According to "Time Series Analysis, Forecasting and Control" (Box, Jenkins), for any process represented by a linear filter $$ z_t = a_t + \psi_1 a_{t-1} + \psi_2 a_{t-2} ... = a_t + \sum_{J=1}^\infty \psi_j a_{t-j} $$ be a valid stationary process, it is necessary for the coefficients $\psi_j$ to be absolutely summable: $\sum_{J=1}^\infty |\psi_j| < \infty$.

For an ARIMA(0, 1, 1) process, $$ (1-B)Y_t = (1-\theta_1 B)a_t\\ Y_t = \frac{(1-\theta_1B)}{(1-B)}a_t = [1 + (1-\theta_1)B + (1-\theta_1)B^2 + (1-\theta_1)B^3 ...] \cdot a_t. $$

Thus $\psi_j = (1-\theta_1)$ for all $j$.

Since, I think, $\sum_{j=1} ^\infty |\psi_j|= \infty$, doesn't this mean that an ARIMA(0, 1, 1) process can't be stationary?


You cannot manipulate the lag operator $B$ as you did when there is unit root. Box and Jenkins should tell you that. $\theta_1$ should be $\theta_1 B$, in your notation.

(The heuristic $$ ``\frac{1}{1-B} = \sum_{h \geq 0} B^h" $$ that your heuristic calculation is based on is incorrect. The precise reason---see, e.g. Box and Jenkins---is that, for a complex polynomial $f(z)$, $\frac{1}{f(z)}$ admit a power series representation $$ \frac{1}{f(z)} = \sum_{h \geq 0} \psi_h z^h $$ on an open neighborhood in the complex plane if and only if the roots of $f(z)$ lie strictly outside the unit circle.)

As to your question, yes, ARIMA(0,1,1) can be stationary. For example, take a white noise. Its first difference is an MA(1) series. Integrate this over-differenced MA(1) series gives the original white noise. (Indeed, if you take the first difference of any stationary series and integrate it, you recover the stationary series.)

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  • $\begingroup$ Thanks for catching the $B$ error. But I left $\theta_1$ arbitrary, so in the above situation, even if there is no unit root to the characteristic polynomial, wouldn't the infinite sum of $\psi_i$ still go to infinity for any $\theta_1$? $\endgroup$ – Frank Feb 15 at 6:33
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    $\begingroup$ Your algebra is still incorrect, Frank: every "$1-\theta_1$" in your final expression for $Y_t$ should be $1-\theta_1B.$ You will need to collect like powers of $B$ in order to determine the $\psi_j$ correctly. $\endgroup$ – whuber Feb 15 at 15:31
  • $\begingroup$ @whuber Not sure what you mean : $\frac{1-\theta B}{1-B} = 1 + \frac{B-\theta B}{1-B} = 1 + \frac{(1-\theta)B}{1-B} = 1 + (1-\theta) B + \frac{(1-\theta)B - [(1-\theta)B (1-B)]}{1-B} = 1 + (1-\theta) B + \frac{(1-\theta)B^2}{1-B}...$ $\endgroup$ – Frank Feb 15 at 18:10
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    $\begingroup$ $$\frac{1-\theta B}{1-B} = (1-\theta B)(1+B+B^2+\cdots+B^n+\cdots) = 1+(1-\theta)B+(1-\theta)B^2+\cdots$$ implying $\psi_j=1-\theta$ for $j\ge1,$ whence $\sum_{j\ge 1}|\psi_j| = 1/\theta$ provided $0\lt \theta \lt 2.$ $\endgroup$ – whuber Feb 15 at 19:22
  • $\begingroup$ @whuber, Thank you for the reply. Take $\theta = 1/2$, then $\sum_{j \geq 1} |1-\frac{1}{2}| = \sum_{j \geq 1} \frac{1}{2}$. You're saying the infinite sum of $\frac{1}{2} = \frac{1}{1/2} = 2$ ? For any $n$, $\sum_1^n \frac{1}{2} = \frac{n}{2}$. You're saying this converges to $2$? $\endgroup$ – Frank Feb 15 at 22:23

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