ARIMA(0, 1, 1) process cant be stationary?

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.)
• 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 $\theta_1$? – Frank Feb 15 at 6:33
• 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. – whuber Feb 15 at 15:31
• @whuber Not sure what you mean : $\frac{1-\theta B}{1-B} = 1 + \frac{B-\theta B}{1-B} = 1 + \frac{(1-\theta)B}{1-B} = 1 + (1-\theta) B + \frac{(1-\theta)B - [(1-\theta)B (1-B)]}{1-B} = 1 + (1-\theta) B + \frac{(1-\theta)B^2}{1-B}...$ – Frank Feb 15 at 18:10
• $$\frac{1-\theta B}{1-B} = (1-\theta B)(1+B+B^2+\cdots+B^n+\cdots) = 1+(1-\theta)B+(1-\theta)B^2+\cdots$$ implying $\psi_j=1-\theta$ for $j\ge1,$ whence $\sum_{j\ge 1}|\psi_j| = 1/\theta$ provided $0\lt \theta \lt 2.$ – whuber Feb 15 at 19:22
• @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 $2$? – Frank Feb 15 at 22:23