \ u_0=\begin{pmatrix}1 \\1 \end{pmatrix}\end{equation*}

For which values of $a$ and $b$ is the above equation a Markov process?

Calculate $u_k$ as a function of $a,b$.

Which conditions do $a,b$ have to satisfy so that $u_k$ approximates a finite limit as $k\rightarrow \infty$ and which is that limit?

I have done the following:

So that the equation is a Markov process the element of the matrix must be non-negativ and the sum of the colums must be $1$.

So, we get $0\leq a,b\leq 1$.

\begin{align*}&u_0=\begin{pmatrix}1 \\ 1\end{pmatrix} \\ &u_1=\begin{pmatrix}a+b \\ 2-(a+b)\end{pmatrix} \\ &u_2=\begin{pmatrix}a^2+2b-b^2 \\ 2-(a^2+2b-b^2)\end{pmatrix}\\ &u_3=\begin{pmatrix}a^3+2ab-b^2a+2b-a^2b-2b^2+b^3 \\ 2-(a^3+2ab-b^2a+2b-a^2b-2b^2+b^3)\end{pmatrix} \\ & \ldots\end{align*}

Let $u_k=(x_k,y_k)^T$ then it holds that \begin{align*}&x_k+y_k=2 \\ & x_{k+1}=(a−b)x_k+2b \\ & x0=y0=1\end{align*}

We have that \begin{align*} x_{k+1}&=(a−b)x_k+2b \\ & =(a-b)^2x_{k-1}+2b(1-b)+2b \\ & = (a-b)^3x_{k-2}+2b(a-b)^2+2b(a-b)+2b \\ & =\ldots \\ & = (a-b)^{k+1}x_0+2b\left (\sum_{i=0}^k(a-b)^i\right )\end{align*}

Therefore, we get \begin{equation*}u_k=\begin{pmatrix}x_k \\ y_k\end{pmatrix}=\begin{pmatrix}(a-b)^{k+1}x_0+2b\left (\sum_{i=0}^k(a-b)^i\right )\\ 2-(a-b)^{k+1}x_0+2b\left (\sum_{i=0}^k(a-b)^i\right )\end{pmatrix} \end{equation*}

So that $u_k$ approximates a finite limit as $k\rightarrow \infty$ it must be $0<a-b<1$.

Is everything correct?