Does $x_{t} = 1-x_{t-1}$ have a stable steady state solution?

Question

At steady state, $x = x_{t} = x_{t-1}$. So I can solve for the steady state value of $x=0.5$.

The general rule of determining the stability of the steady state is that the $|\text{slope}|<1$. But here slope = 1, so can I say that the steady state is unstable?

Graphically the picture looks like this

I have a picture of an oscillation motion around the steady state. But will it converge to the steady state and will it be stable?

Answer 1

The solutions of

$x_t = 1 - x_{t - 1} \tag 1$

are of two kinds, depending on whether or not there exists an $x_t$ such that

$x_t = \dfrac{1}{2} \tag 2$

or not. If (1) holds for some $t$, then it is easy to see that

$\forall t, \; x_t = \dfrac{1}{2}; \tag 2$

if, on the other hand,

$\exists t, \; x_t \ne \dfrac{1}{2} , \tag 3$

then since

$x_{t + 1} = 1 - x_t, \tag 4$

we have

$x_{t + 2} = 1 - (1 - x_t) = x_t, \tag 5$

and every solution is of period $2$. In this case, the oscillating solutions are in fact stable, alternating as they do 'twixt two values, one greater and the other less than $1/2$.

It is worth observing in this context that the somewhat more general dynamic

$x_t = 1 - kx_{t - 1}, \; k \in \Bbb R, \tag 6$

satisfies

$x_t - x_{t - 1} = (1 - kx_{t - 1}) - (1 - kx_{t - 2}) = k(x_{t - 2} - x_{t - 1}), \tag 7$

and has a fixed point satisfying

$x = 1 - kx, \tag 8$

that is,

$x = \dfrac{1}{k + 1}, \; k \ne -1; \tag 9$

also, (7) yields

$\vert x_t - x_{t - 1} \vert = \vert k \vert \vert x_{t - 1} - x_{t - 2} \vert, \tag{10}$

so the distance 'twixt successive iterates grows or shrinks according to whether $\vert k \vert$ is larger or smaller than $1$; furthermore it is easy to see that

$x_t - x = k(x - x_{t - 1}), \tag{11}$

$\vert x_t - x \vert = \vert k \vert \vert x - x_{t - 1} \vert, \tag{12}$

which show the $x_t$ move farther or closer to the fixed point, again as $\vert k \vert > < 1$; from (11);we see that the $x_t$ either oscillate around $x$ or converge/diverge directly to/from $x$, depending on the sign of $k$. When $k = -1$ we find

$x_t = 1 + x_{t - 1}; \tag{13}$

the iterates march off to $+\infty$; there is no equilibrium state. The stability here is thus determined by the magnitude of $\vert k \vert$ relative to $1$.

A rather cute example, this, illustrating stable, decaying, or growing oscillation as well as monotonic decay to or growth from equilibrium, or no equilibrium at all.