Financing the newsvendor: raising the loan limit by insurance contract

Abstract

Banks often control their risks to be below a risk limit through setting a loan limit and the capital constrained newsvendor can make the loan limit increase by buying a guarantee insurance policy. This paper examines the impact of a bank’s risk limit, interest rate setting and initial capital on the newsvendor’s financial and ordering decisions with deductible insurance contract. In the perfectly competitive bank market, it is shown that the newsvendor will restore his profit to the optimal level without capital constraint by buying a full insurance. In the regulated monopolistic bank market, the newsvendor only buys insurance when both his initial capital and the bank’s risk limit are low; for a poorer newsvendor, the bank should require a lower risk limit to force the newsvendor to buy insurance, not just to meet regulatory requirements. It is also shown that the insurance is more useful for a poor newsvendor and a more risk-averse bank.

Appendix

Proof of Proposition 1

Define $A = (1 - \alpha )(wQ + m(b) - B)$, we have $A < pz(Q,b)$ always holds. If the newsvendor borrows money, the bank’s risk is as follows (Fig. 8):

$$\begin{aligned} & \Pr \{ Y(Q,b) \le A\} \\ & \quad = \Pr \left\{ {b < A\left| {0 \le \xi < \frac{b}{p}} \right.} \right\}\Pr \left( {0 \le \xi < \frac{b}{p}} \right) \\ & \quad \quad + \,\Pr \left\{ {p\xi \le A\left| {\frac{b}{p} \le \xi < z(Q,b)} \right.} \right\}\Pr \left( {\frac{b}{p} \le \xi < z(Q,b)} \right) \\ & \quad \quad + \,\Pr \{ (wQ + m(b) - B)(1 + r) \le A|\xi \ge z(Q,b)\} \Pr (\xi \ge z(Q,b)) \\ & \quad = \,\Pr \left\{ {b < A\left| {0 \le \xi < \frac{b}{p}} \right.} \right\}F\left( {\frac{b}{p}} \right) + \Pr \left\{ {p\xi \le A\left| {\frac{b}{p} \le \xi < z(Q,b)} \right.} \right\}\left[ {F(z(Q,b)) - F\left( {\frac{b}{p}} \right)} \right] \\ \end{aligned}$$

• 1. If $\frac{A}{p} < \frac{b}{p}$, that is, $Q < \frac{B}{w} + \frac{b}{w(1 - \alpha )} - \int_{0}^{{\frac{b}{p}}} {\frac{p}{w}F(x)dx} \equiv Q_{2} (b)$, we have $\Pr \{ Y(Q,b) \le A\} = 0$, which means the bank can control risk all the time.

• 2. If $\frac{A}{p} \ge \frac{b}{p}$, that is, $Q \ge \frac{B}{w} + \frac{b}{w(1 - \alpha )} - \int_{0}^{{\frac{b}{p}}} {\frac{p}{w}F(x)dx} \equiv Q_{2} (b)$, we have $\Pr \{ Y(Q,b) \le A\} = F\left( {\frac{b}{p}} \right) + \Pr \left( {\frac{b}{p} < \xi < \frac{A}{p}} \right) = F\left( {\frac{A}{p}} \right)$. Thus, $F\left( {\frac{A}{p}} \right) \le \beta$, that is, $Q \le \frac{{B{ + }L(\beta )}}{w} - \int_{0}^{{\frac{b}{p}}} {\frac{p}{w}F(x)dx} \equiv Q_{3} (b)$, must hold to control the bank’s risk.

Fig. 8

The different location of $\frac{A}{p}$

It is easy to see that $Q_{2} (b)$ increases with $b$, $Q_{3} (b)$ decreases with $b$. Because $Q_{2} (0) < Q_{3} (0)$, there exists a unique insurance coverage level $b_{0} = pF^{ - 1} (\beta )$, such that the newsvendor’s order limit is $Q_{3} (b)$ for $b \le b_{0}$ and $Q_{2} (b)$ for $b > b_{0}$.

