First-Order Linear Difference Equation with Constraint 
Consider the following first order linear difference equation for $y$:
$$y_{t+1} = \alpha * y_{t} + \beta * x_{t-n+1} ~~\forall t \ge n$$
For initial conditions, one could assume that $x_{i} \in \mathbb{R} ~~\forall t = 1, \ldots n$. (except that they cannot all be zero).
The other constraint is that $$y_{n} = \sum_{i=0}^{n-1} x_{n-i}$$ so that the $n$th value of $y$ in the sequence always sums to the previous $n$ values of $x_{i}$.

For example:
$y_{n} = x_{n} + x_{n-1} \ldots + x_{1}$ and process starts at time t = n because y needs to have a value for it to start.
So, for example, if $n=30$, then $y_{30}$ is the first observation for $y$ so the process starts at time t = n = 30. Note that the $x_{t}$ process starts at time $t = 1$ so we need the $x_{i} \forall i= 1,\ldots 30$ before the y process can start.
Similarly, once t is greater than or equal to n, we have
$y_{n+1} = x_{n+1} + x_{n} + \ldots x_{2}$
$y_{n+2} = x_{n+2} + x_{n+1} + \ldots x_{3}$
$\vdots$
Given the assumptions and the constraint, is there a relation between  $\lambda$ and $\beta$ which meets the requirements?
I have no background in difference equations (I do have the book by Mann which talks about such a difference equation but not with that constraint) so if someone knows where an explanation for this type of thing is explained, it would be appreciated.
Essentially, once $t = n$, then, starting from there, the sum constraint needs to hold and the difference equation has to hold also. I'm pretty sure that the answer is that $\beta = -\alpha$ so that the equation reduces to
$$ y_{t+1} = \alpha * (y_{t}  -  x_{t-n+1}) ~~\forall t \ge n$$
and then, in order to satisfy the sum constraint,  $\alpha =  (\frac{1}{2})^{1/n}$. But how to prove this is a different story.
 A: If I understood correctly, the terms in your post can be summarized as
$$
\left\{ \matrix{
  y_{\,t}  - y_{\,t - 1}  = \left( {1 - \lambda } \right)\left( {x_{\,t - 1}  - y_{\,t - 1} } \right)\quad \left| {\;n \le \forall t} \right. \hfill \cr 
  y_{\,t}  = \sum\limits_{1\, \le \,k\, \le \,t} {x_{\,k} }  \hfill \cr}  \right.
$$
Now, the starting point for $t$ is not clear. Supposing it is $0$ then:
$$
\left\{ \matrix{
  y_{\,0}  = 0 \hfill \cr 
  y_{\,1}  = x_{\,1}  \hfill \cr 
  \quad  \vdots  \hfill \cr 
  y_{\,n - 1}  = \sum\limits_{1\, \le \,k\, \le \,n - 1} {x_{\,k} }  \hfill \cr 
  \left\{ \matrix{
  y_n  - y_{\,n - 1}  = \left( {1 - \lambda } \right)\left( {x_{\,n - 1}  - y_{\,n - 1} } \right) \hfill \cr 
  y_{\,n}  = \sum\limits_{1\, \le \,k\, \le \,n} {x_{\,k} }  \hfill \cr}  \right. \hfill \cr 
  \left\{ \matrix{
  y_{\,t}  - y_{\,t - 1}  = \left( {1 - \lambda } \right)\left( {x_{\,t - 1}  - y_{\,t - 1} } \right)\quad  \hfill \cr 
  y_{\,t}  = \sum\limits_{1\, \le \,k\, \le \,t} {x_{\,k} }  \hfill \cr}  \right.\left| {\;n < t} \right. \hfill \cr}  \right.
$$
that is
$$
\left\{ \matrix{
  y_{\,0}  = 0 \hfill \cr 
  y_{\,1}  = x_{\,1}  \hfill \cr 
  \quad  \vdots  \hfill \cr 
  y_{\,n - 1}  = \sum\limits_{1\, \le \,k\, \le \,n - 1} {x_{\,k} }  \hfill \cr 
  \left\{ \matrix{
  y_n  - y_{\,n - 1}  = x_{\,n}  = \left( {1 - \lambda } \right)\left( {x_{\,n - 1}  - y_{\,n - 1} } \right) =  - \left( {1 - \lambda } \right)\left( {\sum\limits_{1\, \le \,k\, \le \,n - 2} {x_{\,k} } } \right) \hfill \cr 
  y_{\,n}  = \sum\limits_{1\, \le \,k\, \le \,n} {x_{\,k} }  \hfill \cr}  \right. \hfill \cr 
  \left\{ \matrix{
  x_{\,t}  =  - \left( {1 - \lambda } \right)\left( {\sum\limits_{1\, \le \,k\, \le \,t - 2} {x_{\,k} } } \right) \hfill \cr 
  y_t  - y_{\,t - 1}  =  - \left( {1 - \lambda } \right)y_{\,t - 2}  \hfill \cr}  \right.\quad \left| {\;n < t} \right. \hfill \cr}  \right.
$$
At this point, if $n=2$, we shall put $x_{\,2}  = 0,\;y_2  = y_1  = x_{\,1} $, is that compatible with your problem ?
In any case for $3 \le n$ we get separated equations for $x$ and $y$, and I do not see any bound related with $\lambda$: so probably I did not catch some point about your problem.
