Liquidity management with decreasing returns to scale and secured credit line

This paper examines the dividend and investment policies of a cash constrained firm, assuming a decreasing-returns-to-scale technology and adjustment costs. We extend the literature by allowing the firm to draw on a secured credit line both to hedge against cash-flow shortfalls and to invest/disinvest in a productive asset. We formulate this problem as a two-dimensional singular control problem and use both a viscosity solution approach and a verification technique to get qualitative properties of the value function. We further solve quasi-explicitly the control problem in two special cases.

1. The spread may be justified by the cost of equity capital for the bank. Indeed, the full commitment to supply liquidity up to the firm’s credit limit prevents bank’s shareholders to allocate part of their equity capital to more valuable investment opportunities.

2. Ly Vath et al. [22] have also studied a reversible investment problem in two alternative technologies for a cash-constrained firm that has no access to external funding.

3. The extension to the case of variable size will be studied in Sect. 4.

4. Under a credit line agreement, banks must block part of their funds to provide liquidity. This prevents banks from seizing new opportunities and this especially as the demand for funds is high. When the banking system is not competitive, banks charge an increasing credit line spread to offset the opportunity costs.

5. This assumption is standard in models with cash. It captures in a simple way the agency costs; see [8, 18] for more details.

The authors gratefully acknowledge the financial support of the research initiative IDEI-SCOR “Risk Market and Creation Value” under the aegis of the risk foundation and the chair Finance and Sustainable Development (IEF sponsored by EDF and CA).

Authors and Affiliations

1. EDF Lab Paris-Saclay, 7 Boulevard Gaspard Monge, 91120, Palaiseau, France

   Erwan Pierre

2. Toulouse School of Economics (CRM-IDEI), Manufacture des Tabacs, 21, Allée de Brienne, 31000, Toulouse, France

   Stéphane Villeneuve

3. EDF Lab Paris-Saclay & FiME, Laboratoire de Finance des Marchés de l’Energie, 7 Boulevard Gaspard Monge, 91120, Palaiseau, France

   Xavier Warin

Corresponding author

Correspondence to Stéphane Villeneuve.

Appendix

1.1 6.1 Proof of Theorem 5.4

1. Supersolution property. Let \((\bar{x},\bar{k}) \in S\) and \(\varphi\in C^{2}(\mathbb{R}_{+}^{2})\) be such that \((\bar{x},\bar{k})\) is a minimum of \(V^{*}-\varphi\) in a neighbourhood \(B_{\varepsilon}(\bar {x},\bar{k})\) of \((\bar{x},\bar{k})\) with \(\varepsilon\) small enough to ensure \(B_{\varepsilon}\subset S\) and \(V^{*}(\bar{x},\bar{k}) = \varphi(\bar{x},\bar{k})\).

First, consider the admissible control \(\hat{\pi} = (\hat{Z}, \hat{I})\), where the shareholders decide to never invest or disinvest, while the dividend policy is defined by \(\hat{Z}_{t} = \eta\) for \(t \geq0\), with \(0 \leq\eta \leq \varepsilon\). Define the exit time \(\tau_{\varepsilon}=\inf\{t \geq0: (X_{t}^{\bar{x}},K_{t}^{\bar{k}}) \notin\overline{B}_{\varepsilon}(\bar {x},\bar{k}) \}\). We notice that \(\tau_{\varepsilon}< \tau_{0}\) for \(\varepsilon\) small enough. From the dynamic programming principle, we have

Applying Itô’s formula to the process \((e^{-r t}\varphi(X_{t}^{\bar{x}},K_{t}^{\bar{k}}))\) between 0 and \(\tau_{\varepsilon}\wedge h\) and taking expectations, we obtain

Combining (6.1) and (6.2), we have

⋆:

First take \(\eta= 0\). We then observe that \(X\) is continuous on \([\![0, \tau_{\varepsilon}\wedge h]\!]\) and only the first term of (6.3) is nonzero. By dividing the above inequality by \(h\) and letting \(h \rightarrow0\), we conclude that \(- \mathcal{L} \varphi(\bar{x},\bar{k}) \geq0\).

