<p>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</p>

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

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

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

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

<p>which is very different compared to the previous case.</p>

CC BY-SA 3.0
2017-05-04T15:56:31.287Z
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.
Richard Hardy
CC BY-SA 3.0
2017-05-04T16:06:52.400Z
Now I have elaborated the problem. May be now it would be easier to answer.
Abhishek Kumar
CC BY-SA 3.0
2017-05-04T16:16:50.830Z
@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).
Alecos Papadopoulos
CC BY-SA 3.0
2017-05-04T14:06:06.263Z
2017-05-04T14:06:06.263Z
Alecos Papadopoulos
<p>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 $|\rho|&lt;1$. 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 $E_t (x_{t+j} - \rho x_{t+j-1}) = x_t - \rho x_{t-1}$.</p>

<hr>

<p>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, $\lim_{j\to\infty} E_t(a_{t+j}) = \lim E_t(x_{t+j}) + \lim E_t(z_{t+j}) = \frac{1}{1-\rho} (x_t-\rho x_{t-1}) + 0$.</p>

<hr>

<p>Above, we <em>assumed</em> 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
$$
\pmatrix{x_t\\ x_{t-1}} = \pmatrix{1+\rho &amp; -\rho\\ 1 &amp; 0} \pmatrix{x_{t-1}\\ x_{t-2}} + \pmatrix{\epsilon_t\\ 0},
$$
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.)</p>

CC BY-SA 3.0
2017-05-14T03:44:27.813Z
2017-05-14T03:44:27.813Z
chan1142
<p>Assume $x_t$ is an unit root process. Write first difference of $x_t$  as</p>

<p>$\Delta x_t$ =$\rho$ $\Delta x_{t-1}$+ $\epsilon_t$</p>

<p>Where $\epsilon_t$ is a white noise process. </p>

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

<p>Suppose:</p>

<p>$a_t$= $x_t$ +$z_t$ </p>

<p>$\Delta x_t$ =$\rho$ $\Delta x_{t-1}$+ $\epsilon_t$</p>

<p>$ z_t$ =$\rho$ $z_{t-1}$+ $\eta_t$</p>

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

<econometrics><applied-econometrics>
Expected Value of a Infinite Process
Abhishek Kumar
# Expected Value of a Infinite Process

(Asked by: username_0 on 2017-05-04)

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 $j \rightarrow \infty$.

**Tags:** econometrics, applied-econometrics


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## 2 answers

### Answer by username_1 on 2017-05-04. Score: 1

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.

> **Comments:**
> **username_2**: 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.  
> **username_0**: Now I have elaborated the problem. May be now it would be easier to answer.  
> **username_1**: @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).  

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### Answer by username_3 on 2017-05-14. Score: 0

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 $E\_t (x\_{t+j} - \rho x\_{t+j-1}) = x\_t - \rho x\_{t-1}$.

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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, $\lim\_{j\to\infty} E\_t(a\_{t+j}) = \lim E\_t(x\_{t+j}) + \lim E\_t(z\_{t+j}) = \frac{1}{1-\rho} (x\_t-\rho x\_{t-1}) + 0$.

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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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