<p>(I suspect that $E_t[x_{t+1}] = E(x_{t+1}\mid I_t)$, where $I_t$ is the Information Set, i.e. $E_t[x_{t+1}]$ is translated as "the expected value of random variable $x_{t+1}$ conditional on Information available at time $t$").  </p>

<p>@FooBar answer implemented the "standard solution algorithm", arrived at a constant solution, and remarked that such a solution violates the limiting condition.<br>
Strictly speaking, a constant solution would satisfy the limiting solution also, if it was equal to zero (which in our case implies $k=0$): If indeed we had as solution</p>

<p>$$x_t = \frac k{1-\beta}$$
then</p>

<p>$$\lim_{t \to \infty} x_t = \lim_{t \to \infty}\frac k{1-\beta} = \frac k{1-\beta} \neq 0$$</p>

<p>except if $k=0$. But this is trivial (or at least, uninteresting).</p>

<p>So I ask, <strong>if $k\neq 0$, what would it take to obtain a solution</strong>? The law of evolution must hold also at the limit so </p>

<p>$$\lim_{t \to \infty} x_t = \lim_{t \to \infty}\Big[\beta E_t[x_{t+1}] + k\Big]=0$$</p>

<p>$$\Rightarrow \lim_{t \to \infty}E_t[x_{t+1}] = -\frac k{\beta} \tag{1}$$</p>

<p>In words, as the random variable tends to zero, its expected value conditional on information available in the previous period, must equal a non-zero <em>constant</em>, while the variable itself tends to zero. Consider some cases, to see whether and when such a thing can happen:  </p>

<p><strong>A)</strong> If $x_t$ was <em>totally unobservable</em>, (not observed directly, and not amenable to indirect estimation though other variables)? In such a case, the information set would not contain any information on $x_t$, meaning that the conditional expected value would equal the unconditional one. So we would have,</p>

<p>$$\lim_{t \to \infty}E_t[x_{t+1}] = \lim_{t \to \infty}E[x_{t+1}] \tag{2}$$</p>

<p>This is economics, meaning that the properties of the mathematical entities should take into account what do they represent in the real world. So it would be meaningless <em>not</em> to assume that $x_{t+1}$ is <em>not</em> bounded (more over, this would mean that the $x$ random variable goes to infinity in finite time and then "comes back" to tend to zero in infinite time -and for those who may think "hyperinflation", hyperinflation is a <em>very very</em> small number compared to infinity). If it is bounded then the Dominated/Bounded Convergence theorem holds and we have that</p>

<p>$$\lim_{t \to \infty}E\left(x_{t+1}\right) =E\left(\lim_{t \to \infty}x_{t+1}\right) =0 \neq -\frac k{\beta} \tag{3}$$</p>

<p>So this is not a case in which we can obtain the required relation $(1)$.</p>

<p><strong>B)</strong> The limiting condition is part of the information sets for finite $t$. Then immediately  </p>

<p>$$E\left(\lim_{t \to \infty}x_{t+1}\mid I_t\right) =0$$</p>

<p>In words, at any point in time we know that the variable will tend eventually to zero. Can this be compatible with relation $(1)$? We could invoke again the boundedness of $x_t$, and a generalization of the Dominated Convergence Theorem, since here the (conditional probability) measure is also time-varying, and again obtain a zero expectation at the limit. But intuitively also, if we know that eventually the variable will tend to zero, at some point in time this must be reflected in $E_t$, most certainly at the limit. But then again, the required relation $(1)$ won't hold.</p>

<p><strong>C)</strong> We do observe perfectly or imperfectly the value of the variable, but "we don't give proper notice": our Expectations are <em>not</em> Rational. In that case, the symbol $E_t$ <em>no longer represents the conditional expectations operator</em>, but it is just a generic symbol for the "anticipation at time $t$" (in Adaptive Learning literature this is usually symbolized by $E^*_t$) and <em>it acquires the value it needs to acquire for the limiting condition to hold</em>.  In other words, <em>given</em> the law of evolution <em>and</em> the limiting condition, we <em>conclude</em> that expectations must not obey the Rational Expectations Hypothesis, <em>not even at the limit</em>.  </p>

<p>In such a case, a solution is obvious: <em>define</em> 
$$E^*_t(x_{t+1}) = -\frac {(1-c\beta^t)k}{\beta}$$</p>

<p>for some constant $c$, This expectations formation rule satisfies $(1)$, and  transforms the law of motion into</p>

<p>$$x_t = \beta \Big[-\frac {(1-c\beta^t)k}{\beta}\Big] + k = ck\beta^t$$</p>

<p>which in turn satisfies the limiting condition.</p>

<p>So we conclude that: <strong>the specific equation, with the specific limiting condition, has a solution if expectations formation does not obey the Rational Expectations Hypothesis, but instead follows the above rule, or some other to the same limiting effect.</strong></p>

