Optimal stock option schemes for managers

Abstract

This paper analyzes which stock option scheme best aligns the interests of a firm’s manager and shareholders when both are risk-averse. We consider granting to the manager a basic fixed salary and one of the following four options: European, Parisian, Asian and American options. Choosing the strike of the options optimally, the shareholders can mostly implement a first best solution with all payoff schemes. The American option scheme best aligns the interests of the manager and the shareholders in the most common case in which the strike price equals the grant-date fair market value.

Appendix

1.1 Partial derivatives of the first two moments of a European call

The derivatives of the first moments with respect to μ and σ are as follows:

$$ \begin{aligned} {\frac{\partial E[ h(S(T))]}{\partial \mu}} &= T \cdot e^{(\mu-\delta) T} \cdot \Upphi(d_1 ) + \left(e^{(\mu-\delta) T} \phi(d_1) - K \phi(d_2)\right) {\frac{\partial d_1}{\partial \mu}} = T \cdot e^{(\mu-\delta) T} \cdot \Upphi(d_1 ) \\ {\frac{\partial E[h(S(T))]}{\partial \sigma}} &= e^{(\mu-\delta) T} \phi(d_1) \left({\frac{\partial d_2}{\partial \sigma}} + \sqrt{T} \right) - K \phi(d_2) {\frac{\partial d_2}{\partial \sigma}} = \sqrt{T} \cdot e^{(\mu-\delta) T} \cdot \phi(d_1). \end{aligned} $$

Hereby we have used ${\frac{\partial d_1}{\partial \mu}}={\frac{\partial d_2}{\partial \mu}}={\frac{\sqrt{T}}{\sigma}};\, {\frac{\partial d_1}{\partial \sigma}}={\frac{\partial d_2}{\partial \sigma}} + \sqrt{T}$; and $\phi(d_1)= K e^{-(\mu-\delta) T} \phi(d_2)$, with $\phi(x)\equiv {\frac{1}{\sqrt{2 \pi} }} e^{-x^2/2}.$ The following two derivatives are:

$$ \begin{aligned} {\frac{\partial E[ h^2(S(T))]}{\partial \mu}} &= 2 T \cdot e^{2 (\mu-\delta) T+\sigma^2 T} \Upphi(d_3) - 2 K T \cdot e^{(\mu-\delta) T} \Upphi(d_1) \\ & \quad + \left(e^{2 (\mu-\delta) T+\sigma^2 T} \phi(d_3) {\frac{\partial d_3}{\partial \mu}} +K^2 \phi(d_2){\frac{\partial d_2}{\partial \mu}} - 2 K e^{(\mu-\delta) T} \phi(d_1) {\frac{\partial d_1}{\partial \mu}} \right) \\ &= 2 T \cdot e^{2 (\mu-\delta) T+\sigma^2 T} \Upphi(d_3) - 2 K T \cdot e^{(\mu-\delta) T} \Upphi(d_1) \\ {\frac{\partial E[h^2(S(T))]}{\partial \sigma}} &= 2 \sigma T \cdot e^{2 (\mu-\delta) T+\sigma^2 T} \Upphi(d_3) \\ & \quad + e^{2 (\mu-\delta) T+\sigma^2 T} \phi(d_3) {\frac{\partial d_3}{\partial \sigma}} +K^2 \phi(d_2){\frac{\partial d_2}{\partial \sigma}} - 2 K \cdot e^{(\mu-\delta) T} \phi(d_1) {\frac{\partial d_1}{\partial \sigma}} \\ &= 2 \sigma T e^{2 (\mu-\delta) T+\sigma^2 T} \Upphi(d_3). \end{aligned} $$

