Financial market globalization, nonconvergence and credit cycles

Abstract

This paper explores the dynamic consequences of variable investment-project size in a global economy consisting of many small open countries that are plagued with domestic credit market frictions. As is customary in the literature, borrowers provide some internal funds, but they also need external funds to implement their investment projects, which are subject to the costly-state-verification problem. Contrary to the literature, the investment-project size increases with the country’s own capital stock. We find that financial market globalization may lead to a process of oscillatory convergence, even in the absence of any exogenous shocks, if the investment-project size is very sensitive to the change in capital stock.

Appendix: Proofs of Propositions

Proof of Proposition 2

To prove (i), note that according to (23),

$$\begin{aligned} \frac{\partial k_{t+1}}{\partial k_{t}}|_{k_{t+1}=k_{t}=k} \equiv \psi ^{'}(k_{t})|_{k_{t+1}=k_{t}=k}= & {} \frac{(\theta -\beta )k^{\theta -\beta -1}_{t}}{[q-(1-\theta )k^{\theta -\beta }_{t}]}\left\{ \frac{\theta q{\bar{z}}(1-\frac{\gamma }{q{\bar{z}}})^2}{2r[q-(1-\theta )k^{\theta -\beta }_{t}]}\right\} ^{\frac{1}{1-\theta }} \\= & {} \frac{(\theta -\beta )k^{\theta -\beta -1}_{t}k_{t+1}}{[q-(1-\theta )k^{\theta -\beta }_{t}]} \\= & {} \frac{(\theta -\beta )k^{\theta -\beta }}{[q-(1-\theta )k^{\theta -\beta }]}. \end{aligned}$$

Obviously, $\psi ^{'}({\hat{k}})|_{k_{t+1} = k_{t}= {\hat{k}}} > 1$ if ${\hat{k}} > (\frac{q}{1-\beta })^{\frac{1}{\theta -\beta }}$ where ${\hat{k}}$ is given by (25). Or equivalently,

$$\begin{aligned} \left\{ \frac{q}{1-\theta }[1-\frac{{\bar{z}}(1-\frac{\gamma }{q{\bar{z}}})^{2}}{2A}] \right\} ^{\frac{1}{\theta -\beta }}> & {} (\frac{q}{1-\beta })^{\frac{1}{\theta -\beta }} \\ {\hat{\beta }} \equiv \frac{\theta - \frac{{\bar{z}}(1-\frac{\gamma }{q{\bar{z}}})^2}{2A}}{1 - \frac{{\bar{z}}(1-\frac{\gamma }{q{\bar{z}}})^2}{2A}} > \beta . \end{aligned}$$

and it is easy to see that ${\hat{\beta }} < \theta$ holds. This condition requires the map of (23) cutting (24) from below with a slope bigger than one at ${\hat{k}}$. Now consider the situation when ${\hat{k}} = (\frac{\theta A}{r})^{\frac{1}{1-\theta }}$ holds with ${\hat{k}}$ given by (25). Solving it for r gives

$$\begin{aligned} {\bar{r}} \equiv r = \frac{\theta A}{{\hat{k}}^{1-\theta }}. \end{aligned}$$

This condition is graphically described by the locus labeled with $r = {\bar{r}}$ in Fig. 4. Decreasing (increasing) r slightly below (above) ${\bar{r}}$ shifts the locus labeled with $r = {\bar{r}}$ upward (downward). Note that, when the value of r is equal to ${\bar{r}}$, ${\hat{k}} = {\tilde{k}}$ holds. However, when $r < {\bar{r}}$, ${\hat{k}} > {\tilde{k}}$ and when $r > {\bar{r}}$, ${\hat{k}} < {\tilde{k}}$.

Obviously, $\frac{\partial k_{t+1}}{\partial k_{t}}|_{k_{t+1}=k_{t}} = \psi ^{'}(k_{t})|_{k_{t+1}=k_{t}} =1$ when $k_{t} = (\frac{q}{1-\beta })^{\frac{1}{\theta - \beta }}$. Substituting $k_{t} = (\frac{q}{1-\beta })^{\frac{1}{\theta - \beta }}$ into (23) yields:

$$\begin{aligned} k_{t+1}= & {} \left\{ \frac{\theta {\bar{z}}q(1-\frac{\gamma }{q{\bar{z}}})^2}{2r[q-\frac{q(1-\theta )}{(1-\beta )}]}\right\} ^{\frac{1}{1-\theta }} \\= & {} \left[ \frac{\theta {\bar{z}}(1-\beta )(1-\frac{\gamma }{q{\bar{z}}})^2}{2r(\theta -\beta )} \right] ^{\frac{1}{1-\theta }}. \end{aligned}$$

