# autoregressive time series issue understanding expectation

Consider the following question How is it obvious that $$\mathbb{E}(X_t) = 0$$? Is it because of the following? Recursively we find that $$\mathbb{E}(X_t) = \phi \mathbb{E}(X_{t-1}) = \phi^2 \mathbb{E}(X_{t-2}) = \phi^{t-1} \mathbb{E}(X_{1})$$. So now we see that for $$\mathbb{E}(X_t) = \mathbb{E}(X_s)$$ for all $$s (a requirement for stationarity) either $$\phi=1$$ or $$\mathbb{E}(X_t) = 0$$ we cannot have the former since $$|\phi|<1$$ thus we have $$\mathbb{E}(X_t) = 0$$.

• Use stationarity and linearity of expectation to compute $0=E[Z_t] = E[X_t-\phi X_{t-1}]$ and solve for $E[X_t].$ (To guarantee a solution you need $\phi\ne 1$ and $|\phi|\gt 1$ is inconsistent with stationarity: that's where the restriction on $\phi$ comes from.) The appeal to recursion falls apart because there is no starting point: $t$ is an arbitrary integer, not just a natural number. – whuber Aug 30 '19 at 13:18
• notice the very important hint from @whuber when it comes to $| \phi | > 1$ that would be inconsistent with stationarity too.. very often this is not well reported in textbooks.. but it is important. – Fr1 Aug 30 '19 at 13:47

## 1 Answer

You explanation seems me admissible. However a more convincing one come from a generalization like this:

$$X_t = \phi_0 + \phi_1 X_{t-1} + Z_t$$

it is possible to demonstrate that

$$E[X_t] = \phi_0 / (1 - \phi_1)$$

than if $$\phi_0 = 0$$ we have that $$E[X_t] = 0$$