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I would just like to check my understanding for this question: Given that you have a generated ARMA(p,q) model, can an ARIMA(p,1,q) model be created?

My thought was that it would be that this is true, but only provided that the AR characteristic polynomial has at a root that is equal to 1.

Also, I know that if this was the other way around, that is, an ARIMA(p,1,q) implies a ARMA(p,q) model, that statement is true as well.

Any comments for the above that could supplement my understanding of this?

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  • $\begingroup$ I was going through my old answers and noticed this one was not accepted. Do you perhaps need further clarification? $\endgroup$ – Richard Hardy Feb 12 '17 at 13:01
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Here is an example for Richard's general statement:

Consider the process $$Y_t=1/2Y_{t-1}+1/2Y_{t-2}+\epsilon_t+1/3\epsilon_{t-1}.$$

Clearly, the process is an $ARMA(2,1)$ process. To check whether it is an $ARIMA(1,1,1)$ we need to look at the $AR$ polynomial $\phi(z)=1-1/2z-1/2z^2$. Calculating the roots of this polynomial, we can write this as $\phi(z)=(1-z)(1+1/2z)$ so that indeed $d=1$.

Therefore, we can write $\phi(L)Y_t=\theta(L)\epsilon_t$ as $$(1+1/2L)\Delta Y_t=\theta(L)\epsilon_t\ \text{or}\ \Delta Y_t=-1/2\Delta Y_{t-1}+\epsilon_t+1/3\epsilon_{t-1},$$ an $ARMA(1,1)$ in $\Delta Y_t$.

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ARMA($p,q$) can be written in term of polynomials of lag operators as

$$ \Phi(L) y_t = \Theta(L) \varepsilon_t $$.

where $\Phi(L)$ is a polynomial in $L$ of order $p$ and $\Theta(L)$ is a polynomial in $L$ of order $q$.
If $\Phi(L)$ has a unit root, then you may write

$$ \Phi'(L)(1-L)y_t = \Theta(L) \varepsilon_t $$

for $\Phi'(L):=\frac{\Phi(L)}{1-L}$ where $\Phi'(L)$ is a polynomial in $L$ of order $p-1$.

Since $(1-L)y_t=y_t-y_{t-1}=\Delta y_t$, you see that the process is an ARIMA($p-1,1,q$) process. Similarly, if there are $n$ unit roots in place of just one, you have

$$ \Phi''(L)(1-L)^n y_t = \Theta(L) \varepsilon_t $$

for $\Phi''(L):=\frac{\Phi(L)}{(1-L)^n}$ where $\Phi''(L)$ is a polynomial in $L$ of order $p-n$ (and $n \leqslant p$).

Since $(1-L)^n y_t=\Delta^n y_t$, you see that the process is an ARIMA($p-n,n,q$) process.

Hence the answer to your question: you cannot write ARMA($p,q$) as ARIMA($p,1,q$) -- but if the AR part has a unit root, then you can write ARMA($p,q$) as ARIMA($p-1,1,q$). More generally, if there are $n$ unit roots in the AR part, you can write ARMA($p,q$) as ARIMA($p-n,n,q$). However, in presence of unit roots one should use the form ARIMA($p-n,n,q$) rather than ARMA($p,q$).

Similarly, you cannot write an ARIMA($p,1,q$) as ARMA($p,q$) -- but you can write it as ARMA($p+1,q$).

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I think when you decided choose ARMA(p,q) model for your time-series data, the data itself is stationary. If it not, it should be differenced number of times to become stationary. The number is d value, and this procedure is called integration. Then you apply ARMA(p, q) to the series which had been already differenced. Consequently, you have ARIMA(p,d,q).

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  • $\begingroup$ Differencing is not called "integration", it is the opposite of it. $\endgroup$ – Richard Hardy Feb 12 '17 at 13:00

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