Proof of Lemma 1

Because $Q_{2} (b_{1} ) = \frac{{B{ + }L(\beta )}}{w}$, we have (10) holds. Using the Implicit Function Theorem of (10), we have

$$\frac{{db_{1} }}{d\beta } = \frac{1}{{f(F^{ - 1} (\beta ))\left[ {\frac{1}{p(1 - \alpha )} - \frac{1}{p}F\left( {\frac{{b_{1} }}{p}} \right)} \right]}} > 0$$

Thus, $b_{1}$ increases with $\beta$.

Proof of Proposition 2

When the newsvendor borrows money from the bank, the bank’s risk can be defined as

$$\begin{aligned} & \Pr \{ Y(Q) \le (1 - \alpha )(wQ - B)\} \\ & \quad = \,\Pr \{ p\xi \le (1 - \alpha )(wQ - B)|0 \le \xi < z(Q)\} \Pr (0 \le \xi < z(Q)) \\ & \quad \quad + \,\Pr \{ (wQ - B)(1 + r(Q)) \le (1 - \alpha )(wQ - B)|\xi \ge z(Q)\} \Pr (\xi \ge z(Q)) \\ & \quad = \,\Pr \left\{ {\left. {\xi \le \frac{(1 - \alpha )(wQ - B)}{p}} \right|0 \le \xi < z(Q)} \right\}F(z(Q)) \\ & \quad = \,F\left( {\frac{(1 - \alpha )(wQ - B)}{p}} \right) \\ \end{aligned}$$

Thus, the bank will set the loan limit to let the newsvendor’s order limit equal $Q_{1}$. Because $Q_{1}$ increases with $B$, the bank’s risk control does not work if $L(\beta ) \ge wQ^{N}$. If $L(\beta ) < wQ^{N}$, the newsvendor’s optimal order quantity is $Q^{*} = \hbox{min} (Q_{1} ,Q^{N} )$.

Proof of Lemma 2

Define

$$M(Q,b) = p\left[ {z(Q,b) - \int_{{\frac{b}{p}}}^{z(Q,b)} {F(x)dx} } \right] - [wQ + m(b) - B]$$

Based on Implicit Function Theorem, we have

$$\frac{\partial r(Q,b)}{\partial b} = - \frac{\partial M(Q,b)/\partial b}{\partial M(Q,b)/\partial r} = - \frac{{F\left( {\frac{b}{p}} \right)(1 + r(Q,b))}}{wQ + m(b) - B} < 0$$

$$\frac{\partial z(Q,b)}{\partial b} = \frac{1 + r(Q,b)}{p}\frac{dm(b)}{db} + \frac{wQ + m(b) - B}{p}\frac{\partial r(Q,b)}{\partial b} = 0$$

which implies the results.

Proof of Proposition 4

If the newsvendor borrows money, the bank’s risk is the same as that in the perfectly competitive bank market. Thus, the bank will set the loan limit such that the newsvendor’s order limit is $Q_{1}$. Using the Implicit Function Theorem of (14), we have

$$\begin{aligned} \frac{{dQ^{LO} }}{dB} & = \frac{{\frac{{w(1 + r)^{2} }}{p}f(z(Q^{LO} ))}}{{\frac{{w^{2} (1 + r)^{2} }}{p}f(z(Q^{LO} )) - pf(Q^{LO} )}} \\ & = \frac{{w(1 + r)^{2} f(z(Q^{LO} ))}}{{p[1 - F(Q^{LO} )][w(1 + r)h(z(Q^{LO} )) - ph(Q^{LO} )]}} < 0 \\ \end{aligned}$$

Thus, $Q^{LO}$ decreases with $B$. When $B = wQ^{M}$, $Q^{LO} = Q^{M}$, so $Q^{LO} > Q^{M}$ for $B < wQ^{M}$.

