representing a recursive difference equation of two variables into one variable equation

Question

suppose the following recursive difference equation ($t$ is time):

$$x_t = \frac{a}{1+a}x_{t-1} + \frac{1}{1+a}x_{t+1}$$

where $0<a<1$ is assumed and all values of $a$ at past times are equally defined (by this, I mean: $x_{t-1} = \frac{a}{1+a}x_{t-2} + \frac{1}{1+a}x_{t}$) and there is no beginning time or ending time - time ranges from $-\infty$ to $\infty$.

I want to represent $x_t$ using purely $x_{t+1}$. What would the equation for $x_t$ be?

Edit:

I want to interpret the above equation as $x_t$ depending on $x_{t+1}$ and $x_{t-1}$, not the other way around.

Edit 2:

Let me rephrase my question. I now get a sort-of feeling that the above equation has to be $x_t = x_{t+1}$. Problem is, I can't prove it rigorously. Can anyone show what the rigorous proof would be?

Answer 1

What you are asking is equivalent to finding an expression for the 2nd order recursive relation $$ x_{k}=(1+a)x_{k-1}-ax_{k-2}, \quad k\in \mathbb Z, $$ This does have a general expression, and it is $$ x_k=c_1\omega_1^k+c_2\omega_2^k, $$ where $c_1,c_2$ constants and $\omega_1,\,\omega_2$ are the solutions of the quadratic equation $$ X^2-(1+a)X+a=0, $$ and these are $1$ and $a$. Hence $$ x_k=c_1+c_2 a^k, \quad \text{for all}\,\, k\in\mathbb Z. $$

Answer 2

I assume what you actually mean is find $x_t$ as a function of $t$. Here are three hints for you: $$ \text{Hint 1}:s= 1 \cdot s=\bigg(\frac{a}{1+a}+\frac{1}{1+a}\bigg)s\\ \text{Hint 2}:x_{t+1}-x_{t}=X_{t+1}\\ \text{Hint 3}:\sum_{k=1}^{t}X_{k}=x_{t}-x_{0} $$