⋆:

Now take \(\eta> 0\) in (6.3). We see that \(\hat{Z}\) jumps only at \(t = 0\) with size \(\eta\), so that

$$\mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}\wedge h} e^{-r t}( - \mathcal{L} \varphi )(X_{t}^{\bar{x}},K_{t}^{\bar{k}}) dt \bigg] - \eta-\big(\varphi(\bar{x} - \eta,\bar{k}) - \varphi(\bar {x},\bar {k})\big) \geq0. $$

By sending \(h \rightarrow0\) and then dividing by \(\eta\) and letting \(\eta\rightarrow0\), we obtain

$$\frac{\partial\varphi}{\partial x}(\bar{x},\bar{k}) - 1 \geq0. $$

Second, consider the admissible control \(\bar{\pi} = (\bar{Z}, \bar{I})\), where the shareholders decide to never pay out dividends, while the investment/disinvestment policy is defined by \(\bar{I}_{t}=\eta\in\mathbb{R}\) for \(t\) ≥ 0, with \(0<|\eta| \leq \varepsilon\). Define again the exit time \(\tau_{\varepsilon}=\inf\{t \geq0: (X_{t}^{\bar{x}},K_{t}^{\bar{k}}) \notin\overline{B}_{\varepsilon}(\bar {x},\bar {k}) \}\). Proceeding analogously as in the first part and observing that \(\bar{I}\) jumps only at \(t=0\), we get

Assuming first \(\eta>0\), by sending \(h \rightarrow0\), and then dividing by \(\eta\) and letting \(\eta\rightarrow0\), we obtain

When \(\eta<0\), we get in the same manner

This proves the required supersolution property.

2. Subsolution property. We prove the subsolution property by contradiction. Suppose that the claim is not true. Then there exist \((\bar{x},\bar{k}) \in S\) and a neighbourhood \(B_{\varepsilon}(\bar {x},\bar{k})\) of \(\bar{x},\bar{k}\), included in \(S\) for \(\varepsilon\) small enough, a \(C^{2}\) function \(\varphi\) with \((\varphi- V^{*})(\bar{x},\bar{k})= 0\) and \(\varphi\geq V^{*}\) on \(B_{\varepsilon}(\bar{x},\bar{k})\), and \(\eta> 0\) such that we have, for all \((x,k) \in B_{\varepsilon}(\bar{x},\bar{k})\),

For any admissible control \(\pi\), the exit time \(\tau_{\varepsilon}= \inf\{ t \geq0: (X_{t}^{\bar {x}},K_{t}^{\bar{k}}) \notin B_{\varepsilon}(\bar{x},\bar{k}) \}\) satisfies \(\tau_{\varepsilon}< \tau_{0}\). Applying Itô’s formula to the process \((e^{-r t}\varphi(X_{t}^{\bar{x}},K_{t}^{\bar{k}}))\) between 0 and \(\tau_{\varepsilon}{-}\), we have

Using (6.4)–(6.7), we obtain

Note that \(\Delta X_{s} = -\Delta Z_{s} - \gamma( \Delta I_{s}^{+} + \Delta I_{s}^{-} )\), \(\Delta K_{s} = \Delta I_{s}^{+} - \Delta I_{s}^{-}\) and by the mean value theorem, there is some \(\theta\in(0,1)\) such that

Because \((X_{s}+\theta\Delta X_{s}, K_{s}+\theta\Delta K_{s}) \in B_{\varepsilon}(\bar {x},\bar{k})\), we use (6.5)–(6.7) again to get

Therefore,

Notice that while \((X_{\tau_{\varepsilon}-}, K_{\tau_{\varepsilon}-}) \in B_{\varepsilon}(\bar {x},\bar{k})\), \((X_{\tau_{\varepsilon}},K_{\tau_{\varepsilon}})\) is either on the boundary \(\partial B_{\varepsilon}(\bar{x},\bar{k})\) or outside of \(\bar{B}_{\varepsilon}(\bar{x},\bar{k})\). However, there is some random variable \(\alpha\) valued in \([0,1]\) such that

is in \(\partial B_{\varepsilon}(\bar{x},\bar{k})\). Proceeding analogously as above, we show that