Because $Q_{1}$ increases with $B$, the bank’s risk control constraint does not work if $L(\beta ) \ge wQ_{ 0}^{LO}$. If $L(\beta ) < wQ_{ 0}^{LO}$, when $B = wQ^{M}$, $Q^{LO} = Q^{M} < Q_{1}$, so there exists a unique $B_{0} \in (0,wQ^{M} )$ satisfies $p[1 - F(Q_{1} (B_{0} ))] = w(1 + r)[1 - F(z(Q_{1} (B_{0} )))]$ such that $Q^{LO} > Q_{1}$ for $B < B_{0}$ and $Q^{LO} < Q_{1}$ for $B > B_{0}$. Besides, the conclusion that $B_{0}$ decreases with $\beta$ can be easily obtained from Lemma 5.

Proof of Lemma 3

The first-order derivatives of $\varPi_{M}^{LI} (Q,b)$ with respect to $b$ and $Q$ are

$$\frac{{\partial \varPi_{M}^{LI} (Q,b)}}{\partial b} = - [1 - F(z(Q,b))]F\left( {\frac{b}{p}} \right)(1 + r) < 0$$

$$\frac{{\partial \varPi_{M}^{LI} (Q,b)}}{\partial Q} = p[1 - F(Q)] - w(1 + r)[1 - F(z(Q,b))]$$

Given $b$, $\frac{{\partial \varPi_{M}^{LI} (Q,b)}}{\partial Q}$ decreases with $Q$ at $Q = Q^{LI} (b)$.

Using the Implicit Function Theorem of (17), we have

$$\frac{{dQ^{LI} (b)}}{db} = \frac{{w(1 + r)^{2} f(z(Q^{LI} (b),b))F\left( {\frac{b}{p}} \right)}}{{p[1 - F(Q^{LI} (b))][ph(Q^{LI} (b)) - w(1 + r)h(z(Q^{LI} (b),b))]}} > 0$$

Thus, $Q^{LI} (b)$ increases with $b$.

Proof of Lemma 4

For $B < B_{0}$,

$$\varPi_{M}^{LI} (Q_{3} (b),b) = p\left[ {Q_{3} (b) - \frac{1 + r}{1 - \alpha }F^{ - 1} (\beta )} \right] - \int_{{\frac{1 + r}{1 - \alpha }F^{ - 1} (\beta )}}^{{Q_{3} (b)}} {pF(x)dx} - B$$

$$\varPi_{M}^{LI} (Q^{LI} (b),b) = p[Q^{LI} (b) - z(Q^{LI} (b),b)] - \int_{{z(Q^{LI} (b),b)}}^{{Q^{LI} (b)}} {pF(x)dx} - B$$

$$\varPi_{M}^{LI} (Q_{2} (b),b) = p\left[ {Q_{2} (b) - \frac{b(1 + r)}{p(1 - \alpha )}} \right] - \int_{{\frac{b(1 + r)}{p(1 - \alpha )}}}^{{Q_{2} (b)}} {pF(x)dx} - B$$

Because

$$\frac{{d\varPi_{M}^{LI} (Q_{3} (b),b)}}{db} = - p[1 - F(Q_{3} (b))]\frac{1}{w}F\left( {\frac{b}{p}} \right) < 0,$$

$$\frac{{d\varPi_{M}^{LI} (Q^{S} (b),b)}}{db} = \frac{{\partial \varPi_{M}^{LI} (Q^{LI} (b),b)}}{{\partial Q^{LI} (b)}}\frac{{dQ^{LI} (b)}}{db} + \frac{{\partial \varPi_{M}^{LI} (Q^{LI} (b),b)}}{\partial b} = \frac{{\partial \varPi_{M}^{LI} (Q^{LI} (b),b)}}{\partial b} < 0$$

Both $\varPi_{M}^{LI} (Q_{3} (b),b)$ and $\varPi_{M}^{LI} (Q^{LI} (b),b)$ decrease with $b$.

