Let $X_t$ be an ARIMA(1,1,1) process and $Y_t = Y_{t-1} + X_t$. What kind of process is $Y_t$?

Q: Let $$X_t$$ be an ARIMA(1,1,1) process and $$Y_t = Y_{t-1} + X_t$$. What kind of process is $$Y_t$$?

$$X_t$$ 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

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

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

The process $$Y_t$$ is the integration of $$X_t$$

$$Y_t = \sum_{-\infty}^t X_t$$

So you have one more integration step and that means that it is an ARIMA(1,2,1) process.

Indeed, the recursive formula can also be rewritten as the recursive function of an ARMA(3,1) process, as you deduced, but it won't be a stable ARMA process. (see for more about that here: ARMA vs ARIMA Models)

• Why can the OP not difference the series twice and then estimate an ARMA(1,1) on that series ? Where does the instability and lack of solution arise ? Thanks. Commented Nov 10, 2023 at 17:19
• @mlofton sure, taking twice the difference $Z_t := (Y_{t} - Y_{t-1}) - (Y_{t-1} - Y_{t-2})$ is an ARMA(1,1) process, but $Y_t$ is not that ARMA(1,1) process in the same way as a random walk is not equivalent to Gaussian white noise (instead it is the sum of that noise). Commented Nov 10, 2023 at 17:30
• Right. It's not. But as far as estimation, can one not take $Z_t$ and estimate the parameters of the ARMA(1,1) process ? Maybe I didn't understand your comment about stable process and no solution. Thanks. Commented Nov 11, 2023 at 2:19
• @mlofton If I have a random walk with the given recursive formula $X_t = X_{t-1} + \epsilon$ then, conditional on the parameters, you do not have a unique distribution for the vector $\mathbf{X}$ (you would need some anchor point as well, e.g. given that $X_0 = 0$, you can describe the distribution of $X_t$ for $t \geq 0$). I am not sure, but I believe that you can always (without problems) solve the parameters for the recursive formula by using a least squares formulation. But it is not possible to solve the equations with maximum likelihood / method of moments. Commented Nov 11, 2023 at 7:43
• And besides using least squares, the parameters can also be 'solved'/found by differencing twice and finding the parameters for the ARMA(1,1) process (as you mentioned in the first comment). I have removed the 'has no solution' part which has been confusing. You do have a solution for finding the parameters in the problem... but, it is in finding the likelihood function that there is no solution. Commented Nov 11, 2023 at 7:52