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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 $k > -1$.

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if $k=0$, you have a term $y_{-1}$, so for $k=1$ the smallest term is $y_0$. Do you have any additional condition on coefficients? – Alex Nov 1 '12 at 10:28
up vote 0 down vote accepted

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.

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Thank you for your answer "did". Does your solution solve also the problem with $y_0 = 1$? The difference equation doesn't satisfy $y_1 \in [0,1]$ if we use $y_{-2} = \frac{1}{3}$. But, you have pointed out that we don't need to use $y_{-2}$, so the problem of satisfying $y_1 \in [0,1]$ is solved. Am I correct? – Gollum Nov 1 '12 at 14:23
As I said, $y_0=1$ is a condition from the start (hence what is the problem with it is not clear to me). For the rest, see the edited version of my answer. – Did Nov 1 '12 at 14:56

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 $y_2$.

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I hope it is clearer now! – Gollum Nov 1 '12 at 11:56

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