When $[ \frac{\theta {\bar{z}}(1-\beta )(1-\frac{\gamma }{q{\bar{z}}})^2}{2r(\theta -\beta )}]^{\frac{1}{1-\theta }} = (\frac{q}{1-\beta })^{\frac{1}{\theta -\beta }}$, the locus tangents to the 45 degree line. Solving this equation for r gives:

$$\begin{aligned} {\underline{r}} \equiv r = \frac{\theta {\bar{z}}(1-\beta )(1-\frac{\gamma }{q{\bar{z}}})^{2}}{2(\theta -\beta )(\frac{q}{1-\beta })^{(\frac{1-\theta }{\theta -\beta })}} \end{aligned}$$

(the locus associated with ${\underline{r}} = r < {\bar{r}}$ is not shown in Fig. 4). If $\psi ^{'}({\hat{k}})|_{k_{t+1}=k_{t}={\hat{k}}} > 1$ holds and for any r satisfying ${\underline{r}}< r < {\bar{r}}$, $\psi (k_{t})$ will cut the 45 degree line at $k^{l}$ and ${\tilde{k}}$. It is easy to verify that $\psi ^{''}(k) > 0$. If $\psi ^{'}(k^{l}) < 1$, $\psi ^{'}(\frac{q}{1-\beta }) = 1$ and $\psi ^{'}({\tilde{k}}) > 1$, $k^{l}< (\frac{q}{1-\beta })^{\frac{1}{\theta -\beta }} < {\tilde{k}}$ must hold. Furthermore, ${\tilde{k}}$ is unstable because $\psi (k_{t})$ cuts the 45 degree line from below at ${\tilde{k}}$ and $k^{l}$ is stable because $\psi (k_{t})$ cuts the 45 degree line from above at $k^{l}$. Furthermore, if $r < {\bar{r}}$, ${\hat{k}} < k^{h}$ and ${\tilde{k}} < {\hat{k}}$ must be satisfied. It immediately follows that $k^{l}< (\frac{q}{1-\beta })^{\frac{1}{\theta -\beta }}< {\tilde{k}}< {\hat{k}} < k^{h}$. Finally, $k^{h}$ is stable because (24) cuts the 45 degree line form above. When $r < {\underline{r}}$, the locus is above the one associated with ${\underline{r}} = r$ and hence the economy produces at ${\bar{k}}^{h}$ which is stable and satisfies ${\bar{k}}^{h} > {\hat{k}}$. On the other hand, when $r > {\bar{r}}$, the locus shifts below the one associated with $r = {\bar{r}}$. As a result, it cuts the 45 degree line at ${\underline{k}}^{l}$ which is also stable and satisfies ${\underline{k}}^{l} < {\hat{k}}$.

To prove (ii), note that when $\theta > \beta \ge {\hat{\beta }}$, ${\hat{k}} \le (\frac{q}{1-\beta })^{\frac{1}{\theta -\beta }}$ holds and $\psi ^{'}({\hat{k}})|_{k_{t+1} = k_{t}={\hat{k}}} \le 1$ follows. Therefore, the map of (23) cuts (24) from below with a slope less than or equal to 1 at ${\hat{k}}$, as shown in Fig. 5. In such case, ${\bar{r}}$ is still equal to $\frac{\theta A}{{\hat{k}}^{1-\theta }}$. Reducing r shifts the locus with $r = {\bar{r}}$ upward and the economy produces at steady state ${\bar{k}}^{h}$. ${\bar{k}}^{h}$ is stable and satisfies ${\bar{k}}^{h} > {\hat{k}}$. To the contrary, increasing r shifts this locus downward and the economy produces at steady state ${\underline{k}}^{l}$. ${\underline{k}}^{l}$ is stable and satisfies ${\underline{k}}^{l} < {\hat{k}}$.

For (iii), the proof is evident from inspection of Fig. 6.