Observe that

Starting from \((X^{(\alpha)},K^{(\alpha)})\), the strategy that consists of investing \((1-\alpha)\Delta I_{\tau_{\varepsilon}}^{+}\) or disinvesting \((1-\alpha)\Delta I_{\tau_{\varepsilon}}^{-}\), depending on the sign of \(K^{(\alpha )}-K_{\tau_{\varepsilon}}\), and paying out \((1-\alpha)\Delta Z_{\tau _{\varepsilon}}\) as dividends leads to \((X_{\tau_{\varepsilon}},K_{\tau_{\varepsilon}})\), and therefore

Using \(\varphi(X^{(\alpha)},K^{(\alpha)}) \geq V^{*}(X^{(\alpha )},K^{(\alpha)})\), we deduce

Hence,

We now claim there is \(c_{0} > 0\) such that for any admissible strategy,

Let us consider the \(C^{2}\) function \(\phi(x,k) = c_{0}(1-\frac{(x-\bar {x})^{2}}{\varepsilon^{2}})\) with

where

This function satisfies

Applying Itô’s formula, we have

Noting that \(\frac{\partial\phi}{\partial x} \leq1\) and \(\frac {\partial\phi}{\partial k}= 0\), we have

Plugging this into (6.10) with \(\phi(X^{(\alpha )},K^{(\alpha )})=0\), we obtain

This proves the claim (6.9). Finally, by taking the supremum over \(\pi\) and using the dynamic programming principle, (6.8) implies \(V^{*}(\bar{x},\bar{k}) \geq V^{*}(\bar{x},\bar{k}) + \eta c_{0}\), which is a contradiction.

3. Uniqueness. Suppose \(u\) is a continuous subsolution and \(w\) a continuous supersolution of (5.1) on \(S\) satisfying the boundary conditions

and the linear growth condition

for some positive constants \(C_{1}\) and \(C_{2}\). We show by adapting some standard arguments that \(u \le w\).

Step 1. We first construct a strict supersolution of (5.1) with a perturbation of \(w\). Set

with

and

and define for \(\lambda\in[0,1]\) on \(S\) the continuous function

Because

and

we have that

which implies that \(w^{\lambda}\) is a strict supersolution of (5.1). To prove this point, one only needs to take \(\bar{x}\) and \(\varphi\in C^{2}\) such that \(\bar{x}\) is a minimum of \(w^{\lambda}- \varphi\) and notice that \(\bar{x}\) is also a minimum of \(w^{\lambda}- \varphi_{2}\) with \(\varphi_{2} = \frac{\varphi- \lambda h}{1-\lambda}\), which allows us to use that \(w\) is a viscosity supersolution of (5.1).

Step 2. In order to prove the strong comparison result, it suffices to show that for every \(\lambda\in[0,1]\),

Assume by way of contradiction that there exists \(\lambda\) such that

Because \(u\) and \(w\) have linear growth, we have

Using the boundary conditions

and the linear growth condition, it is always possible to find \(C_{1}\) in (6.11) such that both expressions above are negative and the maximum in (6.13) is reached inside the domain \(S\). By continuity of the functions \(u\) and \(w^{\lambda}\), there exists a pair \((x_{0},k_{0})\) with \(x_{0} \ge\gamma k_{0}\) such that

For \(\epsilon> 0\), consider the functions

By standard arguments in the comparison principle of the viscosity solution theory (see Pham [26, Sect. 4.4.2]), the function \(\varPhi_{\varepsilon}\) attains a maximum in \((x_{\epsilon},y_{\epsilon},k_{\epsilon},\ell_{\epsilon})\), which converges (up to a subsequence) to \((x_{0}, k_{0},x_{0},k_{0})\) when \(\varepsilon\) goes to zero. Moreover,

