The capital structure of a firm under rate of return regulation: durability and the yield curve

Abstract

I present a generalized dynamic model of firm behavior under rate of return regulation. The modeled firm has access to multiple types of capital which are substitutes (imperfect or perfect) in production. These capital inputs are differentiated based on durability and heterogeneous marginal effects on the firm’s total cost of capital. This approach is kept general but is motivated by the stylized shape of the bond yield curve, wherein high-durability (longer lived) assets command a higher required return on investment (higher bond yield). The results indicate that a regulated firm (relative to an unregulated firm) will over or under-invest in specific assets depending on their durability and the size of the assets’ marginal effects on the cost of capital relative to the regulated rate of return. Two specific sources of distortion in the capital structure are identified. The “yield curve” effect pushes the firm further (relative to the unconstrained case) towards assets with a low marginal contribution to the cost of capital, thus reducing the firm’s average cost of capital. The “duration” effect pushes the firm towards longer lived assets as a means to inflate the steady state capital stock. This is similar to (yet distinct from) the classic Averch and Johnson (Am Econ Rev 52(5):1052–1069, 1962) result.

Appendices

Appendix 1: Dynamics of the system

I present here a sketch for the proof of dynamic stability (convergence to the steady state) for the optimal control problem described by the Hamiltonian given in Eq. (5). I begin this sketch with an illustration that the system is stable in a subspace comprised of the state variables. I then illustrate that, for each asset class ($i \in N$), the ratio of the liability ($B_i$) to the productive value ($k_i$) of capital is known. Finally I provide a simple phase diagram in $k_i$, $B_i$ space illustrating that the system is stable around the optimal steady state equilibrium.

For each element in set $N$ (the total set of distinct asset classes) there are 3 accompanying dynamic variables: $I_{i,t}$, $k_{i,t}$ and $B_{i,t}$. Consider the lower dimension $2N$ subset of the dynamic system composed of Eqs. (3) and (4). The system can be written in the standard matrix form ($\dot{X} = Ax + c$):

$$\begin{aligned} \begin{pmatrix} \dot{k}_{1} \\ \dot{B}_{1} \\ \dot{k}_{2} \\ \dot{B}_{2} \\ \vdots \\ \dot{k}_{N} \\ \dot{B}_{N} \end{pmatrix} = \begin{pmatrix} &{}\quad -\delta _1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 \\ &{}\quad 0 &{}\quad -\alpha _1 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 \\ &{}\quad 0 &{}\quad 0 &{}\quad -\delta _2 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 \\ &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad -\alpha _2 &{}\quad \cdots &{}\quad 0 &{}\quad 0 \\ &{} \quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad -\delta _N &{}\quad 0 \\ &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -\alpha _N \end{pmatrix} \begin{pmatrix} {k}_{1} \\ {B}_{1} \\ {k}_{2} \\ {B}_{2} \\ \vdots \\ {k}_{N} \\ {B}_{N} \end{pmatrix} + \begin{pmatrix} {I}_{1} \\ {I}_{1} \\ {I}_{2} \\ {I}_{2} \\ \vdots \\ {I}_{N} \\ {I}_{N} \end{pmatrix} \end{aligned}$$

The set of eigenvalues for this subspace can be characterized by the set of values $\varLambda$ which satisfy³¹:

$$\begin{aligned} \prod _{i \in N} \left[ \left( \delta _i + \varLambda \right) \left( \alpha _i + \varLambda \right) \right] = 0 \;\; \implies \;\; \varLambda \in \left\{ -\delta _i \right\} \cup \left\{ -\alpha _i \right\} \end{aligned}$$

Thus the set of eigenvalues for the subspace dimension is uniformly negative. The system is therefore stable in both of the state variables for each asset class via the standard stability conditions. For any given level of the control variables $I_i$ the state variables $k_i$ and $B_i$ will converge to a steady state.