Because$\frac{{d\varPi_{M}^{LI} (Q_{2} (b),b)}}{db} = p[1 - F(Q_{2} (b))]\left[ {\frac{1}{w(1 - \alpha )} - \frac{1}{w}F\left( {\frac{b}{p}} \right)} \right] - \left[ {1 - F\left( {\frac{b(1 + r)}{p(1 - \alpha )}} \right)} \right]\frac{1 + r}{1 - \alpha }$, and $b^{S}$ satisfies the first-order optimality condition,

$$\begin{aligned} & \left. {\frac{{d^{2} \varPi_{M}^{LI} Q_{2} (b),b)}}{{db^{2} }}} \right|_{{b = b^{S} }} = - pf(Q_{2} (b^{S} ))\left[ {\frac{1}{w(1 - \alpha )} - \frac{1}{w}F\left( {\frac{b}{p}} \right)} \right]^{2} - [1 - F(Q_{2} (b^{S} ))]\frac{1}{w}f\left( {\frac{{b^{S} }}{p}} \right) + pf(z(Q_{2} (b^{S} ),b))\frac{{(1 + r)^{2} }}{{p^{2} (1 - \alpha )^{2} }} \\ & \quad = - p[1 - F(Q_{2} (b^{S} ))]\left\{ {\frac{{f(Q_{2} (b^{S} ))}}{{1 - F(Q_{2} (b^{S} ))}}\left[ {\frac{1}{w(1 - \alpha )} - \frac{1}{w}F\left( {\frac{{b^{S} }}{p}} \right)} \right]^{2} - \frac{{f(z(Q_{2} (b^{S} ),b))}}{{1 - F(Q_{2} (b^{S} ))}}\frac{{(1 + r)^{2} }}{{p^{2} (1 - \alpha )^{2} }}} \right\} - [1 - F(Q_{2} (b^{S} ))]\frac{1}{w}f\left( {\frac{b}{p}} \right) \\ & \quad = - p[1 - F(Q_{2} (b^{S} ))]\left\{ {h(Q_{2} (b^{S} ))\left[ {\frac{1}{w(1 - \alpha )} - \frac{1}{w}F\left( {\frac{{b^{S} }}{p}} \right)} \right]^{2} } \right. \\ & \quad \quad \left. { - \,h(z(Q_{2} (b^{S} ),b))\frac{(1 + r)}{p(1 - \alpha )}\left[ {\frac{1}{w(1 - \alpha )} - \frac{1}{w}F\left( {\frac{{b^{S} }}{p}} \right)} \right]} \right\} - [1 - F(Q_{2} (b^{S} ))]\frac{1}{w}f\left( {\frac{b}{p}} \right) \\ & \quad = - p[1 - F(Q_{2} (b^{S} ))]\left[ {\frac{1}{w(1 - \alpha )} - \frac{1}{w}F\left( {\frac{{b^{S} }}{p}} \right)} \right] \\ & \quad \quad \quad \left\{ {h(Q_{2} (b^{S} ))\left[ {\frac{1}{w(1 - \alpha )} - \frac{1}{w}F\left( {\frac{{b^{S} }}{p}} \right)} \right] - h(z(Q_{2} (b^{S} ),b))\frac{1 + r}{p(1 - \alpha )}} \right\} - [1 - F(Q_{2} (b^{S} ))]\frac{1}{w}f\left( {\frac{b}{p}} \right) \\ \end{aligned}$$

From (21), we can see that $\frac{1}{w(1 - \alpha )} - \frac{1}{w}F\left( {\frac{{b^{S} }}{p}} \right) \ge \frac{1 + r}{p(1 - \alpha )}$ holds since $Q_{2} (b^{S} ) \ge z(Q_{2} (b^{S} ),b^{S} ) = \frac{{b^{S} (1 + r)}}{p(1 - \alpha )}$. Thus, $\left. {\frac{{d^{2} \varPi_{M}^{LI} (Q_{2} (b),b)}}{{db^{2} }}} \right|_{{b = b^{S} }} < 0$ which means $\varPi_{M}^{LI} (Q_{2} (b),b)$ is concave and obtains the maximal value in $b = b^{S}$.