Hereby we have used the fact that ${\frac{\partial d_1}{\partial \mu}}={\frac{\partial d_2}{\partial \mu}}={\frac{\partial d_3}{\partial \mu}}={\frac{\sqrt{T}}{\sigma}}$; ${\frac{\partial d_3}{\partial \sigma}}={\frac{\partial d_2}{\partial \sigma}} + 2 \sqrt{T}$ and $\phi(d_1)= K e^{-(\mu-\delta) T} \phi(d_2),\,\phi(d_3)=K^2 e^{-2 (\mu-\delta) T- \sigma^{2T}} \phi(d_2)$. The optimal volatility results from numerically solving

$$ \begin{aligned} \,&\left(\left(1+2 \lambda_{eff} E[h] \right)T e^{(\mu-\delta) T} \Upphi(d_1 ) -\lambda_{eff} \left(2 T e^{2 (\mu-\delta) T+\sigma^2 T} \Upphi(d_3) - 2 K T e^{(\mu-\delta) T} \Upphi(d_1)\right)\right) {\frac{\partial \mu}{\partial \sigma }} \\ \,&\quad+\left(\left(1+2 \lambda_{eff} E[h] \right)\sqrt{T} e^{(\mu-\delta) T} \phi(d_1)-\lambda_{eff} \left( 2 \sigma T e^{2 (\mu-\delta) T+\sigma^2 T} \Upphi(d_3)\right)\right)=0. \end{aligned} $$

1.2 First two moments and the corresponding derivatives under Parisian scheme

Using the implied barrier concept implies that

$$ E[h(S(T))] = E\left[\left[{\frac{S(T)}{S(0)}} - K\right]^+ 1_{\{T_B^+<T\}}\right] \approx E\left[\left[{\frac{S(T)}{S(0)}} - K\right]^+ 1_{\{\tau_{B^*} < T\}}\right]. $$

Note that

$$ \begin{aligned} &S_t > B_t^* \Leftrightarrow W_t + m t > {\frac{1}{\sigma}} \ln {\frac{B_0^*}{S_0}} := b \\ &{\frac{S_T}{S_0}} > K \Leftrightarrow W_T + m T > {\frac{1}{\sigma}} \ln (K e^{-g T}) := k \end{aligned} $$

with $m= {\frac{1}{\sigma}} (\mu -\delta -g -\frac{1}{2} \sigma^2).$ Consequently, we can rewrite

$$ \begin{aligned} \,& E\left[\left[{\frac{S(T)}{S(0)}}- K\right]^+ 1_{\{\tau_{B^*} < T\}}\right] \\ &\quad= E\left[\left(\exp\{\sigma (W_T + m T )\} e^{g T}-K \right) 1_{\{W_T + m T > k, \sup_{t \in [0,T]} W_t + m t \geq b \}}\right]\\ &\quad= \tilde{E} \left[\exp\{m Z_T- {\frac{1}{2}} m^2 T\} \left(\exp\{\sigma Z_T \} e^{g T}-K \right) 1_{\{Z_T > k, \sup_{t \in [0,T]} Z_t \geq b \}}\right] \end{aligned} $$

with Z _t  = W _t  + mt and $\tilde{E}$ is the expectation taken under the new probability measure $\tilde{P}$ which is defined by

$$ \frac{{d P}}{{d \tilde{P}}} \Big|_{{{\mathcal{F}}}_t} = \exp\left\{ m Z_t - \frac{1}{2} m^2 t \right\}. $$

We proceed with the calculation of the expected value by distinguishing two cases: k ≥ b (K S ₀ ≥ B ^* _T ) and k < b (K S ₀ < B ^* _T ). For k ≥ b, the above expectation is reduced to

$$ \begin{aligned} \,& \tilde{E} \left[\exp\{m Z_T- {\frac{1}{2}} m^2 T\} \left(\exp\{\sigma Z_T \} e^{g T}-K \right) 1_{\{Z_T > k \}}\right]\\ &\quad = e^{(\mu-\delta) T} \Upphi\left({\frac{\ln{\frac{1}{K}} + (\mu-\delta+\frac{1}{2} \sigma^2 ) T}{\sigma \sqrt{T} }} \right) - K \Upphi\left({\frac{\ln{\frac{1}{K}} + (\mu-\delta-\frac{1}{2} \sigma^2 ) T}{\sigma \sqrt{T} }} \right). \end{aligned} $$