CC BY-SA 3.0
2014-12-20T22:51:39.443Z
2014-12-22T16:29:49.283Z
2014-12-22T16:29:49.283Z
Alecos Papadopoulos
<p>Since there is nothing random on the right-hand side, we know that the left-hand-side must be non-random either. Hence, we can drop the expectations operator</p>

<p>$$x_t = \beta x_{t+1} + k$$
$$x_{t+1} = \beta x_{t+2} + k$$
$$x_t = \beta (\beta x_{t+2} + k) + k$$
$$x_t = k + \beta k + \beta\beta x_{t+2} $$</p>

<p>Where we can now see where this equation is leading towards:</p>

<p>$$x_t = k\sum_{s=0}^\infty \beta^s + \lim_{T\to\infty} \beta^Tx_T$$</p>

<p>where I have used that the limiting term with $T\to\infty$ will converge to zero.</p>

<p>$$x_t = \frac{1}{1-\beta} k$$</p>

<p>Which yields a constant $x$. However, this violates $lim_{t\to\infty} x_t = 0$, as $x_t$ is a constant. To conclude, the given equation has no solution.</p>

CC BY-SA 3.0
2014-12-20T09:37:40.803Z
2014-12-21T13:02:15.897Z
2014-12-21T13:02:15.897Z
FooBar
<p>As FooBar says, you can drop the expectation. Then derive the general $n$-th case for $x_{t}$ in terms of $x_{t+n}$</p>

<p>$$x_{t}=\beta^{n}x_{t+n}+k\sum_{i=0}^{n}\beta^{i}.$$</p>

<p>As $n\rightarrow \infty$, $\beta^{n} \rightarrow 0$ and (we are told) $x_{t+n} \rightarrow 0$.</p>

<p>So, using the above $$x_{t}=\lim_{n\rightarrow \infty}\left[\beta^{n}x_{t+n}+k\sum_{i=0}^{n}\beta^{i}\right]=\frac{k}{1-\beta},$$
where we have used the fact that the geometric sum converges because $0&lt;\beta&lt;1$,
$$\lim_{n\rightarrow \infty}\sum_{i=0}^{n}\beta^{i}=\frac{1}{1-\beta}.$$</p>

CC BY-SA 3.0
2014-12-20T22:53:05.343Z
I ask here too: How does this solution satisfy the limiting condition if $k\neq 0$?
Alecos Papadopoulos
CC BY-SA 3.0
2014-12-20T12:53:19.160Z
2014-12-20T15:12:31.417Z
2014-12-20T15:12:31.417Z
Rusan Kax
<p>How does one solve a linear forward-looking equation $x_t = \beta E_t[x_{t+1}] + k$ where $\lim_{t \to \infty} x_t = 0$ and constant $k,\beta \in \mathbb{R}$, $0&lt;\beta &lt;1$? </p>

<macroeconomics><mathematical-economics>
How to solve a linear forward-looking equation $x_t = \beta E_t[x_{t+1}] + k$ where $\lim_{t \to \infty} x_t = 0$ and $0<\beta < 1$?
memesis
# How to solve a linear forward-looking equation $x_t = \beta E_t[x_{t+1}] + k$ where $\lim_{t \to \infty} x_t = 0$ and $0<\beta < 1$?

(Asked by: username_0 on 2014-12-20)

How does one solve a linear forward-looking equation $x\_t = \beta E\_t[x\_{t+1}] + k$ where $\lim\_{t \to \infty} x\_t = 0$ and constant $k,\beta \in \mathbb{R}$, $0<\beta <1$?

**Tags:** macroeconomics, mathematical-economics


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## 3 answers

### Answer by username_1 on 2014-12-20. Score: 1

(I suspect that $E\_t[x\_{t+1}] = E(x\_{t+1}\mid I\_t)$, where $I\_t$ is the Information Set, i.e. $E\_t[x\_{t+1}]$ is translated as "the expected value of random variable $x\_{t+1}$ conditional on Information available at time $t$").

@username_2 answer implemented the "standard solution algorithm", arrived at a constant solution, and remarked that such a solution violates the limiting condition.  
Strictly speaking, a constant solution would satisfy the limiting solution also, if it was equal to zero (which in our case implies $k=0$): If indeed we had as solution

$$x\_t = \frac k{1-\beta}$$
then

$$\lim\_{t \to \infty} x\_t = \lim\_{t \to \infty}\frac k{1-\beta} = \frac k{1-\beta} \neq 0$$

except if $k=0$. But this is trivial (or at least, uninteresting).

