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Q: If $X_t$ is an ARIMA(1,1,1) process, what is $Y_t = Y_{t-1} + X_t$?

Attempted solution: $X_t$ is an ARIMA(1,1,1), i.e $\nabla X_t = X_t - X_{t-1} = Z_t $ where $Z_t$ is a casual ARMA(1,1) process and satisfies $(1-\phi_1 B)Z_t = (1+\theta_1B)\epsilon_t. $ Since $Z_t$ is casual, we may write $Z_t = \frac{1 + \theta_1 B}{1-\phi_1 B} \epsilon_t $.

Then $X_t = \frac{1}{1-B} \frac{1 + \theta_1 B}{1-\phi_1 B} \epsilon_t$, and

\begin{equation} \begin{split} (1-B)Y_t &= X_t \\ (1-B)Y_t &= \frac{1}{1-B} \frac{1 + \theta_1 B}{1-\phi_1 B} \epsilon_t \\ (1-B)^2(1-\phi_1 B)Y_t &= (1+\theta_1B)\epsilon_t \end{split} \end{equation}

So $Y_t$ is an ARMA(3,1) process?

Oskar
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  • From what you've written, isn't $Y_t$ an $\text{ARIMA}(1,2,1)$ process? I mean, $Y_t$ has to be differenced once to get $X_t$, which has to be differenced once to get $Z_t$ which is a causal $\text{ARMA}(1,1)$ process. So $Y_t$ is differenced twice to get an ARMA process, i.e. the index of integration equals $2$. – Sarvesh Ravichandran Iyer Jun 01 '23 at 14:57
  • Makes sense. How do you formulate it mathematically? – Oskar Jun 01 '23 at 15:08
  • To me at least, the "mathematical" formulation is that an ARIMA process is given by the last differencing equation which you wrote. Namely, if you see here at the last line of the section "Definition", there is an exact formula for the difference equation satisfied by an ARIMA$(p,d,q)$, which is exactly what you wrote with $p=1,d=2,q=1$. – Sarvesh Ravichandran Iyer Jun 01 '23 at 15:31

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