Applying Theorem 3.2 in Crandall et al. [7], we get the existence of symmetric square matrices \(M_{\varepsilon}\), \(N_{\varepsilon}\) of size 2 such that

and

where

and

so that

Equation (6.14) implies that

Because \(u\) and \(w^{\lambda}\) are respectively a subsolution and a strict supersolution, we have

and

We then distinguish the following four cases:

Case 1.:

If \(\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon }+(x_{\epsilon}-x_{0})^{3} - 1 \leq0\), we get from (6.16) that \(\lambda + (x_{\epsilon}- x_{0})^{3} \leq0\), yielding a contradiction when \(\epsilon \) goes to 0.

Case 2.:

If \(\gamma(\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon }+(x_{\epsilon}-x_{0})^{3}) - (\frac{k_{\epsilon}-\ell_{\epsilon}}{\epsilon }+(k_{\epsilon}-k_{0})^{3}) \leq0\), we get from (6.16) that \(\lambda+ \gamma((x_{\epsilon}-x_{0})^{3} -(k_{\epsilon}-k_{0})^{3}) \leq0\), yielding a contradiction when \(\epsilon\) goes to 0.

Case 3.:

If \(\gamma(\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon }+(x_{\epsilon}-x_{0})^{3}) + (\frac{k_{\epsilon}-l_{\epsilon}}{\epsilon }+(k_{\epsilon}-k_{0})^{3}) \leq0\), we get from (6.16) that \(\lambda+ \gamma((x_{\epsilon}-x_{0})^{3} +(k_{\epsilon}-k_{0})^{3} )\leq 0\), yielding a contradiction when \(\epsilon\) goes to 0.

Case 4.:

If

$$\begin{aligned} &-\Big(\beta(k_{\epsilon})\mu- \alpha\big((k_{\epsilon}-x_{\epsilon})^{+}\big)\Big) \bigg(\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon}+(x_{\epsilon}-x_{0})^{3}\bigg) \\ &- \text{tr}\bigg(\frac{\sigma^{2} \beta(k_{\epsilon})^{2}}{2}M_{\epsilon}\bigg) + r u(x_{\epsilon},k_{\epsilon}) \leq0, \end{aligned}$$

we deduce from

$$-\Big(\beta(\ell_{\epsilon})\mu- \alpha\big((\ell_{\epsilon}-y_{\epsilon})^{+}\big)\Big) \frac{x_{\epsilon}-y_{\epsilon}}{\epsilon} - \text{tr}\bigg(\frac{\sigma^{2} \beta(\ell_{\epsilon})^{2}}{2}N_{\epsilon}\bigg) + r w^{\lambda}(y_{\epsilon},\ell_{\epsilon}) \geq\lambda $$

that

$$\begin{aligned} &\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon}\Big(\mu\big(\beta(\ell _{\epsilon})-\beta(k_{\epsilon})\big)+\alpha\big((k_{\epsilon}-x_{\epsilon})^{+}\big)-\alpha\big((\ell_{\epsilon}-y_{\epsilon})^{+}\big)\Big)\\ &-\text{tr}\bigg(\frac{\sigma^{2}\beta(k_{\epsilon})^{2}}{2}N_{\epsilon}\bigg) + \text{tr}\bigg(\frac{\sigma^{2}\beta(k_{\epsilon})^{2}}{2}N_{\epsilon}\bigg) \\ &-\Big(\beta(k_{\epsilon})\mu- \alpha\big((k_{\epsilon}-x_{\epsilon})^{+}\big)\Big) (x_{\epsilon}-x_{0})^{3} \\ &+ r\big(u(x_{\epsilon},k_{\epsilon}) - w^{\lambda}(y_{\epsilon},\ell _{\epsilon})\big)\leq-\lambda. \end{aligned}$$

