Question about ARMA process: Changing a Stochastic Process into a Transfer Function

[e0ne199]

TL;DR Summary

I want to change arma process into transfer function, I have done it but I am unsure if it is correct or not

hello everyone, I have a question about stochastic process (ARMA process) that looks like this :

I would like to change it into a transfer function, so the final result looks like this :

My question is, is this equation correct? if it is not correct, what should I change for this equation? any response is really appreciated, thx.

 

[FactChecker]

I think that's not exactly right, but it's a good try. The right side has single-step signal delays of 1, 2, and 3. So your powers of $z$ (on one side or the other) are off by 1. (I am assuming that the index in your variables are related to time such that k-1 is a step backward in time, a one step delay, $1/z$)
I am not an expert at this, but here is what I would do. I prefer to work with the $z^{-1}$ single-step delay, rather than the $z$ look into the future, but that is just a matter of preference. I would get:
$W(z)(1-1.08z^{-1}+0.81z^{-2})=V(z)(z^{-1}+0.1z^{-2}-0.9z^{-3})$
So $W(z)/V(z) = (z^2+0.1z-0.9)/(z^3-1.08z^2+0.81z)$.

 

[e0ne199]

I think that's not exactly right, but it's a good try. The right side has single-step signal delays of 1, 2, and 3. So your powers of $z$ (on one side or the other) are off by 1. (I am assuming that the index in your variables are related to time such that k-1 is a step backward in time, a one step delay, $1/z$)
I am not an expert at this, but here is what I would do. I prefer to work with the $z^{-1}$ single-step delay, rather than the $z$ look into the future, but that is just a matter of preference. I would get:
$W(z)(1-1.08z^{-1}+0.81z^{-2})=V(z)(z^{-1}+0.1z^{-2}-0.9z^{-3})$
So $W(z)/V(z) = (z^2+0.1z-0.9)/(z^3-1.08z^2+0.81z)$.

hahaha thank you, that explanation is what I have been missing until now