Find the Limit.? What is the limit of
$Y_{n} := \frac{1}{3}Y_{n-1} + \frac{2}{3}Y_{n-2} $
for $n > 2$ and $Y_{1}<Y_{2}$.
I have tried solving, but the limit tends to exist between $y_{1}$ and $y_{2}$ and hence I am not able to evaluate what would be the ratio between which they would lie?
 A: This is a linear difference equation. The general solution for the recursion is
$$
Y_n = \alpha (-2/3)^n + \beta,
$$
where $\alpha, \beta$ are computed from the initial conditions, in this case $Y_1, Y_2$.
$$
Y_1 = -\frac 23 \alpha  + \beta, \quad Y_2 = \frac 49 \alpha + \beta
$$
i.e. $\alpha = \frac{9}{10}(Y_2-Y_1)$ and $\beta = \frac 25 Y_1 + \frac 35 Y_2$. So,
$$
Y_n =  \frac{9}{10}(Y_2-Y_1) (-2/3)^n + \frac 25 Y_1 + \frac 35 Y_2$$
The limit is therefore $ \frac 25 Y_1 + \frac 35 Y_2$, regardless of having $Y_1<Y_2$ or not.
A: $Y_{n} := \frac{1}{3}Y_{n-1} + \frac{2}{3}Y_{n-2}$ is homogenous linear diference equation which has characteristic equation $x^2=\frac{1}{3}x+\frac{2}{3}$ or $3x^2-x-2=0$, $(3x+2)(x-1)=0$. Solution is $$Y_{n}=C_1(-\frac{2}{3})^n+C_2(1)^n=C_1(-\frac{2}{3})^n+C_2.$$
Constants $C_1$, $C_2$ can be found from $Y_1, Y_2$:
$$-\frac{2}{3}C_1+C_2=Y_1$$
$$\frac{4}{9}C_1+C_2=Y_2$$
from where $C_1=\frac{9}{10}(Y_2-Y_1)$ and $C_2=\frac{2}{5}Y_1+\frac{3}{5}Y_2$ and
$$Y_n=\frac{9}{10}(-\frac{2}{3})^n(Y_2-Y_1)+\frac{2}{5}Y_1+\frac{3}{5}Y_2.$$
Therefore, $$\lim_{n\to\infty} Y_n=\frac{2}{5}Y_1+\frac{3}{5}Y_2.$$
A: Let me show you the general method for dealing with these linear recurrences. The idea is to convert the linear recurrence into a matrix recurrence: $v_{n} = Av_{n-1}$ so that $v_n = A^nv_0$.
Here, you have
$$ \begin{pmatrix} Y_n \\ Y_{n - 1} \end{pmatrix} = \begin{pmatrix} \frac13 & \frac23 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} Y_{n - 1} \\ Y_{n - 2} \end{pmatrix} $$
So by induction,
$$ \begin{pmatrix} Y_n \\ Y_{n - 1} \end{pmatrix} = \begin{pmatrix} \frac13 & \frac23 \\ 1 & 0 \end{pmatrix}^{n - 2} \begin{pmatrix} Y_{2} \\ Y_{1} \end{pmatrix}. $$
The matrix
$$ P = \begin{pmatrix} \frac13 & \frac23 \\ 1 & 0 \end{pmatrix}$$
is a regular, row-stochastic matrix. Thus by standard Markov chain theory,
$$ P^n \to \begin{pmatrix} \frac35 & \frac25 \\ \frac35 & \frac25 \end{pmatrix} $$
where $(3/5, 2/5)$ is the probability vector in the 1-eigenspace of $P^T$. So $Y_n \to \frac25Y_1 + \frac 35Y_2$.