So I ask, **if $k\neq 0$, what would it take to obtain a solution**? The law of evolution must hold also at the limit so

$$\lim\_{t \to \infty} x\_t = \lim\_{t \to \infty}\Big[\beta E\_t[x\_{t+1}] + k\Big]=0$$

$$\Rightarrow \lim\_{t \to \infty}E\_t[x\_{t+1}] = -\frac k{\beta} \tag{1}$$

In words, as the random variable tends to zero, its expected value conditional on information available in the previous period, must equal a non-zero *constant*, while the variable itself tends to zero. Consider some cases, to see whether and when such a thing can happen:

**A)** If $x\_t$ was *totally unobservable*, (not observed directly, and not amenable to indirect estimation though other variables)? In such a case, the information set would not contain any information on $x\_t$, meaning that the conditional expected value would equal the unconditional one. So we would have,

$$\lim\_{t \to \infty}E\_t[x\_{t+1}] = \lim\_{t \to \infty}E[x\_{t+1}] \tag{2}$$

This is economics, meaning that the properties of the mathematical entities should take into account what do they represent in the real world. So it would be meaningless *not* to assume that $x\_{t+1}$ is *not* bounded (more over, this would mean that the $x$ random variable goes to infinity in finite time and then "comes back" to tend to zero in infinite time -and for those who may think "hyperinflation", hyperinflation is a *very very* small number compared to infinity). If it is bounded then the Dominated/Bounded Convergence theorem holds and we have that

$$\lim\_{t \to \infty}E\left(x\_{t+1}\right) =E\left(\lim\_{t \to \infty}x\_{t+1}\right) =0 \neq -\frac k{\beta} \tag{3}$$

So this is not a case in which we can obtain the required relation $(1)$.

**B)** The limiting condition is part of the information sets for finite $t$. Then immediately

$$E\left(\lim\_{t \to \infty}x\_{t+1}\mid I\_t\right) =0$$

In words, at any point in time we know that the variable will tend eventually to zero. Can this be compatible with relation $(1)$? We could invoke again the boundedness of $x\_t$, and a generalization of the Dominated Convergence Theorem, since here the (conditional probability) measure is also time-varying, and again obtain a zero expectation at the limit. But intuitively also, if we know that eventually the variable will tend to zero, at some point in time this must be reflected in $E\_t$, most certainly at the limit. But then again, the required relation $(1)$ won't hold.

**C)** We do observe perfectly or imperfectly the value of the variable, but "we don't give proper notice": our Expectations are *not* Rational. In that case, the symbol $E\_t$ *no longer represents the conditional expectations operator*, but it is just a generic symbol for the "anticipation at time $t$" (in Adaptive Learning literature this is usually symbolized by $E^\*\_t$) and *it acquires the value it needs to acquire for the limiting condition to hold*. In other words, *given* the law of evolution *and* the limiting condition, we *conclude* that expectations must not obey the Rational Expectations Hypothesis, *not even at the limit*.

In such a case, a solution is obvious: *define*
$$E^\*\_t(x\_{t+1}) = -\frac {(1-c\beta^t)k}{\beta}$$

for some constant $c$, This expectations formation rule satisfies $(1)$, and transforms the law of motion into

$$x\_t = \beta \Big[-\frac {(1-c\beta^t)k}{\beta}\Big] + k = ck\beta^t$$

which in turn satisfies the limiting condition.

So we conclude that: **the specific equation, with the specific limiting condition, has a solution if expectations formation does not obey the Rational Expectations Hypothesis, but instead follows the above rule, or some other to the same limiting effect.**


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### Answer by username_2 on 2014-12-20. Score: 0

Since there is nothing random on the right-hand side, we know that the left-hand-side must be non-random either. Hence, we can drop the expectations operator

$$x\_t = \beta x\_{t+1} + k$$
$$x\_{t+1} = \beta x\_{t+2} + k$$
$$x\_t = \beta (\beta x\_{t+2} + k) + k$$
$$x\_t = k + \beta k + \beta\beta x\_{t+2} $$

Where we can now see where this equation is leading towards:

$$x\_t = k\sum\_{s=0}^\infty \beta^s + \lim\_{T\to\infty} \beta^Tx\_T$$

where I have used that the limiting term with $T\to\infty$ will converge to zero.

$$x\_t = \frac{1}{1-\beta} k$$

Which yields a constant $x$. However, this violates $lim\_{t\to\infty} x\_t = 0$, as $x\_t$ is a constant. To conclude, the given equation has no solution.


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### Answer by username_3 on 2014-12-20. Score: 0

As username_2 says, you can drop the expectation. Then derive the general $n$-th case for $x\_{t}$ in terms of $x\_{t+n}$

$$x\_{t}=\beta^{n}x\_{t+n}+k\sum\_{i=0}^{n}\beta^{i}.$$

As $n\rightarrow \infty$, $\beta^{n} \rightarrow 0$ and (we are told) $x\_{t+n} \rightarrow 0$.

So, using the above $$x\_{t}=\lim\_{n\rightarrow \infty}\left[\beta^{n}x\_{t+n}+k\sum\_{i=0}^{n}\beta^{i}\right]=\frac{k}{1-\beta},$$
where we have used the fact that the geometric sum converges because $0<\beta<1$,
$$\lim\_{n\rightarrow \infty}\sum\_{i=0}^{n}\beta^{i}=\frac{1}{1-\beta}.$$

> **Comments:**
> **username_1**: I ask here too: How does this solution satisfy the limiting condition if $k\neq 0$?  

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