Financial conditions and supply decisions when firms are risk averse

Abstract

Extending earlier work by Greenwald and Stiglitz (Q J Econ 108:77–114, 1993) on the role of a firm’s equity position and bankruptcy costs in determining its production decision we show that, even if bankruptcy costs are ignored, a firm’s decision makers’ risk aversion, whether they are owner-entrepreneurs or hired managers, can give rise to the same results. What is more, we argue that, in the presence of risk aversion, increased variance of the output price affects a firm’s supply decision as the sum of an impact and an indirect effect. Under reasonable assumptions the impact effect prevails and then output decreases. We show this to hold for risk attitudes represented both by CARA and by CRRA utility functions. Finally, we explore the dynamics of the equity base. We provide examples in which the accumulation of net worth slows down as a consequence of an increase of risk.

Appendix

A)Derivation of condition (13) We have $\pi \ge 0$ if and only if

$$\begin{aligned} u\ge \frac{\left( 1+r\right) w\phi \left( q\right) }{q}-\frac{\left( 1+r\right) a}{q} \end{aligned}$$

(26)

and we know that, for any w and r, the firm always chooses $q\le {\hat{q}} =\left( \phi ^{\prime }\right) ^{-1}\left( \frac{1}{(1+r)w}\right)$. Moreover, by de l’Hospital $\lim _{q\rightarrow 0}\frac{\left( 1+r\right) w\phi \left( q\right) }{q} = \lim _{q\rightarrow 0}\left( 1+r\right) w\phi ^{\prime }\left( q\right) = 0$, while

$$\begin{aligned} \frac{\partial }{\partial q}\left( \frac{\left( 1+r\right) w\phi \left( q\right) }{q}\right) =\frac{q\left( 1+r\right) w\phi ^{\prime }\left( q\right) -\left( 1+r\right) w\phi \left( q\right) }{q^{2}} \end{aligned}$$

is positive if $\phi ^{\prime }\left( q\right) >\phi \left( q\right) /q$ which is true due to $\phi \left( 0\right) =0$ and strict convexity of $\phi$. Thus $\left( 1+r\right) w\phi \left( q\right) /q$ is strictly increasing and on $\left[ 0,{\hat{q}}\right]$ reaches its maximum at ${\hat{q}}.$ Therefore, and since $\left( 1+r\right) a/q\ge 0$,

$$\begin{aligned} supp\left( f\right) \sqsubseteq \left[ \frac{\left( 1+r\right) w\phi \left( \left( \phi ^{\prime }\right) ^{-1}\left( \frac{1}{\left( 1+r\right) w} \right) \right) }{\left( \phi ^{\prime }\right) ^{-1}\left( \frac{1}{\left( 1+r\right) w}\right) },\infty \right) , \end{aligned}$$

i.e. (13), is a sufficient condition for (26) to hold for all $u\in supp\left( f\right)$. Finally, to show

$$\begin{aligned} \frac{\left( 1+r\right) w\phi \left( \left( \phi ^{\prime }\right) ^{-1}\left( \frac{1}{\left( 1+r\right) w}\right) \right) }{\left( \phi ^{\prime }\right) ^{-1}\left( \frac{1}{\left( 1+r\right) w}\right) }<1 \end{aligned}$$

(27)

let $f_{1}\left( q\right) =\left( 1+r\right) w\phi \left( q\right)$ and $f_{2}\left( q\right) =q$. Then $f_{1}\left( 0\right) =0=f_{2}\left( 0\right)$, $f_{1}^{\prime }\left( 0\right) =0<1=f_{2}^{\prime }\left( 0\right)$, $f_{1}^{\prime }\left( {\hat{q}}\right) =1=f_{2}^{\prime }\left( {\hat{q}}\right)$ and $f_{1}^{\prime }\left( q\right) <f_{2}^{\prime }\left( q\right)$ for all q s.t. $0\le q<{\hat{q}}$. But this implies

$$\begin{aligned} f_{1}\left( {\hat{q}}\right) =\int _{0}^{{\hat{q}}}f_{1}^{\prime }\left( q\right) dq<\int _{0}^{{\hat{q}}}f_{2}^{\prime }\left( q\right) dq=f_{2}\left( {\hat{q}} \right) \end{aligned}$$

which is equivalent to (27). $\square$

B)$\textit{Derivation}\, \textit{of}\, sgn\left( \partial q^{\star }/\partial \sigma _{u}^{2}\right) \textit{in}\, \textit{case}\, \textit{CRRA}$ We can calculate $\partial q^{\star }/\partial \sigma _{u}^{2}$ as A/B, with A given by (16) and

