Optimal dividend decisions with capital infusion in a dynamic nonterminal bankruptcy model

Abstract

We develop a stochastic dynamic model of dividend optimization under the conditions of a positive recovery, in which shareholders can recover a portion of their capital, and nonterminal bankruptcy due to private capital infusion or government bailout. In the presence of a recovery, the optimization problem becomes a mixed classical impulse stochastic control problem. We provide a closed-form solution for optimal dividend payout and timing under nonterminal bankruptcy. We take the model to the real data and show that this model explains the dividend puzzle during the financial crisis when the US government bailed out insurance companies and banks.

Appendices

Appendix A: Some properties of the optimal control function

1.1 A.1. Estimate parameter C

In the optimal policies of (56) and (57), the parameter C is an unknown parameter. In the real practical problems, we need to determine it.

1.1.1 A.1.1. Steps for computing parameter C

For the nonterminal model, we assume that the parameters $\mu$, $\sigma$, k, K and $\eta$ are known already. Then, the steps to compute C are given as follows.

• Step 1. Compare $\eta$ with $\frac{\mu }{2}$.

• Step 2. If $0\le \eta <\frac{\mu }{2}$, using the bisection method or Newton method to calculate C such that $I_1(C)=K$.

• Step 3. If $\eta \ge \frac{\mu }{2}$, using the bisection method or Newton method to calculate C such that $I_2(C)=K$.

By these steps, the uncertain parameter C can be obtained. Meanwhile, we can get other parameters, such as $x_1^C$ and $\widetilde{x}^{C}$, which are the left and right roots of equation $CH_1(x)=k$ or $CH_2(x)=k$, respectively.

Appendix B: Property of dividend events

1.1 B.1. Expected first dividend times after capital injections

Let $\phi _{x_1}(x)$ and $\psi _{0,x_1}(x)$ be defined as in (61) and (62) respectively, then the expected time $\psi _0^{x_1}(x)$ first reaches $x_1$ can be given by

$$\begin{aligned}\psi _0^{x_1}(x) = \frac{\psi _{0,x_1}(x)}{\phi _{x_1}(x)}. \end{aligned}$$

Proof

As in Chapter 15 of Karlin and Taylor (1981), the equations satisfied by $\phi _{x_1}(x)$ and $\psi _{0,x_1}(x)$ for $x\in (0, x_1)$ can be presented as,

$$\begin{aligned} \frac{1}{2}\sigma ^2 u^2(x) \frac{d^2 \phi _{x_1} }{dx^2} + \mu u(x) \frac{d \phi _{x_1} }{d x} = 0, \hspace{.5cm}\phi _{x_1}(0) = 0, \hspace{.5cm} \phi _{x_1}(x_1) = 1, \end{aligned}$$

and

$$\begin{aligned} \frac{1}{2}\sigma ^2 u^2(x) \frac{d^2 \psi _{0,x_1} }{dx^2} + \mu u(x) \frac{d \psi _{0,x_1} }{d x} = -1, \hspace{.5cm}\psi _{0,x_1}(0) = 0, \hspace{.5cm} \psi _{0,x_1}(x_1) = 0, \end{aligned}$$

respectively.

To solve these two equations above, we define the function $s:(0,\infty )\rightarrow [0,\infty )$ by

$$\begin{aligned} s(x):=\exp \left\{ -\int _{x_0^+}^x\frac{2\mu u(t)}{\sigma ^2u^2(t)}dt\right\} =\left\{ \begin{array}{lll} \left( \frac{\lambda x+\eta \gamma }{\lambda x_0^++\eta \gamma }\right) ^{2\gamma -2},&{} \quad x<x_0^+,\\ e^{-\frac{2\mu }{\sigma ^2}(x-x_0^+)}, &{}\quad x\ge x_0^+, \end{array}\right. \end{aligned}$$

