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Assume that we are dealing with an $\textrm{ARMA}(1,1)$ model: $$ y_{t} = \theta y_{t-1} + \epsilon_{t} + \alpha \epsilon_{t-1} $$ where $$ \epsilon_{t} \sim\textrm{ WN}(0, \sigma^{2}) $$ Then, we can rewrite the model il lag polynomial: $$ (1-\theta L)y_{t}= (1+\alpha L)\epsilon_{t} $$ from which $$ y_{t} = \frac{1}{1-\theta L} (1+\alpha L)\epsilon_{t} $$ and if $\theta$ = 1 the process is obviously not invertible and we cannot take the expectation of $y_{t}$. However, in some lecture notes I found a random walk process (that is not stationary) written as follows: $$ y_{t} = y_{t-1} + \epsilon_{t} = \sum_{i=1}^{t} \epsilon_{i} + y_{0} $$ from which: $ E[y_{t}]=y_{0} $.

Probably I'm missing something.

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  • $\begingroup$ Could you state a question? You don't really seem to be concerned about existence of an expectation per se, but rather appear to be surprised by something in the six formulas you have exhibited--but what is it? $\endgroup$
    – whuber
    May 6, 2020 at 16:37

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You are right, the random walk with no drift has mean zero, or the starting value if such a value is given. As you said, the random walk is just the sum of i.i.d. random variables $\epsilon_t$ with mean zero and variance $\sigma^2$. As such, if $y_t$ is a random walk then it can be written as $y_t = \sum_{i=1}^t \epsilon_i$, so in fact $y_t\sim N(0,t\sigma^2)$. In your case, if $y_0$ is fixed then $y_t \sim N(y_0,t\sigma^2)$.

Note that even if the process has the same mean throughout, the variance dependes on the time, so it is not even weakly stationary. A process being nonstationary does not mean you cannot take expectations or other kind of operators, just that the (unconditional) distribution changes over time.

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    $\begingroup$ What Jesus says is true ( and I could be getting slightly pedantic ) but there really is no long term unconditional mean in the series with a unit root because the effect of a shock on the value of the response is permanent. It never washes away over time so the expectation of the process is always the current value of the process. So it's not constant but it does exist. Conversely, in the AR(1) ( which is stationary ), the effect of the shock eventually washes away over time so that the long term expectation of process really is zero. $\endgroup$
    – mlofton
    Mar 4, 2023 at 11:40

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