The ratio between $k_i$ and $B_i$ in the steady state is fixed and can be derived for any given level of investment. Consider a case in which investment in an asset class $i$ is fixed through time at a value $\bar{I}$. In this situation, $k_i$ and $B_i$ will each converge to a steady state where:

$$\begin{aligned} \left. \begin{aligned} \dot{k}_i = 0&\implies&0=\bar{i} - \delta _i k_i&\implies&k_i = \frac{\bar{i}}{\delta _i} \\ \dot{B}_i = 0&\implies&0=\bar{i} - \alpha _i B_i&\implies&B_i = \frac{\bar{i}}{\alpha _i} \end{aligned} \right\} \implies \; B_i = \frac{\delta _i}{\alpha _i} k_i \end{aligned}$$

Thus, a feasible path to the steady state exists. Full characterization of this path implies that the firm will choose levels for the control variables $I_{i,t}$ which will ensure that all $k_i$ and $B_i$ converge to the optimal steady state ($\widehat{k}_i , \widehat{B}_i$) following the maximization problem outlined in the main paper. At the optimal steady state $I_i = \delta _i \widehat{k}_i = \alpha _i \widehat{B_i} \; \implies \; \dot{I}_i = 0$.

The phase diagram for the 2 dimensional subset $k_i$, $B_i$ of the system around the steady state (i.e. when $I_i =\delta _i \widehat{k}_i = \alpha _i \widehat{B_i}$) is given in Fig. 2.

Fig. 2

Phase diagram for $B_i$,$k_i$ when $I_i = \delta _i \widehat{k}_i = \alpha _i \widehat{B_i}$

Appendix 2: Mathematical appendix

Proof (Proof of Lemma 1)

Arrow’s sufficiency conditions for the existence of an equilibrium requires the Hamiltonian function to be jointly concave with respect to the state variables. Joint concavity of $k_i$ and $k_j$ requires that:

$$\begin{aligned} | b_1 | = \left| \begin{array}{cc} 0 &{}\frac{\partial H}{\partial k_i} \\ \frac{\partial H}{\partial k_i} &{}\frac{\partial ^2 H}{\partial k_i ^2} \\ \end{array} \right| = - \left( \frac{\partial H}{\partial k_i} \right) ^2 \le 0 \end{aligned}$$

(which is satisfied for any input $k_i$ and function H) and

$$\begin{aligned} | b_2 | = -(1- \lambda ) \left[ \left( \frac{\partial H}{\partial k_i} \right) ^2 \frac{\partial ^2 F}{\partial k_j ^2}+\left( \frac{\partial H}{\partial k_j} \right) ^2 \frac{\partial ^2 F}{\partial k_i ^2} - 2 \frac{\partial H}{\partial k_i} \frac{\partial H}{\partial k_j} \frac{\partial ^2 F}{\partial k_i \partial k_j} \right] \ge 0 \end{aligned}$$

Where $b_1$ and $b_2$ are the first and second principle minors of the bordered Hessian of (5). $\frac{\partial F}{\partial k_i} >0 \;\forall i \in N$ and the set of equations given in (10) imply: $sign \left\{ {\frac{\partial H}{\partial k_i}} \right\} \equiv sign \left\{ \frac{\partial H}{\partial k_j}\right\} \;\; \forall i,j \in N$. Thus: $\frac{\partial H}{\partial k_i}\cdot \frac{\partial H}{\partial k_j} >0 \; \forall i,j \in N$. Imposing $\left\{ \frac{\partial ^2 F}{\partial k_i ^2} < 0 \; ; \; \frac{\partial ^2 F}{\partial k_{i,t} \partial k_{j,t} } \ge 0 \right\} \;\forall \;i \in N$, the above relationship implies $\lambda < 1$. A binding rate of return constraint implies $\lambda > 0$ via the standard Karesh-Kuhn-Tucker conditions. Therefore $\lambda \in (0,1)$. $\square$

Proof (Proof of Lemma 2)

