0
$\begingroup$

Consider the following question

enter image description here

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 $s<t$ (a requirement for stationarity) either $\phi=1$ or $\mathbb{E}(X_t) = 0$ we cannot have the former since $|\phi|<1$ thus we have $\mathbb{E}(X_t) = 0$.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – whuber Aug 30 at 13:18
  • $\begingroup$ 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. $\endgroup$ – Fr1 Aug 30 at 13:47
1
$\begingroup$

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 $E[X_t] = 0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.