the function $S:(0,\infty )\rightarrow [0,\infty )$ by

$$\begin{aligned} S(x):=\int _{x_0^+}^x s(\eta )d\eta =\left\{ \begin{array}{lll} \frac{\lambda x_0^++\eta \gamma }{\lambda (1-2\gamma )}\left[ 1-\left( \frac{\lambda x+\eta \gamma }{\lambda x_0^++\eta \gamma }\right) ^{2\gamma -1}\right] ,&{}\quad x<x_0^+,\\ \frac{\sigma ^2}{2\mu }\left[ 1-e^{-\frac{2\mu }{\sigma ^2}(x-x_0^+)}\right] ,&{}\quad x\ge x_0^+, \end{array}\right. \end{aligned}$$

and the function $m:(0,\infty )\rightarrow [0,\infty )$ by

$$\begin{aligned} m(x):=\frac{1}{\sigma ^2u^2(x)s(x)}=\left\{ \begin{array}{lll} \frac{\lambda ^2\sigma ^2(1-\gamma )^2}{\mu ^2(\lambda x_0^++\eta \gamma )^2}\left( \frac{\lambda x_0^++\eta \gamma }{\lambda x+\eta \gamma }\right) ^{2\gamma },&{}\quad x<x_0^+,\\ \frac{1}{\sigma ^2}e^{\frac{2\mu }{\sigma ^2}(x-x_0^+)},&{}\quad x\ge x_0^+. \end{array}\right. \end{aligned}$$

Then, by the similar method of Karlin and Taylor (1981), it follows that

$$\begin{aligned} \phi _{x_1}(x)= & {} \frac{S(x)-S(0)}{S(x_1)-S(0)}\\= & {} \left\{ \begin{array}{lll} \phi _{0,x_0^+}(x):=\frac{\frac{\lambda x_0^++\eta \gamma }{\lambda (1-2\gamma )}\left[ \left( \frac{\eta \gamma }{\lambda x_0^++\eta \gamma }\right) ^{2\gamma -1}-\left( \frac{\lambda x+\eta \gamma }{\lambda x_0^++\eta \gamma }\right) ^{2\gamma -1}\right] }{\frac{\sigma ^2}{2\mu }\left[ 1-e^{-\frac{2\mu }{\sigma ^2}(x_1-x_0^+)}\right] -\frac{\lambda x_0^++\eta \gamma }{\lambda (1-2\gamma )}\left[ 1-\left( \frac{\lambda \eta \gamma }{\lambda x_0^++\eta \gamma }\right) ^{2\gamma -1}\right] },&{} x<x_0^+,\\ \phi _{x_0^+,x_1}(x):= \frac{\frac{\sigma ^2}{2\mu }\left[ 1-e^{-\frac{2\mu }{\sigma ^2}(x-x_0^+)}\right] -\frac{\lambda x_0^++\eta \gamma }{\lambda (1-2\gamma )}\left[ 1-\left( \frac{\eta \gamma }{\lambda x_0^++\eta \gamma }\right) ^{2\gamma -1}\right] }{\frac{\sigma ^2}{2\mu }\left[ 1-e^{-\frac{2\mu }{\sigma ^2}(x_1-x_0^+)}\right] -\frac{\lambda x_0^++\eta \gamma }{\lambda (1-2\gamma )}\left[ 1-\left( \frac{\eta \gamma }{\lambda x_0^++\eta \gamma }\right) ^{2\gamma -1}\right] },&{}x\ge x_0^+.\nonumber \end{array}\right. \end{aligned}$$

For $x\in (0, x_0^+]$, let

$$\begin{aligned} \varphi _{0,x_0^+}^{-}(x):= & {} \int _0^x(S(t)-S(0))m(t)dt\\= & {} \frac{\sigma ^2(1-\gamma )^2}{(2\gamma -1)^2\mu ^2}\left[ \left( \frac{\eta \gamma }{\lambda x+\eta \gamma }\right) ^{2\gamma -1}+(2\gamma -1)\ln \left( \frac{\lambda x+\eta \gamma }{\eta \gamma }\right) -1\right] ; \end{aligned}$$