To prove (iv), note that

$$\begin{aligned} \frac{\partial k_{t+1}}{\partial k_{t}}= & {} \frac{(\theta -\beta )k^{\theta -\beta }_{t}}{[q-(1-\theta )k^{\theta -\beta }_{t}]}\frac{k_{t+1}}{k_{t}} \end{aligned}$$

from (i). Since $q > (1 - \theta )k^{\theta -\beta }_{t}$ and $\theta < \beta$, $\frac{\partial k_{t+1}}{\partial k_{t}} < 0$ holds. Therefore, the map of $k_{t}$ is decreasing with $k_{t+1}$. Furthermore, as $k_{t} \rightarrow \infty$, $k_{t+1} = [\frac{\theta {\bar{z}}(1-\frac{\gamma }{q{\bar{z}}})^2}{2r}]^{\frac{1}{1-\theta }}$ and as $k_{t} \rightarrow (\frac{1-\theta }{q})^{\frac{1}{\beta -\theta }}$, $k_{t+1} \rightarrow \infty$ according to (23). Hence it must intersect the 45 degree line only once at $k^{v}$, which proves the first half of (iv). To prove the second half, note that using Taylor series expansion yields:

$$\begin{aligned} k_{t+1} \equiv \psi (k_t)= & {} \left\{ \frac{\theta {\bar{z}}q(1-\frac{\gamma }{q{\bar{z}}})^2}{2r[q-(1-\theta )k_{t}^{\theta -\beta }]} \right\} ^{\frac{1}{1-\theta }} \\\approx & {} \psi (k) + \psi ^{'}(k)(k_{t} - k) \end{aligned}$$

where k is the steady state value and

$$\begin{aligned} \psi ^{'}(k) = \frac{\partial k_{t+1}}{\partial k_{t}}|_{k_{t+1}=k_{t}=k} = \frac{(\theta -\beta )k^{\theta -\beta }}{[q-(1-\theta )k^{\theta -\beta }]}. \end{aligned}$$

Since $\beta > \theta$, it follow that $q - (1-\theta )k^{\theta - \beta } > (\theta - \beta )k^{\theta - \beta }$ holds. Therefore, the absolute value of $\psi ^{'}(k)$ is smaller than 1. As a result, the unique steady state, $k^{v}$, is locally stable (see Galor 2007). Obviously, $k^{v} > {\hat{k}}$ is satisfied. $\square$

Proof of Proposition 3

(i) Inspecting (30) gives $f(0) = 0$ and $f({\hat{k}}) = {\hat{k}}$. Furthermore, it is easy to verify that

$$\begin{aligned} \frac{\partial k^{h}}{\partial k^{l}}= & {} \left\{ \frac{2A[q-(1-\theta )(k^{l})^{\theta -\beta }]}{{\bar{z}}q(1-\frac{\gamma }{q{\bar{z}}})^2} \right\} ^{\frac{1}{1-\theta }}\frac{q-(1-\beta )(k^{l})^{\theta -\beta }}{q-(1-\theta )(k^{l})^{\theta -\beta }}\\= & {} \frac{[q-(1-\beta )(k^{l})^{\theta -\beta }]k^{h}}{[q-(1-\theta )(k^{l})^{\theta -\beta }]k^{l}}. \end{aligned}$$

Therefore, when $0< k^{l} < (\frac{q}{1-\beta })^{\frac{1}{\theta -\beta }}$, $\frac{\partial k^{h}}{\partial k^{l}} > 0$ whereas when $(\frac{q}{1-\theta })^{\frac{1}{\theta -\beta }}> k^{l} > (\frac{q}{1-\beta })^{\frac{1}{\theta -\beta }}$, $\frac{\partial k^{h}}{\partial k^{l}} < 0$. The curve of $k^{h} = f(k^{l})$ is shown in Fig. 8 as locus SS. The sets of $(k^{h}, k^{l})$ that satisfy $k^{l}< {\hat{k}} < k^{h}$ are shown as the segment of the curve SS with dash in Fig. 8.

Now what remains is to trace out the locus of Eq. (31). According to (31), we obtain

$$\begin{aligned} \frac{dk^{h}}{dk^{l}} = \frac{(1-\phi )[A(1-\theta )\theta (k^{l})^{\theta -1}-1]}{\phi [1-A(1-\theta )\theta (k^{h})^{\theta -1}]} \end{aligned}$$

and $\frac{dk^{h}}{dk^{l}}|_{k^{h} = k^{l}} = -\frac{1-\phi }{\phi }$.

Define ${\underline{k}}$ as $\frac{dk_{t+1}}{dk_{t}}|_{k_{t} ={\underline{k}}} = 1$ using (21) (see Fig. 2). It follows that ${\underline{k}} = [A\theta (1-\theta )]^{\frac{1}{1-\theta }}$. Obviously, $\frac{dk^{h}}{dk^{l}} \ge 0$ iff $k^{h} \in [{\underline{k}}, \infty ]$ and $k^{l} \in [0, {\underline{k}}]$ or $k^{h} \in [0, {\underline{k}}]$ and $k^{l} \in [{\underline{k}}, \infty ]$. Hence the locus defined by (31) has the configuration as locus RR described in Fig. 8. Point B, which is the intersection of the loci SS and RR defined by (30) and (31) respectively, is the steady state of ($k^{h}$, $k^{l}$). Obviously, $k^{h} > {\hat{k}}$ and $k^{l} < {\hat{k}}$. The value of $k^{*}$ satisfies $k^{l}< (\frac{q}{1-\beta })^{\frac{1}{\theta -\beta }}< k^{*}< {\hat{k}} < k^{h}$.

