# Divergence of a variable in forward-looking models. Blanchard-Khan conditions

(1) This post is, first, regarding the issue from Romer's Advanced Macroeconomics, 4 ed. Chapter 6.4 on nominal rigidity, pp. 264-266 to be precise.

He sets out a simple model with microfounded IS curve, AS curve and MP curve, using standard notation with $\pi$ as an inflation, $y$ as an output and $r$ as an interest rate.

My question is why agents in this model cannot expect $y$ to diverge? Is this due to mathematical and/or economical reasons? Or did I miss some assumption in the text?

Also, does diverge here mean rise ad infinitum or only rise fast enough to offset the convergence of $\phi^n$? [I know, this is a rather math than econ question, but entirely a propos the problem considered].

So here's the model:

(2) Secondly, later in the text a similar issue recurs in chapter 11.5 when analyzing the model of optimal monetary policy in a forward-looking setting. Romer considers cases where self-fulfilling equilibria arise (I guess Blanchard-Khan conditions):

So why do we have to assume that our variable is bounded? Again: is this due to mathematical and/or economical reasons?

(3) Finally, when I elswhere read about Blanchard-Kahn conditions there is stipulated that "Blanchard-Kahn conditions apply to models that add as a requirement that the series do not explode". Why is that? Isn't that possible for an economic system to have such a property? Or why are we ruling out from our analysis models with that property?

Do all these 3 issues touch the same problem and have the same solution or this is just the coincidence that I don't get this assumptions/results(?) in different contexts?

PS This comment is really helpful claryfing some math related issues, but it does not pertain to the reason for the assumption/result(?) in question.

Edit: For the clarity reasons I've included more of Romer's solution to the (1) problem.

QUESTION 1
By inserting $(6.28)$ into $(6.29)$, re-arranging and applying direct forward substitution , we obtain

$$y_t = \lim_{n\rightarrow \infty}\left[\left(\frac {\theta}{\theta + b}\right)^nE_t[y_{t+n}]\right] +\frac {\theta}{\theta+b -\theta \rho}u_t \tag{1}$$

Then forwarding $n$ periods and taking the conditional expectation with respect to information available at $t$ we have

$$E_t[y_{t+n}] = E_t\left(\lim_{n\rightarrow \infty}\left[\left(\frac {\theta}{\theta + b}\right)^nE_{t+n}[y_{t+2n}]\right]\right) +E_{t}\left(\frac {\theta}{\theta+b -\theta \rho}u_{t+n} \right)$$

$$\implies E_t[y_{t+n}] = E_t\left(\lim_{n\rightarrow \infty}\left[\left(\frac {\theta}{\theta + b}\right)^nE_{t+n}[y_{t+2n}]\right]\right) \tag{2}$$

The conditional expectation is an integral. Can we interchange integration and limit? To do that Dominated Convergence & Co should hold (they are only sufficient conditions, but we do not have something better). For our case, Dominated Convergence requires that the expression inside the limit is bounded("dominated"), and that the limit is finite. If it is then, using also the Law of Iterated Expectations, and the fact that under the limit $t+2n$ is equivalent to $t+n$ we have

$$E_t[y_{t+n}] = \lim_{n\rightarrow \infty}\left[\left(\frac {\theta}{\theta + b}\right)^nE_t[y_{t+n}]\right] \tag{3}$$

Note carefully that in the left-hand side, $n$ does not go to infinity. Also, remember that we obtained $(3)$ under the assumption that the limit is finite. So $(3)$ tells us that for every finite $n$, the conditional expectation must be equal to the unique limit, i.e. equal to the same number, i.e. it should be a constant.

Can this constant be anything else than zero? Assume the limit is not zero. Then one can verify that the only other finite value that the limit can take is $1$. But for the limit to be $1$ we must have $E_t[y_{t+n}]\rightarrow \left(\frac {\theta + b}{\theta}\right)^n$, which goes to infinity with $n$. So the left hand side would go to infinity and will not be equal to the limit. Hence we conclude that the limit must be zero, for $(3)$ to hold.
This can happen if we assume that the expectation alone is a non-zero constant, or that it is growing but boundedly (say, it has an asymptote to which it converges from below). But then again, in both cases the left hand side will not be equal to the now zero limit. So we see that the only way that $(3)$ can hold is to assume that the expectation alone is zero.

So I would say that the approach "if we assume that the limit is zero, then we obtain etc" (as presented in the quoted text), is wrong: looking at our solution, which is equation $(1)$, we must first assume/impose that the expectation is zero, to make it all consistent, and finite (note: if the solution $(1)$ included also a constant term in the right-hand side, things would be more flexible).

Sticking with $(1)$, we also see that the only other consistent solution is to assume that the limit goes to infinity -and this can happen only if we assume that the expectation on its own goes to infinity.

QUESTIONS 2 & 3
As I have wrote in another answer, everything in real-world economics is finite -even prices, even expectations. The above mathematical formulations and the mathematical possibility of an infinite solution represent in a colorful way the phenomenon of "bubbles". In real life, bubbles eventually burst -but what we have managed to capture by using the "infinity" concept, is the abnormal inflation of the bubble while it hasn't burst yet. So agents can "expect infinity", in the metaphorical sense just described. But since "bubbles" is not your everyday economic phenomenon, these models usually assume them away in order to study the stable situation.

Marx was the one that argued that capitalism will explode, or rather, implode. So you have an articulated theory that advances the view that a specific economic system possesses the property of (eventual) divergence.