$$\begin{aligned} B=\mu ^{\prime \prime }-\frac{1-\gamma }{2}\left\{ \phantom {\left. -\left( 2-\gamma \right) \frac{q^{\star }}{\mu }\frac{ \mu \left( \mu ^{\prime }+q^{*}\mu ^{\prime \prime }\right) -q^{*}\left( \mu ^{\prime }\right) ^{2}}{\mu ^{2}}\right\} \sigma _{u}^{2}.} \frac{\mu -q^{\star }\mu ^{\prime }}{\mu ^{2}}\left[ 2-\left( 2-\gamma \right) \frac{q^{*}}{\mu } \mu ^{\prime }\right] \right. \nonumber \\ \left. -\left( 2-\gamma \right) \frac{q^{\star }}{\mu }\frac{ \mu \left( \mu ^{\prime }+q^{*}\mu ^{\prime \prime }\right) -q^{*}\left( \mu ^{\prime }\right) ^{2}}{\mu ^{2}}\right\} \sigma _{u}^{2}. \end{aligned}$$

As already seen, A is positive for small $\sigma _{u}^{2}$. Thus the sign of $\partial q^{\star }/\partial \sigma _{u}^{2}$ is the same as that of B . To evaluate the latter, we write it, setting $\mu ^{\prime }/\mu =z^{\prime }$ and $\mu ^{\prime \prime }/\mu =z^{\prime \prime },$ as

$$\begin{aligned} B= & {} \mu ^{\prime \prime }-\frac{1-\gamma }{2}\left\{ \frac{1-q^{*}z^{\prime }}{\mu }\left[ 2-\left( 2-\gamma \right) q^{*}z^{\prime } \right] \right. \nonumber \\&\left. -\left( 2-\gamma \right) \frac{q^{*}}{\mu }\left[ z^{\prime }+q^{*}z^{\prime \prime }-q^{*}\left( z^{\prime }\right) ^{2}\right] \right\} \sigma _{u}^{2} =\mu ^{\prime \prime }-\frac{1-\gamma }{2\mu }C\sigma _{u}^{2} \end{aligned}$$

(28)

where

$$\begin{aligned} C=2\left( 1-q^{\star }z^{\prime }\right) \left[ 1-\left( 2-\gamma \right) q^{*}z^{\prime }\right] -\left( 2-\gamma \right) \left( q^{\star }\right) ^{2}z^{\prime \prime }. \end{aligned}$$

Since $\mu ^{\prime \prime }$ is negative, for B to be negative it is sufficient that $C>0$ or that $\frac{1-\gamma }{2\mu }C\sigma _{u}^{2}$ is small. The latter is certainly the case for $\sigma _{u}^{2}$ small and/or $\gamma$ close to (but not larger than) one.

In the subsequent example we show that in the case $\phi \left( q\right) =q^{2}$ no constraints have to be imposed on $\sigma _{u}^{2}$ and $\gamma$ .

Example 3

Let us assume that the firm’s labour requirement function is $\phi \left( q\right) =q^{2}$ as in Example 2. Then we can evaluate the expression C as follows. Since $\phi ^{\prime }\left( q\right) =2q$ and $\phi ^{\prime \prime }\left( q\right) =2,\phi ^{\prime }\left( q\right) q=\phi ^{\prime \prime }\left( q\right) q^{2}.$ Moreover, $\mu ^{\prime }=1-\left( 1+r\right) w\phi ^{\prime }\left( q\right)$ and $\mu ^{\prime \prime }=-\left( 1+r\right) w\phi ^{\prime \prime }\left( q\right) .$ Thus

$$\begin{aligned} q^{2}z^{\prime \prime }= & {} \frac{-\left( 1+r\right) w\phi ^{\prime \prime }\left( q\right) q^{2}}{\mu }=\frac{-\left( 1+r\right) w\phi ^{\prime }\left( q\right) q}{\mu }=\frac{\left[ 1-\left( 1+r\right) w\phi ^{\prime }\left( q\right) \right] q}{\mu }-\frac{q}{\mu }\\= & {} \frac{\mu ^{\prime }q}{\mu } -\frac{q}{\mu }=qz^{\prime }-z, \end{aligned}$$

with $z=q/\mu .$ Therefore

$$\begin{aligned} C= & {} 2\left( 1-qz^{\prime }\right) \left[ 1-\left( 2-\gamma \right) qz^{\prime }\right] -\left( 2-\gamma \right) \left( qz^{\prime }-z\right) \\= & {} 2\left[ 1-\left( 2-\gamma \right) qz^{\prime }\right] \left( 1-qz^{\prime }\right) -\left( 2-\gamma \right) \left( qz^{\prime }-1+1-z\right) \\= & {} \left\{ 2\left[ 1-\left( 2-\gamma \right) qz^{\prime }\right] +2-\gamma \right\} \left( 1-qz^{\prime }\right) +\left( 2-\gamma \right) \left( z-1\right) \\= & {} \left[ 4-\gamma -2\left( 2-\gamma \right) qz^{\prime }\right] \left( 1-qz^{\prime }\right) +\left( 2-\gamma \right) \left( z-1\right) . \end{aligned}$$