for $x\in [x_0^+,x_1]$, let

$$\begin{aligned} \varphi _{x_0^+,x_1}^{-}(x):= & {} \int _0^x(S(t)-S(0))m(t)dt\\= & {} \frac{\sigma ^2(1-\gamma )^2}{(2\gamma -1)^2\mu ^2}\left[ \left( \frac{\eta \gamma }{\lambda x_0^++\eta \gamma }\right) ^{2\gamma -1}+(2\gamma -1)\ln \left( \frac{\lambda x_0^++\eta \gamma }{\eta \gamma }\right) -1\right] \\{} & {} +\,\left( \frac{\sigma ^2}{4\mu ^2}-\frac{S(0)}{2\mu }\right) \left( e^{\frac{2\mu }{\sigma ^2}(x-x_0^+)}-1\right) -\frac{1}{2\mu }(x-x_0^+); \end{aligned}$$

for $x\in (0,x_0^+]$, let

$$\begin{aligned} \varphi _{0,x_0^+}^{+}(x):= & {} \int _x^{x_1}(S(x_1)-S(t))m(t)dt\\= & {} \frac{(1-\gamma )^2\sigma ^2}{(1-2\gamma )\mu ^2}\left( \frac{\lambda S(x_1)}{\lambda x_0^++\eta \gamma }-\frac{1}{1-2\gamma }\right) \left[ 1-\left( \frac{\lambda x+\eta \gamma }{\lambda x_0^++\eta \gamma }\right) ^{1-2\gamma }\right] \\{} & {} +\,\frac{(1-\gamma )^2\sigma ^2}{(1-2\gamma )\mu ^2}\ln \left( \frac{\lambda x_0^++\eta \gamma }{\lambda x+\eta \gamma }\right) +\left( \frac{S(x_1)}{2\mu }-\frac{\sigma ^2}{4\mu ^2}\right) \left( e^{\frac{2\mu }{\sigma ^2}(x_1-x_0^+)}-1\right) \\{} & {} +\,\frac{1}{2\mu }(x_1-x_0^+); \end{aligned}$$

for $x\in [x_0^+,x_1]$, let

$$\begin{aligned} \varphi _{x_0^+,x_1}^{+}(x):= & {} \int _x^{x_1}\left( S(x_1)-S(t)\right) m(t)dt\\= & {} \left( \frac{S(x_1)}{2\mu }-\frac{\sigma ^2}{4\mu ^2}\right) \left( e^{\frac{2\mu }{\sigma ^2}(x_1-x_0^+)} -e^{\frac{2\mu }{\sigma ^2}(x-x_0^+)}\right) +\frac{1}{2\mu }(x_1-x). \end{aligned}$$

Then, by the formula as in Chapter 15 of Karlin and Taylor (1981), $\psi _{0,x_1}(x)$ can be obtained as

$$\begin{aligned} \psi _{0,x_1}(x)=\left\{ \begin{array}{lll} 2\left( \phi _{0,x_0^+}(x)\cdot \varphi _{0,x_0^+}^+(x)+(1-\phi _{0,x_0^+}(x))\cdot \varphi _{0,x_0^+}^-(x)\right) ,&{}\quad x<x_0^+,\\ 2\left( \phi _{x_0^+,x_1}(x)\cdot \varphi _{x_0^+,x_1}^+(x)+(1-\phi _{x_0^+,x_1}(x))\cdot \varphi _{x_0^+,x_1}^-(x)\right) ,&{}\quad x\ge x_0^+. \end{array}\right. \end{aligned}$$

Let $\psi _{x_1}(x)$ be the expected time for both processes, which reach the boundary $x_1$ prior to 0. Meanwhile, for $\psi _0(x)$, it denotes the expected time the processes reach 0 prior to $x_1$. Then we have

$$\begin{aligned} \psi _{0,x_1}(x)=\phi _{x_1}(x)\psi _{x_1}(x)+(1-\phi _{x_1}(x))\psi _0(x). \end{aligned}$$

(64)

