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I am studying the stationarity and invertibility conditions of AR processes.

Concerning the stationarity condition of the simplest form of an AR(1) process (without constant, drift). Let us say that we convert such an AR process to an infinite MA process. If we take the expected value of the new form of the process (the infinite MA) then it is going to be zero since the error term is zero mean white noise so its mean value is 0. So an infinite sum of zeros will be zero (maybe its here that i am wrong). Also its variance is finite (sigma squared) so if we take the variance of the MA process it will be sigma squared divided by one minus phi squared(as an infinite sum of terms of a geometric progress). So since we can prove that the process has constant mean and finite variance without any condition why Diebold says that in order for the AR(1) process to be covariance stationary the AR coefficient should be less than one in absolute value? To be more specific, Diebold says that the MA representation of for y is convergent if and only if phis is less than 1 in absolute value. Thus this is the condition for covariance stationarity in the simple AR(1) case (without constant, drift). The only though that i have is that the sum of infinite zeros cannot be defined so this is why we need the MA form of the AR process to be convergent. And if it is convergent (if phi less than one in absolute value) then the summation of zeros will be finite and hence the expected value of the series will be constant (0).

Is this the answer or it is something else?

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The problem is that your statement about the variance of the MA process isn't correct. If the AR coefficient of an AR(1) process, say, $\theta$, is equal to or greater than one, the error terms in the infinite series are multiplied by an increasing and unbounded sequence of numbers (as you go back in time), and consequently the variance of the sum is infinite. With $\theta < 1$, the terms in the infinite series are multiplied by a sequence that goes to zero fast enough so that the variance of the sum is finite. With $\theta = 1$, you have an unweighted sum of an infinite number of terms with a constant variance, so the variance of the sum is infinite.

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