Are employee stock option exercise decisions better explained through the prospect theory?

Abstract

In this paper, we introduce a new framework for the analysis of employee stock options exercise decisions. We develop a distorted lattice model where the exercise decision obeys a policy that maximizes the expected value of the exercise outcomes under the Cumulative Prospect Theory. Using a large US dataset of exercise transactions, we show that our framework broadly outperforms the Expected Utility Theory framework in explaining empirical exercise decisions. Interestingly, our empirical estimates of probability weighting are consistent with those from the experimental literature. We argue that this analysis provides a unifying stream for thinking about issues related to the exercise and the valuation of stock options.

Appendices

Appendix 1

Proof of proposition 1

Let $B_{i,j} =\left( {{\begin{array}{l} i \\ j \\ \end{array} }} \right) p^{j}\left( {1-p} \right) ^{i-j}$ be the probability of reaching node (i, j). The proof can be established by recursion as follows:

For $i=0$, one can check that $\overline{{p}}_{0,0} =\psi _a \left( {1-p} \right) =1-p_{0,0}$, which implies $0<\overline{{p}}_{0,0} <1$, and therefore $0<p_{0,0} <1$.

For $i=1$, we have $\overline{{p}}_{1,0} =\frac{\omega _{2,0} }{\zeta _{1,0} }=\frac{\psi _a \left( {B_{1,0} \left( {1-p} \right) } \right) }{\psi _a \left( {B_{1,0} } \right) }$, then $0<\overline{{p}}_{1,0} <1$ since $\psi _a \left( . \right)$ is increasing. We have also $\overline{{p}}_{1,1} =\frac{\omega _{2,1} -\left( {1-\overline{{p}}_{1,0} } \right) \zeta _{1,0} }{\zeta _{1,1} }=\frac{\psi _a \left( {B_{2,1} +B_{2,0} } \right) -\psi _a \left( {B_{1,0} } \right) }{1-\psi _a \left( {B_{1,0} } \right) }$. Given that $B_{2,1} +B_{2,0} >B_{1,0}$, we conclude that $0<\overline{{p}}_{1,1} <1$. Finally, from $p_{1,1} =\frac{\omega _{2,2} }{\zeta _{1,1} }=\frac{1-\psi _a \left( {B_{2,1} +B_{2,0} } \right) }{1-\psi _a \left( {B_{1,0} } \right) }$, we deduce that $0<p_{1,1} <1$.

For $i>1$, assume now that $0<\overline{{p}}_{k,j} <1$ and $0<p_{k,j} <1 \forall k\le i,j\le k$, then prove that the inequalities hold for $k=i+1,\forall j\le i+1$. The inequality holds for $j=0$ since $\overline{{p}}_{i+1,0} =\frac{\omega _{i+1,0} }{\zeta _{i,0} }=\frac{\psi _a \left( {B_{i,0} \left( {1-p} \right) } \right) }{\psi _a \left( {B_{i,0} } \right) }$. For $j>0$, we have:

$$\begin{aligned} \overline{{p}}_{i+1,j}= & {} \frac{\omega _{i+2,j} -\left( {1-\overline{{p}}_{i+1,j-1} } \right) \zeta _{i+1,j-1} }{\zeta _{i+1,j} }=\frac{\sum \nolimits _{l=0}^j {\omega _{i+2,l} } -\sum \nolimits _{l=0}^{j-1} {\omega _{i+1,l} } }{\omega _{i+1,j} }\nonumber \\= & {} \frac{\psi _a \left( {\sum \nolimits _{l=0}^j {B_{i+2,l} } } \right) -\psi _a \left( {\sum \nolimits _{l=0}^{j-1} {B_{i+1,l} } } \right) }{\psi _a \left( {\sum \nolimits _{l=0}^j {B_{i+1,l} } } \right) -\psi _a \left( {\sum \nolimits _{l=0}^{j-1} {B_{i+1,l} } } \right) } \end{aligned}$$

(47)

Given that the binomial distribution increases stochastically as i increases, we have:

$\sum _{l=0}^{j-1} {B_{i+1,l} } <\sum _{l=0}^j {B_{i+2,l} } <\sum _{l=0}^j {B_{i+1,l} } ,$ which implies that $0<\overline{{p}}_{i+1,j} <1$ since $\psi _a \left( . \right)$ is increasing.

In the same way, noting that$p_{i+1,i+1} =\frac{\omega _{i+2,i+2} }{\zeta _{i+1,i+1} }=\frac{1-\psi _a \left( {\sum _{l=0}^{i+1} {B_{i+2,l} } } \right) }{1-\psi _a \left( {\sum _{l=0}^i {B_{i+1,l} } } \right) }$, we obtain $0<p_{i+1,i+1} <1$. This completes the proof. $\square$

Appendix 2

Proof of proposition 2

Let’s proceed by recursion as previously done:

For $i=0$, we have $\overline{{p}}_{0,0} =\psi _a \left( {1-p} \right) =1-p_{0,0}$.