From Eqs. (13) and (14):

$$\begin{aligned}&\displaystyle \widehat{\mu _i} \!=\! max \left\{ \left[ \left( \frac{r_f\!+\!\delta _i}{r_f\!+\!\alpha _i}\right) \cdot \left( r_i \!+\! \alpha _i \!-\! \left( \frac{\lambda }{1- \lambda }\right) \left( S\!-\!r_i\right) \right) \!-\! \frac{\partial F}{\partial k_i} \bigg | _{k_i = 0} \right] (1-\lambda ) \;\;,\;\; 0 \right\} \\&\displaystyle \mu _i ^* = max \left\{ \left[ \left( \frac{r_f+\delta _i}{r_f+\alpha _i}\right) \cdot \left( r_i + \alpha _i \right) - \frac{\partial F}{\partial k_i} \bigg | _{k_i = 0} \right] \;\;,\;\; 0 \right\} \end{aligned}$$

From Lemma 1: $\lambda \in (0,1)$. Thus:

• If $S>r_i$ : $\widehat{\mu _i} > 0 \implies \mu _i ^* > 0$, it follows that $\widehat{k_i} = 0 \; \implies \; k_i ^* = 0$.

• If $S<r_i$ : $\mu _i ^* > 0 \implies \widehat{\mu _i} > 0$, it follows that $k_i ^* = 0 \; \implies \; \widehat{k_i} = 0$.

• If $S=r_i$ : $\mu _i ^* > 0 \Leftrightarrow \widehat{\mu _i} > 0$, it follows that $\widehat{k_i} = 0 \Leftrightarrow k_i ^* = 0$.

$\square$

Proof (Proof of Lemma 3)

Via the definition of $M$: $k_i >0 \;\;\forall i \in M$. Imposing the condition $\left( \frac{\partial F}{\partial k_i} \ge 0 \; \;i \in M \right)$ Eq. (13) implies: $\lambda \le \left( \frac{r_i +\alpha _i}{S+\alpha _i} \right) \; \forall i \in M$ if and only if $\lambda \in (0,1)$. From Lemma 1: $\lambda \;\in (0,1)$. Define $\overline{\lambda }$ as the largest value that satisfies $\left( \overline{\lambda } \le \frac{r_i +\alpha _i}{S+\alpha _i} \right) \; \forall i \in M$. Then $\lambda \in (0,\overline{\lambda })$. $\square$

1.1 2.1

Proof (Proof of Proposition 1)

Subtracting Eq. (14) from Eq. (13) forms a set of equalities given by:

$$\begin{aligned} \widehat{\frac{\partial F}{\partial k_i}} - \frac{\partial F}{\partial k_i}^* = (r_i - S) \left( \frac{\lambda }{1- \lambda } \right) \;\;\quad \forall i \in M \end{aligned}$$

(Recall that $\mu _i =0 \;\; \forall i \in M$.) Taking a linear approximation of the change on the left hand side of the above set of equalities, the set can be given in matrix notation as: $\mathcal {H} K_{\varDelta } = \mathcal {C}$ where $\mathcal {H}$ is the $MxM$ Hessian matrix associated with the capital inputs $i \in M$ of the revenue function ($F(K,L)$), $K_{\varDelta }$ is an $Mx1$ vector composed of elements $\widehat{k_i} - k_i ^*$ and $\mathcal {C}$ is a vector of values $(r_i - S) \left( \frac{\lambda }{1- \lambda } \right)$. As a Hessian matrix, $\mathcal {H}$ is a symmetric Hermitian matrix. The function $F(K,L)$ is assumed jointly concave in all capital inputs such that $\mathcal {H}$ is a negative definite matrix and $x^T \mathcal {H} x <0$ for any real vector x. It follows that: $K_{\varDelta }^T \mathcal {H} K_{\varDelta } = K_{\varDelta }^T \mathcal {C} < 0$

Note that the row and column orderings in the Hessian matrix $\mathcal {H}$ are arbitrary so long as symmetry is maintained. As a negative definite matrix, all upper left symmetric sub-matrices of $\mathcal {H}$ are themselves negative definite. Therefore the above equality holds for any sub-matrix of $\mathcal {H}$ and the inequality set implies that:

$$\begin{aligned} \left( \widehat{k_i} - k_i ^* \right) \cdot \left( (r_i - S) \left( \frac{\lambda }{1- \lambda } \right) \right) < 0 \;\; \quad \forall i \in M \end{aligned}$$

