# Does the unconditional mean of a non stationary ARMA process exist?

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.

• 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?
– whuber
May 6, 2020 at 16:37

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 $$y_t \sim N(y_0,t\sigma^2)$$.