For the processes defined in (59) and (60), both of them have positive probability to reach zero points. But if the zero points are touched, then some capitals from outside will be injected to save the company. So, the expected time $\psi _0^{x_1}(x)$ for both processes, which finally reach the bound $x_1$, even they reach the lower bound 0 first, is complicated to calculate. If the process starting from bankruptcy state, then after a huge capital injection it would recover to the position x in a stochastic time period $\tau _s$, which is assumed to be a random variable with small mean and in simplification be neglected in our simulation examples. So, it follows that

$$\begin{aligned} \psi _0^{x_1}(x)=\phi _{x_1}(x)\psi _{x_1}(x)+(1-\phi _{x_1}(x))(\psi _0(x)+\psi _0^{x_1}(x)). \end{aligned}$$

(65)

From (64) and (65), we have that

$$\begin{aligned} \psi _0^{x_1}(x)=\frac{\psi _{0,x_1}(x)}{\phi _{x_1}(x)}. \end{aligned}$$

So, the closed form of $\psi _0^{x_1}(x)$ is obtained. $\square$

1.2 B.2. Variance of $\tau _{x_1}^*$

Before X(t) reaches the upper boundary $x_1$, it may arrive at the lower boundary 0 first, and in this case we assume that the process X(t) would rebound to the original position x in a negligible time after bailouts. Similar to (64), it follows that

$$\begin{aligned} E_x\left( (\tau _{x_1}^*)^2 \right) = P_x(\tau _{x_1} < \tau _0)\cdot E_x(\tau _{x_1}^2)+ P_x(\tau _{x_1} > \tau _0)\cdot \left[ E_x(\tau _0^2)+ E_x((\tau _{x_1}^*)^2 )\right] . \end{aligned}$$

Then, we have that

$$\begin{aligned} E_x((\tau _{x_1}^*)^2 ) = \frac{E_x(\tau _{0, x_1}^2)}{P_x(\tau _{x_1} < \tau _0)}. \end{aligned}$$

Consequently,

$$\begin{aligned} \textrm{Var}_x(\tau _{x_1}^*) = E_x((\tau _{x_1}^*)^2 ) - (E_x(\tau _{x_1}^* ))^2 = \frac{E_x(\tau _{0, x_1}^2)}{P_x(\tau _{x_1}< \tau _0)} -\frac{(E_x(\tau _{0, x_1}))^2}{(P_x(\tau _{x_1} < \tau _0))^2}. \end{aligned}$$

(66)

To derive $E_x(\tau _{0, x_1}^2)$, let $U(x) = E_x(\tau _{0, x_1}^2)$, then as in Chapter 15 of Karlin and Taylor (1981), the function U(x) satisfies

$$\begin{aligned} \frac{1}{2} \sigma ^2 u^2(x) U''(x)+\mu u(x) U'(x) + 2 \cdot \psi _{0, x_1}(x) = 0, \hspace{.5cm} U(0) = 0, \hspace{.5cm} U(x_1) = 0. \end{aligned}$$

As in Appendix B.1, the solution of the equation above can be derived and given as follows

$$\begin{aligned} U(x)&= 4\phi _{x_1}(x) \cdot \int _x^{x_1} [S(x_1)-S(t)]m(t) \psi _{0, x_1}(t)dt \\&\quad + 4(1-\phi _{x_1}(x)) \cdot \int _0^x [S(t)-S(0)]m(t) \psi _{0, x_1}(t) dt, \end{aligned}$$

where $\phi _{x_1}(x)$, S(t), m(t) and $\psi _{0, x_1}(t)$ are presented in Appendix B.1 and we don’t list them out again. Then the variance of $\tau _{x_1}^*$ can be obtained from (66) and

$$\begin{aligned} \sigma _{\tau _{x_1}}^2:=\textrm{Var}_x(\tau _{x_1}^*) = \frac{U(x)}{\phi _{x_1}(x)} - \frac{\psi ^2_{0,x_1}(x)}{\phi ^2_{x_1}(x)}. \end{aligned}$$

(67)