I have a difference function of the form $ z_{t+1} = (1-\beta)z_t + \lambda(1-z_t) $ where $ 0 \leq z_0 \leq 1, 0 < \beta < 1 , 0 < \lambda < 1$. Find conditions such that the sequence ${z_t}$ converges in a monotone manner.
I can see that this sequence ${z_t}$ converges to $ \frac{\lambda}{\lambda + \beta}$, but how do I find conditions which guarantee that the sequence converges in a monotone manner?
@sucksatmath: The difference equation:
[1] $~~~z_{t+1} = (1-\beta) \times z_{t} + \lambda \times (1-z_{t})$ can be re-written as
[2] $~~~z_{t+1} = (1 - \beta - \lambda) \times z_{t} + \lambda$.
( The relation above is very similar to the AR(1) with non-zero mean but I won't go into that here. See Hamilton's "Time Series Analysis" for a more detailed discussion).
Now, let L denote the lag operator, so that $L(x_{t+1}) = x_{t}$. Then, the previous difference equation can be re-written as
$z_{t+1} \times (1 - (1-\beta-\lambda)L) = \lambda$
Now, in order for the series, $\{z_t\}$, to converge to something pointwise, we write $z_{t+1}$ by itself and obtain
$z_{t+1} = \frac{\lambda}{(1-(1-\beta-\lambda)L)}$
Letting $\rho = (1-\beta-\lambda)$ just to simplify the notation, we have:
$z_{t+1} = \frac{\lambda}{(1- \rho L)}$
Notice that the RHS is an infinite geometric series in $\rho$. The lag operator L does not end up playing a role because $\lambda$ is not a function of time.
So, we can re-write the infinite geometric series without the lag operator. This results in:
$z_{t+1} = \frac{\lambda}{(1-\rho)} = \lambda \sum_{i=0}^{\infty} \rho^{i}$. Obviously this converges as long as $0 \lt \rho \lt 1.0.$
In summary: Given the original conditions stated for $\lambda$ and $\beta$, if we add the condition that $ 0 \lt (1-\beta-\lambda) \lt 1.0$ then $z_{t+1}$ decreases monotonically to the limit $\frac{\lambda}{1-(1-\beta-\lambda)} = \frac{\lambda}{(\lambda + \beta)}$.
Finally, notice that the limiting value is EXACTLY what you claimed it was in your original question so I apologize for that too. I'm not clear how you obtained that but I'd be interested.
Notice that equation [2] illustrates ( "proves" if you will ) that the $z_{t}$ sequence will be monotonic decreasing as long as the condition $0 \lt (1-\beta-\lambda) \lt 1.0$ holds.