If an equilibrium exists, then via Lemma 1: $\lambda \in (0,1)$, thus:

$$\begin{aligned} \left\{ \begin{aligned} r_i < S \;\; \implies (r_i - S) \left( \frac{\lambda }{1- \lambda } \right) < 0 \;\; \implies \widehat{k_i} > k_i ^* \\ r_i > S \;\; \implies (r_i - S) \left( \frac{\lambda }{1- \lambda } \right) > 0 \;\; \implies \widehat{k_i} < k_i ^* \\ r_i = S \;\; \implies (r_i - S) \left( \frac{\lambda }{1- \lambda } \right) = 0 \;\; \implies \widehat{k_i} = k_i ^* \\ \end{aligned} \right\} \quad \forall i \in M \end{aligned}$$

To this point the proof is established for capital inputs $i \in M$. For the set $\left\{ i \in N | i \notin M \right\}$; following directly from Lemma 2 and the definition of the set $N$:

$$\begin{aligned} \left\{ \begin{aligned} r_i < S \;\;&\implies \widehat{k_i} > k_i ^* = 0\\ r_i > S \;\;&\implies k_i ^* >\widehat{k_i} =0\\ \end{aligned} \right\} \quad \forall i \in N | i \notin M \end{aligned}$$

Thus the set of inequalities holds for all $i \in N$:

$$\begin{aligned} \left\{ \begin{aligned} r_i < S \;\;&\implies \widehat{k_i} > k_i ^* \\ r_i > S \;\;&\implies \widehat{k_i} < k_i ^* \\ r_i = S \;\;&\implies \widehat{k_i} = k_i ^* \\ \end{aligned} \right\} \quad \forall i \in N \end{aligned}$$

$\square$

1.2 2.2

Proof (Proof of Proposition 2)

$$\begin{aligned} \begin{aligned} \left. \begin{aligned} r_i > r_j \\ \alpha _i = \alpha _j = \alpha \end{aligned} \right\}&\implies \;\; S r_j + \alpha r_j < S r_i + \alpha r_i \\&\implies \;\; S r_j + S \alpha -r_i r_j - \alpha r_i < S r_i + S \alpha -r_i r_j - \alpha r_j \\&\implies \;\; (S -r_i)( r_j + \alpha ) < (S- r_j) (r_i + \alpha ) \\ \end{aligned} \end{aligned}$$

From Lemma 3, if an equilibrium exists: $\lambda \in (0, \overline{\lambda })$.

$$\begin{aligned} \begin{aligned} \left. \begin{aligned} (S \!-\!r_i)( r_j \!+\! \alpha ) < (S\!-\! r_j) (r_i \!+ \!\alpha ) \\ \lambda \in (0, \overline{\lambda }) \end{aligned} \right\}&\implies \;\; \!-\! \lambda (S \!-\!r_i)( r_j \!+ \alpha ) > \!-\! \lambda (S- r_j) (r_i \!+\! \alpha ) \\&\implies \;\; \frac{(r_i + \alpha )(1- \lambda ) - \lambda (S - r_i)}{(r_j + \alpha )(1- \lambda ) - \lambda (S - r_j)}> \frac{r_i + \alpha }{r_j + \alpha } \\&\implies \;\; \frac{r_i + \alpha _i - \lambda (S + \alpha _i)}{r_j + \alpha _j - \lambda (S + \alpha _j)}> \frac{r_i + \alpha _i}{r_j + \alpha _j} \end{aligned} \end{aligned}$$

