Determine order of an ARIMA process

Question

My problem: ${X_t}$ is a stationary process where ${X_t={\phi}X_{t-1}+Z_{t}+Z_{t-2} }$ with $Z_{t}$ being the error term aka white noise(0,$\sigma^2$). We are given the process ${Y_t=Y_{t-1}+X_{t}-{\phi^2}X_{t-2}}$. We know that ${Y_t}$ is an $ARIMA(p,d,q)$ process and we are asked to identify $(p,d,q)$. \
My attempt has led to an $ARIMA(0,1,3)$ process and my solution is the following: Since ${X_t}$ is clearly an $ARMA(1,2)$ process, I start by differentiating the equation in order to get eventually an expression for ${\phi^2}X_{t-2}$. Differentiation gives: ${X_t={2\phi}X_{t-1}-{\phi^2}X_{t-2}+Z_{t}+{\theta}Z_{t-1}+Z_{t-2}+{\theta}Z_{t-3} }$\
Now I substitute ${X_t}$ with ${X_t={\phi}X_{t-1}+Z_{t}+Z_{t-2} }$ and plug it in the differentiated one and finally get a better expression for ${\phi^2}X_{t-2}$, where ${\phi^2}X_{t-2}={\phi}X_{t-1}+{\theta}Z_{t-1}+{\theta}Z_{t-3}$ \
Plugging it into ${Y_t=Y_{t-1}+X_{t}-{\phi^2}X_{t-2}}$ and knowing that ${X_t={\phi}X_{t-1}+Z_{t}+Z_{t-2} }$ finally gives: ${Y_t=Y_{t-1}+Z_{t}-\theta}Z_{t-1}+{Z_{t-2}-{\theta}Z_{t-3}}$ aka the $ARIMA(1,0,3)$. Is this the right solution? Did I miss something? I feel like a more compact solution with not so many substitutions could be achieved. I would appreciate your help!

Answer 1

I was mistaken in terms of what I said about adding the noise terms. It's allowed but, for this problem, it's best to keep them seperate because the goal is to end up with an ARIMA model so that we can identify $p$, $d$, and $q$. So, here is what I did. We have two equations.

(1) $X_t = \phi X_{t-1} + Z_t + Z_{t-2}$

(2) $Y_t = Y_{t-1} + X_t - \phi^2 X_{t-2}$

So, since 2) does not contain an error term and has $X$ in it, we want to eliminate $X$ from the second equation so that we can be left with $Y$ and error terms ( which of course are the $Z$ ). We don't want the $X_t$ in the second equation because then we don't have an ARIMA model. ( that would be called an auto-regressive distributed lag model ).

So, the first step is to write $X_t$ in terms of the $Z$ in 1). Hopefully, you are familiar with the lag operator, $L$. This is an operator that can be used to do algebra with time series. It is defined so that, $L(x_t) = x_{t-1}$ which is why it is called the lag operator.

In our case, 1) implies that

(3) $(1-\phi L) X_t = Z_t + Z_{t-2}$ so that $X_t = \frac{Z_{t}}{1-\phi L} + \frac{Z_{t-2}}{1 - \phi L}$

Now, since we have $X_t$ as a function of the $Z_t$, we can substitute for $X_t$ in 2) which leads to ( after some algebra )

(4) $(1 - \phi L) (Y_{t} - Y_{t-1}) = Z_{t} + Z_{t-2} - \phi^2(Z_{t-2} + Z_{t-4})$

Simplifying gives,

(5) $ (1 - \phi L) (Y_{t} - Y_{t-1}) = Z_{t} + (1-\phi^2)Z_{t-2} - \phi^2 Z_{t-4}$

Giving each MA coefficient on the RHS seperate parameters, $\phi_{i}$, results in an ARIMA(p, d, q) with constraints on the MA coefficients:

(6) $(1 - \phi L) (Y_{t} - Y_{t-1}) = Z_{t} + \phi_{1} Z_{t-1} + \phi_2 Z_{t-2} + \phi_3 Z_{t-3} - \phi_4 Z_{t-4}$

Note that, in the ARIMA model above, p = 1, d = 1 and q = 4 with the MA parameter constraints $\phi_1 = 0$, $\phi_3 = 0$, $\phi_2 = (1 - \phi^2)$ and $\phi_4 = \phi_2 - 1$. Note that I left the AR parameter as $\phi$ so that i is seperated from the MA parameters. Also, recall that $Z_t \sim N(0, \sigma^2)$.

If you have any questions, just let me know. Also, go over the algebra because it's quite possible that I could have made a mistake somewhere but the idea is correct. Lag operators are covered in some reasonable manner in hamilton, box and jenkins, granger and newbold and any other decent time series text.