Using (6.15), we get

$$\begin{aligned} &\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon}\Big(\mu\big(\beta(\ell _{\epsilon})-\beta(k_{\epsilon})\big)+\alpha\big((k_{\epsilon}-x_{\epsilon})^{+}\big)-\alpha\big((\ell_{\epsilon}-y_{\epsilon})^{+}\big)\Big)\\ &-\Big(\beta(k_{\epsilon})\mu- \alpha\big((k_{\epsilon}-x_{\epsilon})^{+}\big)\Big) (x_{\epsilon}-x_{0})^{3} + r\big(u(x_{\epsilon},k_{\epsilon}) - w^{\lambda}(y_{\epsilon},\ell_{\epsilon})\big) \\ &\leq-\lambda+ \frac{3\sigma^{2}}{2\epsilon}\big(\beta(k_{\epsilon})^{2} - \beta(\ell_{\epsilon})^{2}\big) + \frac{9\sigma^{2} \beta(k_{\epsilon})^{2}}{2} (x_{\epsilon}- x_{0})^{2}\big(1+\epsilon(x_{\epsilon}- x_{0})^{2}\big). \end{aligned} $$

By sending \(\varepsilon\) to zero and using the continuity of \(u\), \(w^{\gamma}_{i}\), \(\alpha\) and \(\beta\) we obtain the required contradiction, namely \(r M \leq-\lambda\).

We have thus shown for every \(\lambda\in[0,1]\) that (6.12) holds, i.e., \(\sup_{S} (u - w^{\lambda}) \leq0\). This implies by letting \(\lambda\to0\) the strong comparison result \(u \leq w\) for any subsolution \(u\) and supersolution \(w\). Clearly, this strong comparison result implies uniqueness. This ends the proof of Theorem 5.4.  □

1.2 6.2 Proof of Proposition 5.5

Because \(\beta\) is concave and \(\beta'\) goes to 0, the existence of \(a\) is equivalent to assuming

Let us define the function \(w_{A}\) for \(A>0\) as the unique solution on \((a, \infty)\) of the Cauchy problem

with \(w_{A}(x)=Ax^{\delta}\) for \(0 \le x \le a\) and \(w_{A}\) differentiable at \(a\).

Remark 6.1

The above Cauchy problem is well defined with the condition that \(w_{A}\) is differentiable at \(a\). Moreover, it is easy to check, using the definition of \(a\), that the function \(w_{A}\) is also \(C^{2}\). Because the spread \(\alpha\) is high, the shareholders optimally choose not to tap the credit line, but rather adjust costlessly their level of investment.

Lemma 6.2

For every \(A>0\), the function \(w_{A}\) is increasing.

Proof

Clearly, \(w_{A}\) is increasing and therefore positive on \([0,a]\). If we define \(c\) by \(c=\min\{x> a:w'_{A}(c)=0\}\), then \(w_{A}(c)>0\) because \(w_{A}\) is increasing and positive in a left neighbourhood of \(c\). Thus, according to the differential equation, we have \(w_{A}''(c)\ge0\) which implies that \(w_{A}\) is also increasing in a right neighbourhood of \(c\). Therefore, \(w'_{A}\) cannot become negative. □

Lemma 6.3

For every \(A>0\), there is some \(b_{A}\) such that \(w''_{A}(b_{A})=0\) and \(w_{A}\) is a concave function on \((a, b_{A})\).

Proof

Assume by way of contradiction that \(w_{A}''\) does not vanish. Using (5.6) and (5.5), we have

Therefore, we equivalently assume that \(w_{A}''<0\). This implies that \(w_{A}'\) is strictly decreasing and bounded below by 0 by Lemma 6.2; therefore \(w_{A}\) is an increasing concave function. Therefore, \({\lim_{x\to\infty}}w_{A}'(x)\) exists and is denoted by \(\ell \). Letting \(x \to\infty\) in the differential equation, we obtain, because \(\beta\) has a finite limit,

Therefore, either \(\lim_{x\to\infty}w_{A}(x)\) is \(+\infty\), from which we get a contradiction, or finite, from which we get \({\lim_{x\to \infty }}w_{A}''(x)=0\) by the mean value theorem. In the second case, differentiating the differential equation, we have

Proceeding analogously, we obtain that \({\lim_{x\to\infty }}w_{A}'''(x)=0\) and thus \(\ell=0\). Coming back to the differential equation, we get

contradicting the fact that \(w_{A}\) is increasing. Setting \(b_{A}:=\inf\{x \ge a:w_{A}''(x)=0\}\) allows us to conclude. □

Lemma 6.4

There exists \(A^{*}\) such that \(w'_{A^{*}}(b_{A^{*}})=1\).

