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Assume $x_t$ is an unit root process. Write first difference of $x_t$ as

$\Delta x_t$ =$\rho$ $\Delta x_{t-1}$+ $\epsilon_t$

Where $\epsilon_t$ is a white noise process.

How can we calculate $ E_t [x_{t+j}]$ when $j \rightarrow \infty$.

Suppose:

$a_t$= $x_t$ +$z_t$

$\Delta x_t$ =$\rho$ $\Delta x_{t-1}$+ $\epsilon_t$

$ z_t$ =$\rho$ $z_{t-1}$+ $\eta_t$

How can we calculate $ E_t [a_{t+j}]$ when $j \rightarrow \infty$.

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You are asking about the expected value of the leading term of a sequence. There can be no general answer given the assumptions you make, since for example

$$x_{t+1} = x_{t} + v_{t+1}, \;\;\;E(v_{t+1}) = 0,\;\;\; x_0 \;\; \text{given}$$

is a unit root process, and its expected value is equal to $E(x_{t+1})=x_0,\;\;\; \forall t$, while

$$x_{t+1} = a +x_{t} + v_{t+1}, \;\;\;E(v_{t+1}) = 0,\;\;\; x_0 \;\; \text{given}$$

is also a unit-root process, but $E(x_{t+1}) = x_0 + a\cdot(t+1)$

which is very different compared to the previous case.

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  • $\begingroup$ I think the process class given by the OP is definite enough to give a more definite answer. Your second example does not seem to fit in the process class given by the OP. $\endgroup$ – Richard Hardy May 4 '17 at 15:56
  • $\begingroup$ Now I have elaborated the problem. May be now it would be easier to answer. $\endgroup$ – Abhishek Kumar May 4 '17 at 16:06
  • $\begingroup$ @AbhishekKumar Any autoregressive process without constant terms or deterministic trends or purely exogenous non-zero mean variables, is eventually a linear function of its building block only, which is assumed here to be white noise processes, and so... (Also, in your elaboration I do not see any indication that we have a unit root here -which in any case, would not change the result -unit roots affect the variance in the standard case, not the expected value). $\endgroup$ – Alecos Papadopoulos May 4 '17 at 16:16
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The question is about conditional expectation (expectation given information at time $t$). You have $z_{t+j} = \rho^j z_t + \sum_{k=0}^{j-1} \rho^k \eta_{t+j-k}$, and thus $E_t(z_{t+j}) = \rho^j z_t \to 0$ as $j\to\infty$ if $|\rho|<1$. The equation for $x_t$ is more complicated, but it can be written as $(x_t - \rho x_{t-1}) = (x_{t-1} - \rho x_{t-2}) + \epsilon_t$. That is, $x_t - \rho x_{t-1}$ is a random walk. When $w_t$ is a random walk $E_t(w_{t+j}) = w_t$. Substituting $w_t = x_t - \rho x_{t-1}$, you have $E_t (x_{t+j} - \rho x_{t+j-1}) = x_t - \rho x_{t-1}$.


Now, if $E_t(x_{t+j})$ converges as $j\to\infty$, then $\lim_{j\to\infty} E_t(x_{t+j}) = \lim_{j\to\infty} E_t (x_{t+j-1})$, and thus $$ \lim_{j\to\infty} E_t(x_{t+j} - \rho x_{t+j-1}) = (1-\rho) \lim_{j\to\infty} E_t(x_{t+j}) = x_t - \rho x_{t-1}. $$ As a result, $\lim_{j\to\infty} E_t(a_{t+j}) = \lim E_t(x_{t+j}) + \lim E_t(z_{t+j}) = \frac{1}{1-\rho} (x_t-\rho x_{t-1}) + 0$.


Above, we assumed that $E_t(x_{t+j})$ converges as $j\to\infty$. Now we have to show (or check) it. The simplest way would be to write $$ \pmatrix{x_t\\ x_{t-1}} = \pmatrix{1+\rho & -\rho\\ 1 & 0} \pmatrix{x_{t-1}\\ x_{t-2}} + \pmatrix{\epsilon_t\\ 0}, $$ that is $W_t = A W_{t-1} + \xi_t$, where $W_t = (x_t, x_{t-1})'$, $\xi_t = (\epsilon_t,0)'$ and $A$ is the $2\times 2$ matrix on the right-hand side. You then have $W_{t+j} = A^j W_t + \sum_{k=0}^{j-1} A^k \xi_{t+j-k}$, from which $E_t (W_{t+j}) = A^j W_t$. Thus, $E_t(x_{t+j})$ converges if $A^j$ converges. The two eigenvalues of $A$ are 1 and $\rho$, which are real and no greater than unity in magnitude. And the associated two eigenvectors are both real (one proportional to $(1,1)'$ and the other to $(\rho,1)'$). These suffice. (Because $A = V\Lambda V^{-1}$, where $\Lambda$ is the diagonal matrix of 1 and $\rho$, and $V$ is the matrix of eigenvectors, we have $A^j = V \Lambda^j V^{-1}$, which converges as $j\to\infty$ because $\Lambda^j$ converges.)

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