(ii) If $\theta > \beta \ge {\hat{\beta }}$ and each small open economy produces at $\bar{k^{h}} > {\hat{k}}$, the world economy steady state equilibrium will be determined by (31) and (32). Therefore, $k^{h} = k^{l} = k^{*} > {\hat{k}}$ must satisfy and then we can use (29) to find out the world interest rate. On the other hand, if the world economy produces at $\underline{k^{l}} < {\hat{k}}$, the steady state equilibrium is described by (31) and (33). In Fig. 9, (31) is represented as locus RR and (33) is locus NN. To derive point E in this figure, we totally differentiate (33) to obtain:

$$\begin{aligned} \frac{dk^{h}}{dk^{l}} = \frac{(k^{l})^{-\theta }[q-(1-\theta )(k^{l})^{\theta -\beta }]-(\theta -\beta )(k^{l})^{-\beta }}{(k^{h})^{-\theta }[q-(1-\theta )(k^{h})^{\theta -\beta }]-(\theta -\beta )(k^{h})^{-\beta }}. \end{aligned}$$

Let $\frac{dk^{h}}{dk^{l}} = -1$ (to represent the downward sloping 45 degree line) and assume $k^{h} = k^{l}$ (the upward sloping and the downward sloping 45 degree lines intersect at point E). Then it is easy to obtain $k^{h} = k^{l} = (\frac{q}{1-\beta })^{\frac{1}{\theta -\beta }}$ at point E. Furthermore, $(\frac{q}{1-\beta })^{\frac{1}{\theta -\beta }} > k^{*}$ must hold because ${\hat{k}} > k^{h} = k^{l} = k^{*}$ and based on $\theta > \beta \ge {\hat{\beta }}$, we have ${\hat{k}} \le (\frac{q}{1-\beta })^{\frac{1}{\theta -\beta }}$. Of course, the world interest rate can be found through (29).

(iii) If $\beta = \theta > {\hat{\beta }}$, the steady state equilibrium of the world economy is determined by (31) and (33) with $\beta = \theta$. Apparently, all countries have the same steady state capital stock per capita $k^{*}$ and the world interest rate is given by (28) with $\beta = \theta$.

(iv) When $\beta> \theta > {\hat{\beta }}$, all countries produce the same level of k at steady state and hence $k^{h} = k^{l} = k^{*}$ must satisfy. Likewise, the world economy equilibrium is determined by (31) and (33) under the assumption that $\beta$ is bigger than $\theta$. Figure 10 is used to show this equilibrium graphically. It is straightforward to show that $k^{h} = k^{l} = (\frac{1-\beta }{q})^{\frac{1}{\beta - \theta }}$ at point F. Moreover, $k^{*} > (\frac{1-\beta }{q})^{\frac{1}{\beta - \theta }}$ must hold in this case because $k^{v} = k^{*} \ge (\frac{1-\theta }{q})^{\frac{1}{\beta - \theta }} > (\frac{1-\beta }{q})^{\frac{1}{\beta -\theta }}$. To show that the steady state of this world economy is locally stable and is approached cyclically, we need to work with the following two equations:

$$\begin{aligned} \frac{{\bar{z}}q\theta (k^{h}_{t+1})^{\theta -1}(1-\frac{\gamma }{q{\bar{z}}})^{2}}{2[q-(1-\theta )(k^{h}_{t})^{\theta -\beta }]}= & {} \frac{{\bar{z}}q\theta (k^{l}_{t+1})^{\theta -1}(1-\frac{\gamma }{q{\bar{z}}})^{2}}{2[q-(1-\theta )(k^{l}_{t})^{\theta -\beta }]} \end{aligned}$$

and

$$\begin{aligned} \phi k_{t+1}^{h} + (1-\phi )k_{t+1}^{l}= & {} A(1-\theta )[\phi (k_{t}^{h})^{\theta } + (1-\phi )(k_{t}^{l})^{\theta }]. \end{aligned}$$

