Difference equation with strange initial condition I have the difference equation $y_k = \frac{2}{7} y_{k+1} + \frac{2}{7} y_{k-1} + \frac{3}{7} y_{k-2}$. The $y_k$s are probabilities. The equation doesn't hold for $k = 0$. This is because I know that $y_0 = 1$, $y_{-1} = \frac{1}{3}$ and $y_{-2} = \frac{1}{9}$ so this would imply $y_1 > 1$ (while $y_1 \in [0,1])$.  Than I'm asked to solve it for $k \geq -1$ using the initial conditions. And I get the solution $y_k = \frac{7}{9} + \frac{2}{9}\big(-\frac{1}{2} \big)^k $. For $k = 0$ this gives me $y_0 = 1$.
What confuses me is that we have solved the difference equation using an initial condition ($y_0 = 1$) that cannot be satisfied by the equation itself and we get out a solution that is valid for every $k > -1$.
 A: Sure, why not?  The equation they gave you is valid for $k \geq 1$.  They have also told you that $y_0 = 1$.
Plugging in $k = 1$, we find that
\begin{align*}
y_1 &= \alpha y_2 + \beta y_0 + \gamma y_0 \\
&= \alpha y_2 + \beta + \gamma.
\end{align*}
What other initial conditions are you given?  If we are told the value of $y_1$, then we can solve for $y_2$.
A: Call $(E_k)$ the equation $y_k=\frac27y_{k+1}+\frac27y_{k-1}+\frac37y_{k-2}$. You are asked to find a bounded solution $(y_k)_{k\geqslant-1}$ of the system $(E_k)_{k\geqslant1}$ with the initial condition $y_{-1}=\frac13$, $y_0=1$. Nowhere does this involve a condition on $y_{-2}$.
A priori,  the system $(E_k)_{k\geqslant1}$ has a unique solution $(y_k)_{k\geqslant-1}$ for every initial condition $(y_{-1},y_0,y_1)$ hence the missing initial condition $y_1$ is specified by the additional condition that $(y_k)_{k\geqslant-1}$ must stay bounded. Here is how.
As every linear system, this one can be reformulated as $Y_{k+1}=AY_k$ for every $k\geqslant1$, where $Y_k=(y_k,y_{k-1},y_{k-2})^T$, for some $3\times3$ matrix $A$, with the initial condition $Y_1=(y_1,1,\frac13)$. The eigenvalues of $A$ are $1$, $3$, and $a=-\frac12$, hence the fact that $(y_k)_{k\geqslant-1}$ stays bounded is equivalent to the fact that $Y_1$ has no component on the eigenvector of the eigenvalue $3$. Writing $Y_1$ as a linear combination of the eigenvectors $U=(1,1,1)^T$ for the eigenvalue $1$ and $V=(1,-2,4)^T$ for the eigenvalue $a$ imposes the condition $y_1=\frac23$. 
Once $y_1$ is known, $Y_1$ is known as $Y_1=(\frac23,1,\frac13)^T=\frac79U-\frac19V$ and $Y_k=A^{k-1}Y_1$ hence $y_k=(1,0,0)A^{k-1}Y_1=(1,0,0)(\frac79U-\frac19a^{k-1}V)=\frac79-\frac19a^{k-1}=\frac79+\frac29a^{k}$ $(*)$.
Naturally, $(*)$ provides the values of $y_0$ and $y_1$ we started from, namely, $y_{-1}=\frac13$ and $y_0=1$, since these were part of our initial conditions. And $(*)$ also provides the unique value of $y_1$ which makes $(y_k)_{k\geqslant-1}$ stay bounded for these $y_{-1}$ and $y_0$, namely, $y_1=\frac23$. But none of this uses $y_{-2}$ and one could very well have indicated that $y_{-2}=\frac1{\sqrt6}$ or any other value without modifying the rest.