Substituting Eqs. (15) and (16) for the left and right hand side of the above inequality respectively, the inequality can be rewritten as: $\left\{ \widehat{\left( \frac{\partial k_j}{\partial k_i} \right) } > \left( \frac{\partial k_j}{\partial k_i} \right) ^* \right\} \forall i \ne j \in M$ (recall that $\widehat{\mu _i} = \mu _i ^* = 0$ if and only if $i \in M$). Imposing declining marginal revenue product for both inputs:

$$\begin{aligned} \left. \begin{aligned} \widehat{\left( \frac{\partial k_j}{\partial k_i} \right) } > \left( \frac{\partial k_j}{\partial k_i} \right) ^* \\ \frac{\partial F}{\partial k_i}>0 \; \quad \forall i,j\in N \\ \frac{\partial ^2 F}{\partial k_i ^2}<0 \; \quad \forall i,j\in N \end{aligned} \right\} \;\; \implies \;\; \left\{ \frac{\widehat{k_i}}{\widehat{k_j}} < \frac{k_i^*}{k_j^*} \right\} \quad \forall i \ne j \in M \end{aligned}$$

Following directly from Lemma 2 and the definition of the set $N$:

$$\begin{aligned} \left\{ \frac{k_j^*}{k_i^*} < \frac{\widehat{k_j}}{\widehat{k_i}} \right\} \forall \left\{ j \in N| j \notin M \right\} \;,\; \quad \forall i \in M \end{aligned}$$

although the fraction $\frac{k_j^*}{k_i^*}$ is undefined $\forall \left\{ i,j \in N | i,j \notin M \right\}$ Therefore, $S>r_i>r_j$ and $\alpha _i = \alpha _j \; \implies \;\left\{ \frac{k_j^*}{k_i^*} < \frac{\widehat{k_j}}{\widehat{k_i}} \right\} \forall i \ne j \in N$ if and only if $k_i ^* + k_j ^* >0$ . $\square$

1.3 2.3

Proof (Proof of Proposition 3)

$$\begin{aligned} \left. \begin{aligned} r = r_i = r_j \\ \alpha _i < \alpha _j \end{aligned} \right\} \implies \;\; (S-r)(r+ \alpha _j) > (S-r)(r + \alpha _i) \end{aligned}$$

From Lemma 3, if an equilibrium exists $\lambda \in (0, \overline{\lambda })$:

$$\begin{aligned} \begin{aligned} \left. \begin{aligned} (S\!-r)(r\!+\! \alpha _j) \!>\! (S-r)(r \!+\! \alpha _i) \\ \lambda \in (0, \overline{\lambda }) \end{aligned} \right\}&\implies \;\; \!-\lambda (S-r)(r\!+ \alpha _j) \!<\! -\!\lambda (S-r)(r \!+\! \alpha _i) \\&\implies \;\; \frac{(r+ \alpha _i) (1-\lambda ) - \lambda (S-r)}{(r+\alpha _j)(1-\lambda ) - \lambda (S-r)}< \frac{r + \alpha _i}{r+ \alpha _j} \\&\implies \;\; \frac{r_i + \alpha _i - \lambda (S + \alpha _i)}{r_j + \alpha _j - \lambda (S + \alpha _j)}< \frac{r_i + \alpha _i}{r_j + \alpha _j} \end{aligned} \end{aligned}$$

Substituting Eqs. (15) and (16) for the left and right hand side of the above inequality respectively, the inequality can be rewritten as: $\left\{ \widehat{\left( \frac{\partial k_j}{\partial k_i} \right) } < \left( \frac{\partial k_j}{\partial k_i} \right) ^* \right\} \forall i \ne j \in M$. Imposing declining marginal revenue product for both inputs:

$$\begin{aligned} \left. \begin{aligned} \widehat{\left( \frac{\partial k_j}{\partial k_i} \right) } < \left( \frac{\partial k_j}{\partial k_i} \right) ^* \\ \frac{\partial F}{\partial k_i}>0 \; \quad \forall i,j\in N \\ \frac{\partial ^2 F}{\partial k_i ^2}<0 \;\quad \forall i,j\in N \end{aligned} \right\} \;\;\implies \;\; \left\{ \frac{\widehat{k_i}}{\widehat{k_j}} > \frac{k_i^*}{k_j^*} \right\} \quad \forall i \ne j \in M \end{aligned}$$

