Causal unstable system turn into stable anticausal? I would appreciate it very much if someone would be able to provide some clarity, help or comment on this problem.
I have been reading several papers on time series identification such as


https://www.jstor.org/stable/24305996
https://www.jstor.org/stable/24307363
https://www.jstor.org/stable/24308842


and system identification 


https://doi.org/10.1016/S1474-6670(17)56646-X
https://doi.org/10.1109/78.258141
https://doi.org/10.1109/9.827371


among several others.

The question is, given an unstable causal system, is it possible to turn it into an anticausal but stable system in order to simulate it? (assuming that I have the complete input sequence).

I think it might have to do with the time reversal property of the Z-Transform, i.e. 

$\mathcal{Z} \left\{ x[n] \right \} = X(z)$ with it's ROC given by $r_2&lt;|z|&lt;r_1$, then

$\mathcal{Z} \left\{ x[-n] \right \} = X(z^{-1})$ with $\frac{1}{r_1}&lt;|z|&lt;\frac{1}{r_2}$

however i can't figure it out.

In the Forssell and Hjalmarsson paper says that backward filtering is common in off-line signal processing applications, like the zero-phase filtering (filtfilt in Matlab) and the implementation of non-causal Wiener filters. The relevant (at least for my question) equations are the (31), (32) and (33).

To illustrate the problem, please consider the following system given by

$y[n] = \frac{B(z)}{A(z)} u[n] = \frac{(z-0.1)}{(z-0.5)(z-5)} u[n]$;

assume that the complete input sequence $\left\{ u[n] \right\}^{N}_{n=0}$ is known (available).

Then, it is possible to separate the system as follows

$y[n] = \frac{B(z)}{A_s(z)} \frac{1}{A_a(z)} u[n],$

where $A_s(z)$ and $A_a(z)$ represents the stable and anti-stable monic parts of $A(z)$ respectively. Thus we can obtain the causal part of $y_{\text{C}}[n]$ by computing

$y_{\text{C}}[n] = \frac{B(z)}{A_s(z)} u[n] = \frac{(z-0.1)}{(z-0.5)} u[n]$

followed by

$y[n] = \frac{1}{A_a(z)} y_{\text{C}}[n] = \frac{1}{(z-5)} y_{\text{C}}[n]$

So here is my problem. The approach given in Forssell and Hjalmarsson is the following

in order to obtain $y[n]$ backward-filter the reversed sequence given by $\left\{ y_{\text{C}}^R[n] \right\}_{n=0}^{N}$ through the filter $\frac{1}{A_a(z)}$ as follows

$y^R[n] = \frac{1}{A_a(z)} y_{\text{C}}^R[n]$

the desired signal $y[n]$ is then finally obtained by reversing the sequence $\left\{ y^R[n] \right\}_{n=0}^{N}$

I apologize in advance for the length of the question and the orthographic mistakes.
Any comments, suggestions and help will be greatly appreciated. Thanks
An unstable system can be stablized by all-pass decomposition for exampe, but simply constructing the anti-causal impulse respone extension will not provide the same output; so I think it's not possible, but I'm not rigorous at this point. Following just shows why.

Assume that your unstable and causal IIR filter has the impulse response $h_+[n]$ and by evaluating it into $ n &lt; 0$ domain you construct a stable but anti-causal impulse response denoted by $h_{-}[n]$.

Your true output given by the unstable filtering is 
$$ y[n] = h_{+}[n] \star x[n]$$

Now you construct alternate output from a reversed input and the anti-causal filter as
$$ y_{-}[n] = h_{-}[n] \star x[-n] $$

And by reversing $y_{-}[n]$ , do you get $y[n]$ ? 

No as seen:

$$y_{-}[-n] = h_{-}[-n] \star x[n] \neq h_+[n] \star x[n] $$

that's to say unless $h_{-}[-n] = h_+[n]$ , they wont be the same. And indeed the reversed anti-causal impuse response will not be equal to the causal one, since one is stable and the other is not.