Let $d_{t} \equiv \frac{k^{h}_{t}}{k^{l}_{t}}$. Then, the above two equations can be rewritten as:

$$\begin{aligned} d_{t+1}= & {} \left[ \frac{q-(1-\theta )(k^{l}_{t})^{\theta -\beta }}{q-(1-\theta )(d_{t}k^{l}_{t})^{\theta -\beta }}\right] ^{\frac{1}{1-\theta }} \\ k^{l}_{t+1}= & {} \frac{A(1-\theta )(k^{l}_{t})^{\theta }[\phi d^{\theta }_{t} + (1-\phi )]}{\phi d_{t+1} + (1-\phi )}. \end{aligned}$$

To proceed, we linearize these two equations in a neighborhood of the steady state to obtain:

$$\begin{aligned} (k^{l}_{t+1} - k^{l}, d_{t+1} - d)' = J (k^{l}_{t} - k^{l}, d_{t} - d) \end{aligned}$$

where $k^{l}$ and d are the steady state values and J is the Jacobian matrix:

$$\begin{aligned} J = \left( \begin{array}{cc} \frac{\partial k^{l}_{t+1}}{\partial k^{l}_{t}} &{} \frac{\partial k^{l}_{t+1}}{\partial d_{t}} \\ \frac{\partial d_{t+1}}{\partial k^{l}_{t}} &{} \frac{\partial d_{t+1}}{\partial d_{t}} \end{array} \right) \end{aligned}$$

with partial derivatives evaluated at the steady state. Now we display the expressions for these partial derivatives at the steady state in below:

$$\begin{aligned} \frac{\partial k^{l}_{t+1}}{\partial k^{l}_{t}}= & {} \theta - \frac{\phi q(\theta -\beta )(k^{l})^{\theta -\beta }(1-d^{\theta -\beta })d}{[(1-\phi )+\phi d)][q-(1-\theta )(k^{l})^{\theta -\beta }][q-(1-\theta )(dk^{l})^{\theta -\beta }]},\\ \frac{\partial k^{l}_{t+1}}{\partial d_{t}}= & {} \frac{\phi \theta d^{\theta -1}k^{l}}{[(1-\phi )+\phi d^{\theta }]} - \frac{\phi (\theta -\beta )(k^{l})^{\theta -\beta +1}d^{\theta -\beta }}{[(1-\phi )+\phi d][q-(1-\theta )(dk^{l})^{\theta -\beta }]}, \\ \frac{\partial d_{t+1}}{\partial k^{l}_{t}}= & {} \frac{-q(\theta - \beta )d(k^{l})^{\theta -\beta -1}(1-d^{\theta - \beta })}{[q-(1-\theta )(k^{l})^{\theta -\beta }][q-(1-\theta )(dk^{l})^{\theta -\beta }]}, \\ \frac{\partial d_{t+1}}{\partial d_{t}}= & {} \frac{(\theta -\beta )(k^{l})^{\theta -\beta }d^{\theta -\beta }}{[q-(1-\theta )(dk^{l})^{\theta -\beta }]}. \end{aligned}$$

At the (symmetric) steady state, $d = 1$ and $k^{l} = k^{*}$. Substituting them into the above expressions gives:

$$\begin{aligned} \frac{\partial k^{l}_{t+1}}{\partial k^{l}_{t}}= & {} \theta , \\ \frac{\partial k^{l}_{t+1}}{\partial d_{t}}= & {} \phi k^{*}[\theta - \frac{(\theta -\beta )(k^{*})^{\theta -\beta }}{q-(1-\theta )(k^{*})^{\theta -\beta }}],\\ \frac{\partial d_{t+1}}{\partial k^{l}_{t}}= & {} 0, \\ \frac{\partial d_{t+1}}{\partial d_{t}}= & {} \frac{(\theta -\beta )(k^{*})^{\theta -\beta }}{q-(1-\theta )(k^{*})^{\theta -\beta }}. \end{aligned}$$

The eigenvalues of the matrix J are given by $\lambda _{1} = \frac{\partial k^{l}_{t+1}}{\partial k^{l}_{t}}$ and $\lambda _{2} = \frac{\partial d_{t+1}}{\partial d_{t}}$. It is easy to see that $\lambda _{1}$ is positive and less than one but $\lambda _{2}$ is negative (recall that $\beta > \theta$ and $k_{t} > (\frac{1-\theta }{q})^{\frac{1}{\beta - \theta }}$) and less than one in absolute value. Therefore, this dynamical system is locally stable in the neighborhood of the (symmetric) steady state and it is approached oscillatory. $\square$