This corresponds to the formula obtained in the case of the European call. For k < b, we can use the equivalence of the following events:

$$ \begin{aligned} & 1_{\left\{Z_T > k, \sup_{t \in [0,T]} Z_t \geq b \right\}} \\ &\quad = 1_{\left\{\sup_{t \in [0,T]} Z_t \geq b \right\}} -1_{\left\{Z_T \leq k, \sup_{t \in [0,T]} Z_t \geq b\right\}} \\ &\quad= 1_{\left\{Z_T \geq b, \sup_{t \in [0,T]} Z_t \geq b \right\}}+1_{\left\{Z_T < b, \sup_{t \in [0,T]} Z_t \geq b \right\}} -1_{\left\{Z_T \leq k, \sup_{t \in [0,T]} Z_t \geq b \right\}}. \\ \end{aligned} $$

The above decomposition combined with the reflection principle leads to

$$ \begin{aligned} \,& \tilde{E} \left[\exp\{m Z_T- {\frac{1}{2}} m^2 T\} \left(\exp\{\sigma Z_T \} e^{g T}-K \right) 1_{\{Z_T > k, \sup_{t \in [0,T]} Z_t \geq b \}}\right]\\ &\quad = \tilde{E} \left[\exp\{m Z_T- {\frac{1}{2}} m^2 T\} \left(\exp\{\sigma Z_T \} e^{g T}-K \right) 1_{\{Z_T > b\}}\right]\\ &\quad +\tilde{E} \left[\exp\{m (2 b -Z_T)- {\frac{1}{2}} m^2 T\} \left(\exp\{\sigma (2 b -Z_T) \} e^{g T}-K \right) 1_{\{Z_T > b\}}\right]\\ &\quad -\tilde{E} \left[\exp\{m (2 b -Z_T)- {\frac{1}{2}} m^2 T\} \left(\exp\{\sigma (2 b -Z_T) \} e^{g T}-K \right) 1_{\{Z_T > 2 b- K \}}\right]\\ &\quad = e^{(\mu-\delta) T} \Upphi\left(-{\frac{b}{\sqrt{T}}} +(m+\sigma )\sqrt{T} \right) - K \Upphi\left(-{\frac{b}{\sqrt{T}}} +m \sqrt{T} \right)\\ &\quad +e^{(\mu-\delta) T} \left({\frac{B_0^*}{S_0}}\right)^{{\frac{2(\mu-\delta-g)}{\sigma^2}}+{1}} \Upphi\left(-{\frac{b}{\sqrt{T}}} -(m+\sigma )\sqrt{T} \right) - K \left({\frac{B_0^*}{S_0}}\right)^{{\frac{2m}{\sigma}}} \Upphi\left(-{\frac{b}{\sqrt{T}}} -m \sqrt{T} \right) \\ &\quad - e^{(\mu-\delta) T} \left({\frac{B_0^*}{S_0}}\right)^{{\frac{2(\mu-\delta-g)}{\sigma^2}}+{1}} \Upphi\left(-{\frac{2 b- k}{\sqrt{T}}} -(m+\sigma )\sqrt{T} \right) +K \left({\frac{B_0^*}{S_0}}\right)^{{\frac{2m}{\sigma}}} \Upphi\left(-{\frac{2 b- k}{\sqrt{T}}} -m \sqrt{T} \right). \end{aligned} $$

(7.1)

