Exercise 12.1
Transversality Condition In Collateral-Constrained Economies
Problem
Consider the economy of Section 12.1. Show that if sequences \(c_t,\) \(d_{t+1},\) and \(k_{t+1}\) satisfy optimality conditions (12.1)-(12.8) but not the transversality condition (12.9), then there exists a welfare dominating set of sequences that is also feasible(i.e., it satisfies (12.1)-(12.3)).
Answer
The proof is by contradiction. Let
\[ Z_t \equiv \frac{\kappa q_t k_{t+1} - d_{t+1}}{(1+r)^t} \]
Assume that contrary to the claim (12.9) does not hold. In particular, assume that
\[ \lim_{t\rightarrow \infty} Z_t = Z >0. \]
Then for every \(\epsilon>0\), there exists a \(T_{\epsilon}\) such that
\[ -\epsilon < Z_t - Z < \epsilon \]
for all \(t \ge T_{\epsilon}\).
Pick \(\epsilon>0\) such that \(0<\epsilon<Z\). Then we have that \[ Z_t = \frac{\kappa q_t k_{t+1} - d_{t+1}}{(1+r)^t} > Z - \epsilon >0 \]
for all \(t\ge T_{\epsilon}\). It follows that the collateral constraint holds with a strict inequality for all \(t \ge T_{\epsilon}\), that is, \(\mu_t = 0\) for all \(t\ge T_{\epsilon}\).
Now consider the following alternative consumption and debt paths whereby we increase consumption in period \(T_{\epsilon}\) but leaving it unchanged in all other periods. We finance this increase in consumption in period \(T_{\epsilon}\) by issuing debt and then rolling over this additional debt forever. And we leave the path for \(k_{t+1}\) unchanged. We want to know if this alternative path for \(d_{t+1}\) is feasible. Let the change in \(d_{T_{\epsilon}+1}\) be equal to \(z>0\), that is, \(\tilde{d}_{T_{\epsilon}+1} = d_{T_{\epsilon}+1} + z\). Then consumption increases in period \(T_{\epsilon}\) by \(z/(1+r)>0\).
This strategy is feasible in period \(T_{\epsilon}\), that is, it does not violate the collateral constraint (12.3), as long as,
\[ z< (1+r)^{T_{\epsilon}} (Z-\epsilon). \]
We need to show that this new path of debt also does not violate the collateral constraint for any \(t>T_{\epsilon}\). The new level of debt in any period \(t>T_{\epsilon}\), denoted \(\tilde{d}_{t+1}\), can be found by subtracting the sequential budget constraint for any period \({T_{\epsilon}+j}\) under the original plan and the alternative plan. Note that for \(t>T_{\epsilon}\), \(c_t\) and \(k_{t}\) are the same under the original and alternative plans. This yields:
\[ \tilde{d}_{t+1} - d_{t+1} = (1+r) (\tilde{d}_{t} - d_{t}) \]
or
\[ \tilde{d}_{T_{\epsilon}+j+1} - d_{T_{\epsilon}+j+1} = (1+r)^j (\tilde{d}_{T_{\epsilon}+1} - d_{T_{\epsilon}+1}) \]
Now recall from above that for any \(t\ge T_{\epsilon}\)
\[ \begin{eqnarray*} \kappa q_t k_{t+1} - d_{t+1} &>& {(1+r)^t}[Z - \epsilon]\\ \kappa q_t k_{t+1} - \tilde{d}_{t+1} + (\tilde{d}_{t+1} - d_{t+1}) &>& {(1+r)^t}[Z - \epsilon] \\ \kappa q_t k_{t+1} - \tilde{d}_{t+1} &>& (1+r)^t[Z - \epsilon] - (\tilde{d}_{t+1} - d_{t+1}) \\ &=& (1+r)^t[Z - \epsilon - (1+r)^{-T_{\epsilon}} (\tilde{d}_{T_{\epsilon}+1} - d_{T_{\epsilon}+1})]\\ &>& 0 \end{eqnarray*} \]
The last inequality follows from the assumption that
\[ (1+r)^{-T_{\epsilon}} (\tilde{d}_{T_{\epsilon}+1} - d_{T_{\epsilon}+1}) = (1+r)^{-T_{\epsilon}} z < Z - \epsilon. \]
This completes the proof that the alternative sequence \(\tilde{d}_{t+1}\) if feasible. Clearly it is associated with higher welfare because consumption in period \(T_{\epsilon}\) is higher. Therefore, the original sequence could not have been a solution to the household maximization problem. Thus we have arrived at a contradiction. Therefore, it must be the case that under household maximization the transversality condition (12.9) hold.