Notice that

$$\begin{aligned} z-1=\frac{q-\mu }{\mu }=\frac{\left( 1+r\right) w\phi \left( q\right) -\left( 1+r\right) a}{\mu }>0. \end{aligned}$$

Therefore, a sufficient condition for $C>0\,$is that the expression in square brackets is positive which occurs if

$$\begin{aligned} \frac{4-\gamma }{2\left( 2-\gamma \right) }>qz^{\prime }. \end{aligned}$$

This inequality is always satisfied in the case of non-negative relative risk aversion. In fact, the LHS is larger than or equal to 1 while the RHS is smaller than 1. To see the latter, recall that

$$\begin{aligned} qz^{\prime }=\frac{q-\left( 1+r\right) w\phi ^{\prime }\left( q\right) q}{ q-\left( 1+r\right) w\phi \left( q\right) +\left( 1+r\right) a}. \end{aligned}$$

But $\left( 1+r\right) w\phi ^{\prime }\left( q\right) q > \left( 1+r\right) w\phi \left( q\right) - \left( 1+r\right) a$ whenever $\left( 1+r\right) w \times \left[ \phi ^{\prime }\left( q\right) q-\phi \left( q\right) \right] +\left( 1+r\right) a$ is positive which is true since $\phi \left( q\right)$ is convex and $\phi \left( 0\right) =0.$ Therefore $C>0.$ From this follows that the RHS of (28) and thus B and $\partial q^{\star }/\partial \sigma _{u}^{2}$ are negative.

C)Derivation of condition (19) From (17) and (18) the first-order condition is

$$\begin{aligned}&\mu ^{\gamma -1}\left( \frac{1}{\sigma _{u}}-2\left( 1+r\right) w\frac{ \sigma }{\sigma _{u}^{2}}\right) \\&\quad -\frac{1-\gamma }{2}\frac{\mu ^{2-\gamma }2\sigma -\sigma ^{2}\left( 2-\gamma \right) \mu ^{1-\gamma }\left( \frac{1}{ \sigma _{u}}-2\left( 1+r\right) w\frac{\sigma }{\sigma _{u}^{2}}\right) }{ \mu ^{4-2\gamma }}=0 \end{aligned}$$

which is equivalent to

$$\begin{aligned}&\mu ^{\gamma -1}\frac{1}{\sigma _{u}}\left( 1-2\left( 1+r\right) w\frac{ \sigma }{\sigma _{u}}\right) \\&\qquad -\frac{1-\gamma }{2}\mu ^{\gamma -3}\sigma \left[ 2\mu -\left( 2-\gamma \right) \frac{\sigma }{\sigma _{u}}\left( 1-2\left( 1+r\right) w\frac{\sigma }{\sigma _{u}}\right) \right] =0 \end{aligned}$$

$\Leftrightarrow$

$$\begin{aligned} \mu ^{2}\frac{1}{\sigma _{u}}\left( 1-2\left( 1+r\right) w\frac{\sigma }{ \sigma _{u}}\right) -\frac{1-\gamma }{2}\sigma \left[ 2\mu -\left( 2-\gamma \right) \frac{\sigma }{\sigma _{u}}\left( 1-2\left( 1+r\right) w\frac{\sigma }{\sigma _{u}}\right) \right] =0 \end{aligned}$$

$\Leftrightarrow$

$$\begin{aligned}&\mu ^{2}\frac{1}{\sigma _{u}}\left( 1-2\left( 1+r\right) w\frac{\sigma }{ \sigma _{u}}\right) \\&\qquad -\left( 1-\gamma \right) \sigma \mu +\frac{\left( 1-\gamma \right) \left( 2-\gamma \right) }{2}\frac{\sigma }{\sigma _{u}} \left( 1-2\left( 1+r\right) w\frac{\sigma }{\sigma _{u}}\right) =0 \end{aligned}$$

$\Leftrightarrow$ (19). $\square$