Using the Implicit Function Theorem of (21), we have

$$\frac{{db^{S} }}{dB} = \frac{{ph(Q_{2} (b^{S} ))\left[ {\frac{1}{w(1 - \alpha )} - \frac{1}{w}F\left( {\frac{{b^{S} }}{p}} \right)} \right]}}{{ - f\left( {\frac{{b^{S} }}{p}} \right) + \left[ {\frac{1}{1 - \alpha } - F\left( {\frac{{b^{S} }}{p}} \right)} \right]\left\{ {\frac{1 + r}{1 - \alpha }h\left( {\frac{{b^{S} (1 + r)}}{p(1 - \alpha )}} \right) - \left[ {\frac{1}{w(1 - \alpha )} - \frac{1}{w}F\left( {\frac{{b^{S} }}{p}} \right)} \right]ph(Q_{2} (b^{S} ))} \right\}}}$$

Because $Q_{2} (b^{S} ) \ge z(Q_{2} (b^{S} ),b^{S} ) = \frac{{b^{S} (1 + r)}}{p(1 - \alpha )}$ and $\frac{1}{w(1 - \alpha )} - \frac{1}{w}F\left( {\frac{{b^{S} }}{p}} \right) \ge \frac{1 + r}{p(1 - \alpha )}$, $\frac{{db^{S} }}{dB} < 0$ always holds.

Define $H(b) = \frac{{d\varPi_{M}^{LI} (Q_{2} (b),b)}}{db}$ and we have

$$\begin{aligned} H(b_{2} ) & = p[1 - F(Q_{2} (b_{2} ))]\left[ {\frac{1}{w(1 - \alpha )} - \frac{1}{w}F\left( {\frac{{b_{2} }}{p}} \right)} \right] - [1 - z(Q_{2} (b_{2} ),b_{2} )]\frac{1 + r}{1 - \alpha } \\ & = (1 + r)\left[ {1 - F\left( {\frac{{b_{2} (1 + r)}}{p(1 - \alpha )}} \right)} \right]\left[ {\frac{1}{1 - \alpha } - F\left( {\frac{{b_{2} }}{p}} \right)} \right] - \left[ {1 - F\left( {\frac{{b_{2} (1 + r)}}{p(1 - \alpha )}} \right)} \right]\frac{1 + r}{1 - \alpha } \\ & = - \left[ {1 - F\left( {\frac{{b_{2} (1 + r)}}{p(1 - \alpha )}} \right)} \right](1 + r)F\left( {\frac{{b_{2} }}{p}} \right) < 0 \\ \end{aligned}$$

Thus, $b^{S} < b_{2}$ holds. Similarly, we have $H(0) = p\left[ {1 - F\left( {\frac{B}{w}} \right)} \right]\frac{1}{w(1 - \alpha )} - \frac{1 + r}{1 - \alpha }$. When $B < B_{0}$, we have $B < wQ^{M}$, $H(0) > 0$. Thus, $b^{S} > 0$ holds.

Proof of Lemma 5

The first-order derivative of $\varPi_{M}^{LO} (Q_{ 1} )$ with respect to $\beta$ is

$$\frac{{d\varPi_{M}^{LO} (Q_{ 1} )}}{d\beta } = \frac{p}{{(1 - \alpha )f(F^{ - 1} (\beta ))}}G(\beta )$$

where

$$G(\beta ) = \frac{p}{w}[1 - F(Q_{1} )] - (1 + r)\left[ {1 - F\left( {\frac{1 + r}{1 - \alpha }F^{ - 1} (\beta )} \right)} \right]$$

$\beta_{1}$ is the solution of $G(\beta ) = 0$. Because

$$\begin{aligned} G^{{\prime }} (\beta )|_{{\beta = \beta_{1} }} & = \frac{1}{{(1 - \alpha )f(F^{ - 1} (\beta ))}}\left\{ { - \frac{{p^{2} }}{{w^{2} }}f(Q_{1} ) + (1 + r)^{2} f\left( {\frac{1 + r}{1 - \alpha }F^{ - 1} (\beta )} \right)} \right\} \\ & = \frac{{(1 + r)[1 - F(Q_{1} )]}}{{(1 - \alpha )f(F^{ - 1} (\beta ))}}\left\{ {(1 + r)h(Q_{1} ) - \frac{p}{w}h\left( {\frac{1 + r}{1 - \alpha }F^{ - 1} (\beta )} \right)} \right\} < 0 \\ \end{aligned}$$

we have

$$\frac{{d\varPi_{M}^{LO} (Q_{1} )}}{d\beta }|_{{\beta = \beta_{1} }} = \left[ {\frac{p}{{(1 - \alpha )f(F^{ - 1} (\beta_{1} ))}}} \right]^{{\prime }} G(\beta_{1} ) + \frac{p}{{(1 - \alpha )f(F^{ - 1} (\beta_{1} ))}}G^{{\prime }} (\beta_{1} ) < 0$$

