I need to solve the following recurrence equation
$p_i =\begin{cases} r(p_{i-1}+p_{i+2}) &\mbox{if } i \text{ is odd} \\ (1-r)(p_{i-1}+p_{i}) & \mbox{if } i \text{ is even} \end{cases} i \in \mathbb{N}$
This equation came up trying to find an invariant measure vector for a given stochastic matrix, in fact $0\leq r \leq 1$.
I have no clue how to solve this, any help will be great :)
If statement is correct:
$$ p_n =\begin{cases} r(p_{n-1}+p_{n+2}) &\mbox{if } n \text{ is odd} \\ (1-r)(p_{n-1}+p_{n}) & \mbox{if } n \text{ is even} \end{cases} n \in \mathbb{N} $$
then for $r\ne 0$ $$ \Downarrow $$
$$ \begin{cases} rp_{n+2} = p_{n} - rp_{n-1}, &\mbox{if } n \text{ is odd} \\ rp_{n} = (1-r)p_{n-1} & \mbox{if } n \text{ is even} \end{cases} n \in \mathbb{N}\tag{1} $$
$$ \Downarrow $$
$$ \begin{cases} rp_{n+2} = p_{n} - (1-r)p_{n-2}, &\mbox{if } n \text{ is odd} \\ rp_{n} = (1-r)p_{n-1} & \mbox{if } n \text{ is even} \end{cases} n \in \mathbb{N}\tag{2} $$
Now consider only odd $n$: $n=2k-1$. Search $p_{2k-1}$ in the form $$ p_{2k-1} = ab^{k}. $$ Then $(2) \implies$ $$ rab^{k+1} = ab^{k}-(1-r)ab^{k-1}, $$ $$ rb^{2} = b-(1-r), $$ $$ rb^{2} - b+(1-r)=0, $$ this quadratic (on $b$) equation has solutions:
$$ b=1, \qquad b=\dfrac{1-r}{r}. $$
If $b=1$, then solution of $(1)$ has form $p_{2k-1}=a, p_{2k}=\dfrac{1-r}{r}a$.
If $b=\dfrac{1-r}{r}$, then $$ p_{2k-1}=a\left(\dfrac{1-r}{r}\right)^k, $$ $$ p_{2k}=a\left(\dfrac{1-r}{r}\right)^{k+1}. $$
$a$ is any real value (if $p_1$ would defined, then $a$ would defined too).
The case $r=0$ is obvious.