Proof

For every \(A>0\), we have

Let \(A_{1}=\frac{\mu\bar{\beta}}{ra^{\delta}}\). Lemma 6.2 yields

Therefore, (6.19) yields \(w'_{A_{1}}(b_{A_{1}})\ge1\). On the other hand, let \(A_{2}=\frac{a^{1-\delta}}{\delta}\). By construction, \(w'_{A_{2}}(a)=1\) and thus \(w'_{A_{2}}(b_{A_{2}}) \le1\) by the concavity of \(w_{A}\) on \((0,b_{A})\). Thus, there is some \(A^{*} \in[\min (A_{1},A_{2}),\max(A_{1},A_{2})]\) such that \(w'_{A^{*}}=1\). □

Hereafter, we set \(b=b_{A^{*}}\).

Lemma 6.5

We have \(\mu\beta'(b)\le r\).

Proof

Differentiating the differential equation for \(w_{A}\) and plugging in \(x=b\), we get

Because \(w''_{A^{*}}\) is increasing in a left neighbourhood of \(b\), we have \(w_{A}'''(b) \ge0\), implying the result. □

Let us define

We are in a position to prove the following result.

Proposition 6.6

The shareholders’ value is \(v\).

Proof

We have to check that \((v,b)\) satisfies the HJB free boundary problem

By construction, \(v\) is a \(C^{2}\) concave function on \((0, \infty)\) satisfying \(v'\ge1\). It remains to check that \(\max_{k}\mathcal {L}_{k}v(x)\le0\). For \(x >b\), we have

If \(k\le x\), concavity of \(\beta\) and Lemma 6.5 imply

If \(k \ge x\), we differentiate \(\mathcal{L}_{k}v(x)\) with respect to \(k\) and obtain, using again concavity of \(\beta\) and convexity of \(\alpha\), that

Therefore, \(\mathcal{L}_{k}v(x) \le\mathcal{L}_{x}v(x)\le0\).

Let \(x < b\). Because \(v\) is concave, the same argument as in the previous lines shows that

and therefore

The first order condition gives for \(0 \leq k < x\) that

Thus for \(0< x< a\), we have

which gives

Therefore the maximum \(k^{*}(x)\) of \(\mathcal{L}_{k}v(x)\) lies in the interior of the interval \([0,x]\) and satisfies

Hence, for \(x \le a\), we have by construction

Now fix \(x \in(a,b)\). Note that \({\frac{\partial}{\partial k}}(\mathcal{L}_{k}v)(x)\) has the same sign as \(\mu v'(x) + \sigma^{2}\beta(k)v''(x) \) because \(\beta\) is strictly increasing. Moreover, since \(v\) is concave and \(\beta\) increasing, we have

Thus, it suffices to prove \(\mu v'(x) + \sigma^{2}\beta(x)v''(x) \geq0\) for \(x \in(a,b)\), or equivalently, because \(\beta\) is a positive function, that the function \(\phi\) defined as

is positive. We make a proof by contradiction, assuming there is some \(x\) such that \(\phi(x) < 0\). As \(\phi(a) = 0\) by (5.5) and \(\phi(b) > 0\), there is some \(x_{1} \in[a,b]\) such that

Using the differential equation (6.18) satisfied by \(v'\), we obtain

from which we deduce that

But \(x_{1} \geq a\) and thus \(\beta'(x_{1}) \leq\beta'(a)\). Moreover, by the definition of \(a\), we have \(\sigma^{2}\beta'(a) \le\frac{ \mu }{1-\delta}\). Therefore, (5.6) yields

which is a contradiction. □

To complete the characterization of the shareholders’ value when the spread is high, we have to study the optimal policy when (6.17) is not fulfilled. We expect that \(a=0\) in that case, which means that for all \(x\), the manager should invest all the cash in the productive asset. Thus we are interested in the solutions to

such that \(w(0)=0\).