Thus, $\varPi_{M}^{LO} (Q_{1} )$ obtains the maximal value at $\beta = \beta_{1}$. Using the Implicit Function Theorem of (22), we have

$$\frac{{d\beta_{1} }}{dB} = \frac{{\frac{1}{w}h(Q_{1} )}}{{\frac{1}{{(1 - \alpha )f(F^{ - 1} (\beta_{1} ))}}\left\{ {(1 + r)h\left( {\frac{1 + r}{1 - \alpha }F^{ - 1} (\beta )} \right) - \frac{p}{w}h(Q_{1} )} \right\}}} < 0$$

Thus, $\beta_{1}$ decreases with $B$. When $B = wQ^{M}$, $\beta_{1} = 0$, so $\beta_{1} > 0$ for $B < B_{0}$ since $B_{0} < wQ^{M}$.

Proof of Proposition 7

When $\beta = 0$, let

$$T(B) = \varPi_{M}^{LO} (Q_{1} ) - \varPi_{M}^{LI} (Q_{2} (b^{S} ),b^{S} ) = pS\left( {\frac{B}{w}} \right) - B - \varPi_{M}^{LI} (Q_{2} (b^{S} ),b^{S} )$$

and we have

$$\begin{aligned} \frac{dT(B)}{dB} & = \frac{p}{w}\left[ {1 - F\left( {\frac{B}{w}} \right)} \right] - 1 - \frac{{\partial \varPi_{M}^{LI} (Q_{2} (b^{S} ),b^{S} )}}{{\partial b^{S} }}\frac{{db^{S} }}{dB} - \frac{p}{w}[1 - F(Q_{2} (b^{S} ))] + 1 \\ & = \frac{p}{w}\left[ {F(Q_{2} (b^{S} )) - F\left( {\frac{B}{w}} \right)} \right] > 0 \\ \end{aligned}$$

Thus, $T(B)$ increases with $B$. When $B = wQ^{M}$, $b^{S} = 0$, so $T(B) = 0$. Thus, $T(B) < 0$ always holds for $B < B_{0}$ since $T(0) < 0$ and $B_{0} \le wQ^{M}$. That is to say, $\varPi_{M}^{LO} (Q_{1} ) < \varPi_{M}^{LI} (Q_{2} (b^{S} ),b^{S} )$ holds for $\beta = 0$.

When $\beta = \beta_{1}$, i.e. $B = B_{0}$, $Q_{ 1} = Q^{LO}$, so we have

$$\varPi_{M}^{LO} (Q_{1} ) = \varPi_{M}^{LO} (Q^{LO} ) > \varPi_{M}^{LO} (Q_{2} (b^{S} )) > \varPi_{M}^{LI} (Q_{2} (b^{S} ),b^{S} )$$

Because $\varPi_{M}^{LI} (Q_{2} (b^{S} ),b^{S} )$ is independent of $\beta$ and $\varPi_{M}^{LO} (Q_{1} )$ increases with $\beta$ for $B < B_{0}$, there exist a unique value $\beta_{2} \in (0,\beta_{1} )$, such that $\varPi_{M}^{LO} (Q_{1} ) < \varPi_{M}^{LI} (Q_{2} (b^{S} ),b^{S} )$ for $\beta < \beta_{2}$ and $\varPi_{M}^{LO} (Q_{1} ) > \varPi_{M}^{LI} (Q_{2} (b^{S} ),b^{S} )$ for $\beta < \beta_{2}$.