Risk incentives of executive stock options for undiversified managers

Abstract

The starting point of this paper is twofold. First, managers are often undiversified. Second, an increase in systematic risk could increase the market’s discount rate and consequently effect a contemporaneous change in the underlying stock’s market price. The paper makes comparative static analyses of these circumstances by using Meulbroek’s (Financ Manag 30:5–44, 2001) executive stock option model together with the dividend discount model, and shows that options do not provide incentive to increase the proportion of systematic risk to firm-specific risk, as commonly argued. The paper also demonstrates that the option’s value to the manager can be monotone decreasing, but may also show an inverted U-shape with respect to firm-specific risk. The option’s value exhibits a similar pattern against the total risk. In addition, the study finds that total risk incentives may under some conditions lower the shareholder value; executive options may thus encourage managers to act against principals’ interests.

Appendix

Let N denote the standard cumulative normal distribution function, and N′, the derivative of N, the density function of standard normal distribution. Moreover, let $\Upsilon(x)={N^{\prime}(x)}/{N(x)}$ . Then $\Upsilon(x)$ is the strictly decreasing function of x for all $x \in {\mathbb{R}}$ such that $\Upsilon(x)=\infty$ as x→−∞, and $\Upsilon(x)=0$ as x→∞. This means that the cumulativne standard normal distribution function N(x) is a logarithmically convex function.

Proof

We understand that

$$\begin{aligned} \Upsilon^{\prime}(x)&=\frac{N^{\prime\prime}(x)N(x)-N^{\prime}(x)^2}{N(x)^2}\\ &=-\frac{N^{\prime}(x)\left[xN(x)+N^{\prime}(x)\right]}{N(x)^2}, \end{aligned} $$

and hence require that

$$ xN(x)+N^{\prime}(x)> 0 $$

for all $x \in {\mathbb{R}}$ . The above holds because

$$\begin{aligned} \lim_{x\to-\infty}\left[xN(x)+N^{\prime}(x)\right]&=\lim_{x\to-\infty} \left[\frac{x}{N(x)^{-1}}+N^{\prime}(x)\right]\\ &\stackrel{l^{\prime}H}{=}\lim_{x\to-\infty}\left[- \frac{N(x)^2}{N^{\prime}(x)}+N^{\prime}(x)\right] \\ &\stackrel{l^{\prime}H}{=}\lim_{x\to-\infty}\left[2\frac{N(x)}{x}+N^{\prime}(x) \right] \\ &= 0, \end{aligned} $$

and because

$$ \frac{d}{dx}\left[xN(x)+N^{\prime}(x)\right]=N(x) $$

that is strictly positive for all $x\in{\mathbb{R}}$ . Thus $\Upsilon^{\prime}(x)<0$ for all $x\in{\mathbb{R}}$ . Finally, it is clear that

$$ \lim_{x\to\infty}\Upsilon(x)=0, $$

because N′(x) = 0 and N(x) = 1 as x→∞. Using again l’Hospital’s rule, we obtain

$$ \lim_{x\to-\infty}\Upsilon(x)\stackrel{l^{\prime}H}{=}\lim_{x\to\infty}-x, $$

which completes the proof.□