To derive the second moment of h(S(T)), we can follow the same steps as above. First note that

$$ \begin{aligned} \,& E\left[\left(\left[{\frac{S(T)}{S(0)}}- K\right]^+\right)^2 1_{\{\tau_{B^*} < T\}}\right] \\ &\quad = E\left[\left(e^{2 \sigma (W_T + m T )} e^{2 g T}+K^2 -2 K \exp\{\sigma (W_T + m T )\} e^{g T} \right) 1_{\{W_T + m T > k, \sup_{t \in [0,T]} W_t + m t \geq b \}}\right] \\ &\quad = \tilde{E} \Big[e^{m Z_T- {\frac{1}{2}} m^2 T} \left(\exp\{2 \sigma Z_T\} e^{2 g T}+K^2 -2 K \exp\{\sigma Z_T\} e^{g T} \right) \cdot 1_{\{Z_T > k, \sup_{t \in [0,T]} Z_t \geq b \}}\Big]. \end{aligned} $$

Again we distinguish between k ≥ b and k < b. For k ≥ b, we obtain the same formula as in the case of the European call:

$$ \begin{aligned} & E\left[\left(\left[{\frac{S(T)}{S(0)}}- K\right]^+\right)^2 1_{\{\tau_{B^*} < T\}}\right] \\ &= e^{2 ( \mu-\delta) T+ \sigma^2 T} \Upphi\left({\frac{\ln{\frac{1}{K}} + (\mu-\delta+\frac{3}{2} \sigma^2) T}{\sigma \sqrt{T} }} \right) + K^2 \Upphi\left({\frac{\ln{\frac{1}{K}} + (\mu-\delta-\frac{1}{2} \sigma^2 ) T}{\sigma \sqrt{T} }} \right) \\ &\quad - 2 K e^{(\mu-\delta) T} \Upphi\left({\frac{\ln{\frac{1}{K}} + (\mu-\delta+\frac{1}{2} \sigma^2 ) T}{\sigma \sqrt{T} }} \right). \end{aligned} $$

For k < b, we can use the same decomposition of the indicator function which leads to the following results:

$$ \begin{aligned} \,& \tilde{E} \left[e^{m Z_T- {\frac{1}{2}} m^2 T} \left(\exp\{2 \sigma Z_T\} e^{2 g T}+K^2 -2 K \exp\{\sigma Z_T\} e^{g T} \right) \cdot 1_{\{Z_T > k, \sup_{t \in [0,T]} Z_t \geq b \}}\right] \\ &\quad = -2 K e^{(\mu-\delta) T} \Upphi\left(-{\frac{b}{\sqrt{T}}} +(m+\sigma )\sqrt{T} \right) + K^2 \Upphi\left(-{\frac{b}{\sqrt{T}}} +m \sqrt{T} \right) \\ & \quad - 2 K e^{(\mu-\delta) T} \left({\frac{B_0^*}{S_0}}\right)^{{\frac{2(\mu-\delta-g)}{\sigma^2}}+{1}} \Upphi\left(-{\frac{b}{\sqrt{T}}} -(m+\sigma )\sqrt{T} \right)+ K^2 \left({\frac{B_0^*}{S_0}}\right)^{{\frac{2m}{\sigma}}} \Upphi\left(-{\frac{b}{\sqrt{T}}} -m \sqrt{T} \right) \\ &\quad + 2 K e^{(\mu-\delta) T} \left({\frac{B_0^*}{S_0}}\right)^{{\frac{2(\mu-\delta-g)}{\sigma^2}}+{1}} \Upphi\left(-{\frac{2 b- k}{\sqrt{T}}} -(m+\sigma )\sqrt{T} \right) -K^2 \left({\frac{B_0^*}{S_0}}\right)^{{\frac{2m}{\sigma}}} \Upphi\left(-{\frac{2 b- k}{\sqrt{T}}} -m \sqrt{T} \right) \\ &\quad +e^{2 (\mu-\delta) T + \sigma^2 T} \Upphi \left(-{\frac{b}{\sqrt{T}}} + (m+2 \sigma)\sqrt{T} \right) + e^{2 (\mu-\delta) T + \sigma^2 T} \left({\frac{B_0^*}{S_0}}\right)^{{\frac{2(m+2 \sigma)}{\sigma}}} \Upphi\left(-{\frac{b}{\sqrt{T}}} -(m+2 \sigma )\sqrt{T} \right) \\ &\quad - e^{2 (\mu-\delta) T + \sigma^2 T} \left({\frac{B_0^*}{S_0}}\right)^{{\frac{2(m+2 \sigma)}{\sigma}}} \Upphi\left(-{\frac{2 b- k}{\sqrt{T}}} -(m+2 \sigma )\sqrt{T} \right). \end{aligned} $$