From Lemma 2 if $S=r_i$ then $i \in M$ by definition. By extension, the only set of elements $i$ that satisfies $S=r_i$ and $i \in N | i \notin M$ is the null set. Therefore: $r = r_i = r_j$ and $\alpha _i < \alpha _j \; \implies \; \; \left\{ \frac{\widehat{k_i}}{\widehat{k_j}} > \frac{k_i^*}{k_j^*} \right\} \forall i \ne j \in N$. $\square$

1.4 2.4

Proof (Proof of Proposition 4)

$$\begin{aligned} \begin{aligned} (r_j + \alpha _j) - (r_i + \alpha _i) < \frac{r_i \alpha _j - r_j \alpha _i}{S} \;\;&\implies \;\; S r_j + S \alpha _j +\alpha _i r_j\\&\quad + \alpha _i \alpha _j < S r_i + S \alpha _i +r_i \alpha _j + \alpha _i \alpha _j \\&\implies \;\; (S\!+\!\alpha _i) (r_j\! +\! \alpha _j) \!<\! (S+\alpha _j) (r_i + \alpha _i) \end{aligned} \end{aligned}$$

From Lemma 1, if an equilibrium exists $\lambda \in (0, \overline{\lambda })$:

$$\begin{aligned} \begin{aligned}&\left. \begin{aligned} (S+\alpha _i) (r_j + \alpha _j) < (S+\alpha _j) (r_i + \alpha _i) \\ \lambda \in (0,\overline{\lambda }) \end{aligned} \right\} \\&\quad \implies \;\; (r_i + \alpha _i)(r_j + \alpha _j)- \lambda (S+\alpha _i) (r_j + \alpha _j)\\&\qquad > (r_i + \alpha _i)(r_j + \alpha _j)- \lambda (S+\alpha _j) (r_i + \alpha _i) \\&\quad \implies \;\; \frac{r_i + \alpha _i - \lambda (S + \alpha _i)}{r_j + \alpha _j - \lambda (S + \alpha _j)}> \frac{r_i + \alpha _i}{r_j + \alpha _j} \end{aligned} \end{aligned}$$

Substituting Eqs. (15) and (16) for the left and right hand side of the above inequality respectively, the inequality can be rewritten as:$\left\{ \widehat{\left( \frac{\partial k_j}{\partial k_i} \right) }> \left( \frac{\partial k_j}{\partial k_i} \right) ^* \right\} \forall i \ne j \in M$. Imposing declining marginal revenue product for both inputs:

$$\begin{aligned} \left. \begin{aligned} \widehat{\left( \frac{\partial k_j}{\partial k_i} \right) } > \left( \frac{\partial k_j}{\partial k_i} \right) ^* \\ \frac{\partial F}{\partial k_i}>0 \; \quad \forall i,j\in N \\ \frac{\partial ^2 F}{\partial k_i ^2}<0 \; \quad \forall i,j\in N \end{aligned} \right\} \;\;\implies \;\; \left\{ \frac{\widehat{k_i}}{\widehat{k_j}} < \frac{k_i^*}{k_j^*} \right\} \quad \forall i \ne j \in M \end{aligned}$$

The alternative effect; $(r_j + \alpha _j) - (r_i + \alpha _i) < (r_i \alpha _j - r_j \alpha _i) / S \;\; \implies \;\; \frac{\widehat{k_i}}{\widehat{k_j}} > \frac{k_i^*}{k_j^*} \;\; \forall i \ne j \in M$ follows from a parallel proof wherein the direction of the inequities is reversed. The null effect; $(r_j + \alpha _j) - (r_i + \alpha _i) = (r_i \alpha _j - r_j \alpha _i) / S \;\; \implies \;\; \frac{\widehat{k_i}}{\widehat{k_j}} = \frac{k_i^*}{k_j^*} \;\; \forall i \ne j \in M$ follows from a parallel proof wherein an equality is substituted for the inequalities. Via Lemma 2, using the same logic as in the preceding proofs, these results can be generalized to elements $i \ne j \in N$ unless $k_i ^* = k_j ^* = 0$. $\square$