I want to solve the following recurrences system:
$$p_t=\frac{n-2}{2(n-1)}p_{t-1}+\frac{1}{2(n-1)}q_{t-1}+\frac{1}{2}r_{t-1}$$ $$q_t=\frac{1}{2}p_{t-1}+\frac{1}{2}s_{t-1}$$ $$r_t=\frac{n-2}{2(n-1)}r_{t-1}+\frac{1}{2(n-1)}s_{n-1}+\frac{1}{2}p_{t-1}$$ $$s_t=\frac{1}{2}r_{t-1}+\frac{1}{2}q_{t-1}$$
with the following initial conditions:
$p_0=0, p_1=\frac{1}{2(n-1)}$
$q_0=1, q_1=0$
$r_0=0, r_1=0$
$s_0=0,s_1=\frac{1}{2}$
Is there any intelligent way to tackle this problem?
Thanks!
This recursion can be solved in closed form by the following technique:
Start from the defining equations for $\{p_k,q_k,r_k,s_k\}$. Work instead (trick 1) with variables $$ m_k = p_k +r_k \\n_k =p_k -r_k \\ f_k = q_k+s_k \\ g_k = q_k-s_k $$ so that if we obtain closed forms for the new variables, $p_k = \frac12(m_k+n_k)$ and so forth. Call these four new recursions the $mnfg$ recursions.
Simultaneously solve (trick 2) the second and fourth equations (that came from the $q_k$ and $s_k$ recursions) for $f_{k-1}$ and $g_{k-1}$. Use those values of $f_{k-1}$ and $g_{k-1}$ to eliminate those variables in the first and third equations, which now involve various indices of $m_{k-i}$ and $m_{k-i}$ but no longer have any $f$ and $g$ variables except $f_k$ and $g_k$.
Simultaneously solve those new first and third equations for $f_k$ and $g_k$, which gives $$ f_k = (2-n)m_{k-1} + (n-1) m_k\\ g_k = (1-n)n_k $$ Now comes trick 3: replace $k$ with $k-1$ in those $f_k,g_k$ relations to acquire the relations $$ f_{k-1} = (2-n)m_{k-2} + (n-1) m_{k-1}\\ g_{k-1} = (1-n)n_{k-1} $$ Now you can use those $f$ and $g$ expressions to eliminate all the $f$ and $g$ variables from the $mnfg$ recursions. The fourth of those new recursions (which we will cal the $mn$ recursions) reads $$ (3n-4)m_{k-1} = (n-2) m_{k-2}+(2n-2)m_k +n\,n_{k-1} + (2n-2)n_k $$ Solve this for $n_{k-1}$ and substitute that into all of the $mn$ recursions; of course the fourth one becomes an identity, but actually, the first one also is identically satisfied with this expression for $n_{k-1}$, and the second and third recursions are just multiples of each other. That tells us we have made a wise choice for the new variables.
We are left with one second order recursion in $m_k$: $$ m_k = \frac{3n-4}{2n-2} m_{k-1} - \frac{n-2}{2n-2} m_{k-2} $$
This is solved by the usual technique (trick 4) of assuming a solution in the form $m_k= C\alpha^K$; the recursion then gives $$ \alpha^2 - \frac{3n-4}{2n-2}\alpha + \frac{n-2}{2n-2} = 0 $$ which has two solutions, $\alpha = 1$, $\alpha = \frac{n-2}{2n-2}$. So $$ m_k = C + D \left(\frac{n-2}{2n-2}\right)^k $$
Applying the initial conditions we fix $C$ and $D$: $$ m_k = \frac1n \left[ 1 - \left(\frac{n-2}{2n-2}\right)^k \right] $$
Next, we have $f_k$ in terms of $m_k$, and this becomes $$ f_k= \frac1n\left[ 1+(n-1)\left(\frac{n-2}{2n-2}\right)^k\right] $$ To find $n_k$ it is best to go back to the $mnfg$ recursions, applying the close forms we now have for $m_k$ and $f_k$. We get four identical equations, which boil down to $$ n_k = -\frac{n}{2n-2}n_{k-1} $$ The solution, respecting the initial value for $n_1$ is $$ n_k = -\frac1n\left(\frac{n}{2-2n}\right)^k $$ (This form is is not valid for $k=0$ but it is for all $k>0$.)
$g_k$ is easy once we have $n_k$; a bit of simplification gives:
$$ g_k = -2^{-k}\left(\frac{n}{1-n}\right)^{k-1} $$
So the final answer for $\{p_k,q_k,r_k,s_k\}$, valid for all $k>0$, is: $$ p_k = \frac12\left( \frac1n \left[ 1 - \left(\frac{n-2}{2n-2}\right)^k \right] - \frac1n\left(\frac{n}{2-2n}\right)^k \right) \\ q_k = \frac12\left( \frac1n\left[ 1+(n-1)\left(\frac{n-2}{2n-2}\right)^k\right] -2^{-k}\left(\frac{n}{1-n}\right)^{k-1}\right)\\ r_k = \frac12\left( \frac1n \left[ 1 - \left(\frac{n-2}{2n-2}\right)^k \right] + \frac1n\left(\frac{n}{2-2n}\right)^k \right) \\ s_k = \frac12\left( \frac1n\left[ 1+(n-1)\left(\frac{n-2}{2n-2}\right)^k\right] +2^{-k}\left(\frac{n}{1-n}\right)^{k-1}\right) $$
Failing any easier solution, you can follow the method I outlined in this answer. It will be a little cumbersome doing the calculations but Mathematica can find the diagonalization for the transition matrix
$$\begin{pmatrix}a & b & 0.5 & 0.5\\0.5 & 0 & 0 & 0.5\\0.5 &0 & a & b\\0 & 0.5 & 0.5 & 0\end{pmatrix} $$
where $a = \frac{n-2}{2(n-1)}$ and $b= \frac{1}{2(n-1)}$.