The derivatives of the expectation with respect to μ or σ are given by the sum of the derivatives of each component in (7.1) with respect to μ or σ. Taking the derivative of the third component with respect to μ as an example, we obtain

$$ \begin{aligned} \,& {\frac{\partial}{\partial \mu}} \left(e^{(\mu-\delta) T} \left({\frac{B_0^*}{S_0}}\right)^{\frac{{2(m+\sigma)}}{ {\sigma}}} \Upphi\left(-{\frac{b}{\sqrt{T}}} -(m+\sigma )\sqrt{T} \right)\right) \\ & = \Upphi\left(-{\frac{b}{\sqrt{T}}} -(m+\sigma )\sqrt{T} \right) \Bigg[T e^{(\mu-\delta) T} \left({\frac{B_0^*}{S_0}}\right)^{{\frac{2(\mu-\delta-g)}{\sigma^2}}+1}\\ &\quad + e^{(\mu-\delta) T}\left({\frac{B_0^*}{S_0}}\right)^{{\frac{2(\mu-\delta-g)}{\sigma^2}}+1} \cdot \left(-{\frac{d^{3/2} e^{-m^2 d/2 } \sqrt{2 \pi} (\mu-\delta-g) m }{\sigma}}+ {\frac{2 \log {\frac{B_0^*}{S_0}}}{\sigma^2}}\right) \Bigg] \\ &\quad + e^{(\mu-\delta) T} \left({\frac{B_0^*}{S_0}}\right)^{{\frac{2(\mu-\delta-g)}{\sigma^2}}+1} \phi\left(-{\frac{b}{\sqrt{T}}} -(m+\sigma )\sqrt{T} \right) \left(-{\frac{\sqrt{T}}{\sigma}} + {\frac{m d^{3/2}e^{-m^2 d/2 } \sqrt{{\frac{\pi}{2}}} }{\sigma \sqrt{T}}} \right). \end{aligned} $$

The derivatives of the other components can be computed similarly. Note that the fact that the implied barrier B ^*₀ is a function of μ and σ shall be taken into consideration when determining the corresponding derivatives.

Here the detailed calculations for the derivatives of the second moment are neglected. As we have learned from the European case, all the derivatives result in ${\frac{\partial U_{manager}}{\partial \mu}}$ and ${\frac{\partial U_{manager}}{\partial \sigma}}.$ With the help of

$$ {\frac{\partial U_{manager}}{\partial \mu}} {\frac{\partial \mu}{\partial \sigma}} + {\frac{\partial U_{manager}}{\partial \sigma}} =0, $$

we determine the optimal σ-level numerically.

1.3 American ESO scheme

Applying Barone-Adesi and Whaley (1987) straightforwardly in our context, we obtain the expected terminal payoff for the American option case by distinguishing between the following two cases:

$$ \begin{aligned} E \left[h^a(S_T)\right]= \left\{\begin{array}{ll} e^{r T}(1 -K), & \hbox{if } S_0 > S^* \\ E[h(S_T)] + e^{r T } {\frac{1}{S_0 }} A_2 \left(\frac{{S_0 }}{ {S^*}}\right)^{q_2}, & \hbox{if } S_0 < S^* \end{array}\right. \end{aligned} $$

