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 $k,\beta \in \mathbb{R}$, $0<\beta <1$?
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3 Answers
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(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 $t$").
@FooBar 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.
answered Dec 20, 2014 at 22:51
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As FooBar 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 $0<\beta<1$, $$\lim_{n\rightarrow \infty}\sum_{i=0}^{n}\beta^{i}=\frac{1}{1-\beta}.$$
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$\begingroup$ I ask here too: How does this solution satisfy the limiting condition if $k\neq 0$? $\endgroup$ Commented Dec 20, 2014 at 22:53
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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.
