# All ARIMA Processes unit root?

My question is are all ARIMA processes also unit root processes? My guess is yes because $\{X_t\}$ is ARIMA(p, d, q) if

$(1-B)^dX_t = a(B)\epsilon_t$

is stationary ARMA(p, q).

The characteristic function for $(1-B)^dX_t = a(B)\epsilon_t$ is $(1-B)^d$, which clearly is unit root.

Also, I suppose not all unit root processes are ARIMA? Is there a good counterexample?

My third and final question is are all FARIMA processes stationary? My guess is not because it really depends on the roots of the characteristic equation. I am asking this because I see FARIMA often used as an example for a "stationary long memory" process.

• @Taylor Why not? For example, if $d=1$ then $(1-B)^dX_t = a(B)\epsilon_t \Rightarrow X_t = X_{t-1} + a(B)\epsilon_t$. Commented Mar 28, 2017 at 2:05
• also, $(1-B)^dX_t = a(B)\epsilon_t$ is IMA(d,q) or ARMA(1,q), but it is not ARIMA(p,d,q) if $p \ne 0$. Commented Mar 28, 2017 at 2:30
• no, you can't "double-count" it's lag-1 dependence Commented Mar 28, 2017 at 2:38
• yes, and you could change the if to an iff. Commented Mar 28, 2017 at 3:07
• Regarding FARIMA (or ARFIMA), stationarity depends on the order of integration. If I remember correctly, when the order is up to 0.5, it is stationary, when above, then nonstationary. But you may double-check that. Commented Mar 28, 2017 at 6:24

If $X_t$ is ARIMA(p,d,0) with $p,d > 0$, then yes, it is also a unit root process. If your AR model with a unit root is $$\Phi_1(B)X_t = \epsilon_t,$$ then you can factor your characteristic polynomial $$\Phi_1(B) = \Phi_2(B)(1-B)^d.$$ That's what a unit root is. It's just polynomials.
The reason this is more confusing than it should be is because when we say "Let $X_t$ be ARIMA", we're not saying what the restrictions on p,d and q are, and we're not saying what the restrictions on the coefficients are.