Proposition 6.7

Suppose that the functions \(x \mapsto\frac{x}{\beta(x)}\) and \(x \mapsto\frac{x^{2}}{\beta(x)^{2}}\) are analytic in 0 with a radius of convergence \(R\). Then the solutions \(w\) to (6.21) such that \(w(0)=0\) are given by

with

where the functions \(p\) and \(q\) are

the function \(I\) is given by

and \(y_{1}\) is the positive root of \(I\) given by

The radius of convergence of \(w\) is at least equal to \(R\).

Proof

This result is given by the Fuchs’ theorem [1, Sect. 9.5]. □

Note that choosing \(A_{0} > 0\) is enough to characterize a unique solution \(w\) of (6.21) because the \((A_{k})_{k\geq1}\) are given by a recurrence relation. Moreover, because \(\mu\beta'(0) \geq r\), we have \(y_{1} < 1\). As a consequence, if we choose \(A_{0}>0\), we have

Thus, proceeding analogously as in Lemma 6.3, we can prove the existence of \(b\) such that \(w''(b) = 0\) for all \(A_{0} > 0\). Now, observe that \(b\) does not depend on \(A_{0}\) so that we can choose \(A_{0} = A^{*}\) in order to have \(w'(b) = 1\). Hence, we have built a concave solution \(w^{*}\) to (6.21) with \(w^{*}(0) = 0\), \((w^{*})'(b) = 1\) and \((w^{*})''(b) = 0\). We extend \(w^{*}\) linearly on \([b,\infty)\) as usual to obtain a \(C^{2}\) function on \([0,\infty)\).

Proposition 6.8

The shareholders’ value is \(w^{*}\).

Proof

It suffices to check that \(w^{*}\) satisfies the free boundary problem (6.20). By construction, \(w^{*}\) is a \(C^{2}\) concave function on \((0,\infty)\). Because \((w^{*})'(b) = 1\), we have

and

On \([b, \infty)\), we have

Using that \(\beta\) is concave increasing, \(\alpha\) is convex and \(\alpha '(0+)> \mu\beta'(0+)\), we have

Then using the concavity of \(\beta\),

It remains to show that for every \(x< b\)

Using that \(\beta\) is concave, \(\alpha\) is convex, \(\alpha'(0)> \mu \beta '(0)\) and \(w^{*}\) is concave increasing, we have for all \(k>x\) that

Thus,

Moreover, for \(0 < k < x\),

We expect for all \(x \in(0,b]\) and all \(k \leq x\) that

Notice that \(\beta'(k) \geq0\) and

because \((w^{*})''(x) \leq0\) and \(\beta\) is increasing. Thus it is enough to prove that for every \(x< b\),

or equivalently, using \(\beta\geq0\),

We make a proof by contradiction, assuming the existence of \(x\) such that \(\phi(x) < 0\). In a neighbourhood of 0, we have

and

From this we deduce, because \(\beta(x)x^{y_{1}-1}\le\beta'(0)x^{y_{1}}\), that

yielding

But \(\phi(b) > 0\); thus there is \(x_{1} \in(0,b)\) such that

To get the expression of \(\phi'\), we differentiate (6.21) to obtain

Then using that \(\phi''(x_{1}) = 0\), we get that

from which we deduce

Now, remember that \(x_{1} > 0\) and thus, by the concavity of \(\beta\), we have \(\beta'(x_{1}) \leq\beta'(0)\). Furthermore, \(\beta'(0) \leq\frac{\mu^{2}+2r \sigma^{2}}{\sigma^{2}\mu}\) when (5.5) is not fulfilled. Hence,

which yields a contradiction and ends the proof. □

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