(7.2)

with

$$ \begin{aligned} q_2 &=\frac{1}{2} \left(-\left({\frac{2 (\mu-\delta)}{\sigma^2}} -1\right) + \sqrt{\left({\frac{2 (\mu-\delta) }{\sigma^2}}-1\right)^2 + {\frac{8 r}{\sigma^2 (1-e^{-r T} )}} } \right) \\ A_2&= (S^*/q_2) \left(e^{(\mu-r-\delta) T}-e^{(\mu-r-\delta) T} \Upphi(d_1(S^*))\right) \\ d_1(S^*)&= {\frac{\ln {\frac{S^*}{S_0 K}}+(\mu -\delta +\frac{1}{2} \sigma^2 )T }{\sigma \sqrt{T}}}. \end{aligned} $$

And finally S ^* is determined implicitly by solving

$$ S^{*} - K \cdot S_0 = S_0 \left(e^{(\mu-r -\delta) T} \Upphi(d_1(S_0)) - K e^{-r T} \Upphi(d_2(S_0))\right) + (1-e^{(\mu-r -\delta) T} \Upphi(d_1(S^*) )) S^* /q_2. $$

In order to determine the second moment of the American option, we can first decompose the square of the European call option into the following two parts:

$$ \begin{aligned} \left(\left[S_T / S_0 - K \right]^+\right)^2 &= \left(\left(S_T/S_0\right)^2 +K^2 - 2 K S_T/S_0 \right) 1_{\left\{S_T > K \cdot S_0 \right\}} \\ &= \left[\left(S_T/S_0\right)^2 - K^2 \right]^+ - 2 K \left[S_T/S_0 - K \right]^{+}. \end{aligned} $$

The second term on the right-hand side is already determined, and now we can use the same technique to calculate the first term $\left[\left(S_T/S_0\right)^2 - K^2 \right]^+.$ The expected payoff from the European option $\left[\left(S_T/S_0\right)^2 - K^2 \right]^+$ is given by

$$ \begin{aligned} E\left[ \left[\left(S_T/S_0\right)^2 - K^2 \right]^+\right] &= e^{(2 (\mu-\delta)+\sigma^2 ) T} \Upphi\left({\frac{1/K^2 +(2(\mu-\delta)T+3\sigma^2 )T}{2 \sigma \sqrt{T} }} \right) \\ &\quad - K^2 \Upphi\left({\frac{1/K^2 +(2(\mu-\delta)T-\sigma^2 )T}{2 \sigma \sqrt{T} }} \right). \end{aligned} $$

(7.3)

For δ > 0, we need to distinguish between two cases:

$$ \begin{aligned} \,& E \left[\left[\left(S_T/S_0\right)^2 - K^2 \right]^+\right]\\ & \quad = \left\{\begin{array}{ll} e^{ r T}(1 -K^2), & \hbox{if } S_0 > S^* \\ E\left[ \left[\left(S_T/S_0\right)^2 - K^2 \right]^+\right] + {\frac{1}{S_0^2 }} \tilde{A}_2 \left({\frac{S_0^2 }{(S^*)^2}}\right)^{q_2}, & \hbox{if } S_0 < S^{*} \end{array}\right. \end{aligned} $$

(7.4)

with

$$ \begin{aligned} q_2 &=\frac{1}{2} \left(-\left({\frac{( 2(2\mu+\sigma^2-2\delta)}{4 \sigma^2}} -1\right) + \sqrt{\left({\frac{( 2(2\mu+\sigma^2-2\delta)}{4 \sigma^2}} -1\right)^2 + {\frac{8 r }{4 \sigma^2 (1-e^{-r T} )}} } \right) \\ \tilde{A}_2&= (S^*)^2/q_2 (e^{(2 \mu-2 \delta+\sigma^2-r) T}-e^{(2 \mu-2 \delta+\sigma^2-r) T} \Upphi(d_1(S^*))) \\ d_1(S^*)&= {\frac{\ln {\frac{(S^*)^2}{S_0^2 K^2 }}+(2(\mu -\delta) +3 \sigma^2 )T }{2 \sigma \sqrt{T}}}. \end{aligned} $$

And finally S ^* is determined implicitly by solving

$$ \begin{aligned} & (S^*)^2 - K^2 \cdot S_0^2 \\ & \quad = S_0^2 e^{-r T}E\left[ \left[\left(S_T/S_0\right)^2 - K^2 \right]^+\right] + (1-e^{(2 \mu+\sigma^2 -2 \delta-r ) T} \Upphi(d_1(S^*) )) (S^*)^2 /q_2. \end{aligned} $$

Finally, the price of the American option is given by

$$ \begin{aligned} E_{Q}[e^{-r T}h^a(S_T)]= \left\{\begin{array}{ll} 1-K, & \hbox{if } S_0 > S^* \\ E_Q[e^{-r T} h(S_T)] + {\frac{1}{S_0}} A_2 \left({\frac{S_0 }{S^*}}\right)^{q_2}, & \hbox{if } S_0 < S^{*} \end{array}\right. \end{aligned} $$

(7.5)

with

$$ \begin{aligned} q_2 &=\frac{1}{2} \left(-\left({\frac{2 (r-\delta)}{\sigma^2}} -1\right) + \sqrt{\left({\frac{2 (r-\delta) }{\sigma^2}}-1\right)^2 + {\frac{8 r }{\sigma^2 (1-e^{-r T} )}} } \right) \\ A_2&= (S^*/q_2) (1-e^{-\delta T} \Upphi(d_1(S^*))) \\ \tilde{d}_1(S^*)&= {\frac{\ln {\frac{S^*}{S_0 K}}+(r -\delta +\frac{1}{2} \sigma^2 )T }{\sigma \sqrt{T}}}. \end{aligned} $$

And finally S ^* is determined implicitly by solving

$$ S^* - K \cdot S_0 = S_0 \left( e^{-\delta T} \Upphi\left(\tilde{d}_1(S_0)\right) - K \Upphi\left(\tilde{d}_2(S_0)\right)\right) + (1-e^{-\delta T} \Upphi\left(\tilde{d}_1(S^*))\right) S^* /q_2. $$

1.4 Concavity of U _share

Proposition 1

Assume $T \geq {\frac{1}{2}},\, \mu=r+\theta \cdot \sigma > \delta,\, r, \delta, \sigma, \theta \geq 0$ and λ > 0. Then

$$ U_{share} \left(r + \theta \cdot \sigma,\sigma\right) $$

is strictly concave.

Proof

The expected utility is given by

$$ U_{share} (\mu,\sigma) = e^{(\mu-\delta) T} - \lambda_S e^{2( \mu-\delta) T} \left(e^{\sigma^2 T} -1\right) \quad \hbox{with } \mu = r +\theta \cdot \sigma $$

It is sufficient to show that the second derivative of this expression with respect to σ is smaller than 0, which is equivalent to

$$ \begin{aligned} & - e^{- \delta} T \left ( e^{r + \sigma \cdot \theta}\theta^2 - 4 e^{\delta + 2 (-\delta + r + \sigma \cdot \theta)T} \lambda \theta^2 T \right) \\ &-2 e^{- \delta} T \left(e^{\delta + \sigma^2 T + 2 (-\delta + r + \sigma \cdot \theta)T} \lambda \left( 1 + 2 \sigma^2 + 4 \sigma \cdot \theta \cdot T + 2\theta^2 T \right) \right) <0. \end{aligned} $$

This is equivalent to

$$ 2 e^{\sigma^2T} \left ( 1 + 2 \sigma^2 +4 \sigma \cdot \theta \cdot T + 2 \theta^2 T \right) + 4 \theta^2 T > e^{( \sigma \cdot \theta + r - \delta) (1-2T) } \theta^{2}. $$

Using the above assumptions on the parameters we can show that the LHS is larger than θ ² and this is also larger than the RHS, which proves the proposition. □

The above proposition implies the following lemma:

Corollary 1

For a given Sharpe ratio θ there exists a unique $\tilde \sigma(\theta),$ that maximizes the utility of the shareholders:

$$ \tilde \sigma(\theta) =\hbox{argmax}_{\sigma} U_{share} (r + \theta \cdot \sigma, \sigma). $$