For i=1, one may check that $\overline{{p}}_{1,1} +p_{1,1} =1$.

For $i>1$, assume the relationship is verified for i, then prove it for $i+1$. The transition decision weights at the top node verify:

$$\begin{aligned} 1-\overline{{p}}_{i+1,i+1} =1-\frac{\psi _a \left( {\sum \nolimits _{l=0}^{i+1} {B_{i+2,l} } } \right) -\psi _a \left( {\sum \nolimits _{l=0}^i {B_{i+1,l} } } \right) }{1-\psi _a \left( {\sum \nolimits _{l=0}^i {B_{i+1,l} } } \right) }=p_{i+1,i+1} \end{aligned}$$

Proposition 2 states that transition decision weights are unique and may interpret as distorted (i.e. skewed) transition probabilities. There exist at least two different sets of transition decision weights at a given level of the tree leading to the weighted probabilities at the next level (i.e. allow connecting both levels of the tree). The first set is the one retrieved by the algorithm from (5) to (20). The second set is given by a modified version of this algorithm that consists in departing from the top nodes ($j=i)$ towards the bottom nodes ($j=0)$, which boils down to taking $\overline{{p}}_{i,i} =1-p_{i,i}$ instead of (20), and $\overline{{p}}_{i,0} =\frac{\omega _{i+1,1} -\overline{{p}}_{i,1} \zeta _{i,1} }{\zeta _{i,0} }$ instead of (16). Proposition 2 implies that the two sets are identical. This completes the proof. $\square$

Appendix 3

Proof of proposition 3

Consider a given state (i, j) where the option is in the money, which means that $j>\left( {\frac{i}{2}1\hbox {I}_{\left\{ {i\hbox { mod }2=0} \right\} } +\frac{i+1}{2}1\hbox {I}_{\left\{ {i\hbox { mod }2=1} \right\} } } \right)$. In order to show that the probability of exercise increases in the degree of probability weighting (Resp. decreases in a), it suffices to prove that this is true at each node (k, l) leading to node (i, j) where the option is exercisable, which also boils down to showing that the state subjective value is decreasing in a. Let’s denote the exercise probability at the states (k, l) by $\pi _{k,l} =P\left\{ {Ce_{k,l} <h_{k,l} } \right\}$. Let’s also leave aside the exit state for sake of simplicity ($p_e =0$) since it does not affect the sensitivity of $Ce_{k,l}$to a. This allows restating the subjective value as follows:

$$\begin{aligned} Ce_{k,l} =v_{\theta _k }^{-1} \left( {p_{k,l} v_{\theta _k } \left( {V_{k+1,l+1}^0 } \right) +\overline{{p}}_{k,l} v_{\theta _k } \left( {V_{k+1,l}^0 } \right) } \right) e^{-r\delta t} \end{aligned}$$

Given that the value function and its reciprocal in (35) are both monotonically increasing, a sufficient condition for $Ce_{k,l}$ to decrease in a is that $p_{k,l}$ is a decreasing function of a for $l>\left( {\frac{k}{2}1\hbox {I}_{\left\{ {k\hbox { mod }2=0} \right\} } +\frac{k+1}{2}1\hbox {I}_{\left\{ {k\hbox { mod }2=1} \right\} } } \right)$. We have from (47):

$$\begin{aligned} p_{k,l} \left( a \right) =\frac{\psi _a \left( {\sum \nolimits _{h=0}^l {B_{k,h} } } \right) -\psi _a \left( {\sum \nolimits _{h=0}^l {B_{k+1,h} } } \right) }{\psi _a \left( {\sum \nolimits _{h=0}^l {B_{k,h} } } \right) -\psi _a \left( {\sum \nolimits _{h=0}^{l-1} {B_{k,h} } } \right) }=\frac{\psi _a \left( {A_{k,l} } \right) -\psi _a \left( {A_{k+1,l} } \right) }{\psi _a \left( {A_{k,l} } \right) -\psi _a \left( {A_{k,l-1} } \right) }, \hbox { with } A_{k,l-1} <A_{k+1,l} <A_{k,l} . \end{aligned}$$

A second order Taylor series expansion of $\psi _a \left( . \right)$ around $A_0 \ge 0.5$ gives for some $a<b<1$:

$$\begin{aligned} p_{k,l} \left( a \right) -p_{k,l} \left( b \right) \approx D_{k,l} \left( {{\psi }'_a \left( {A_0 } \right) {\psi }''_b \left( {A_0 } \right) -{\psi }'_b \left( {A_0 } \right) {\psi }''_a \left( {A_0 } \right) } \right) \end{aligned}$$

where $D_{k,l} =\frac{\left( {A_{k,l} -A_{k,l-1} } \right) \left( {A_{k,l} -A_{k+1,l} } \right) \left( {A_{k+1,l} -A_{k,l-1} } \right) }{2\left( {\psi _a \left( {A_{k,l} } \right) -\psi _a \left( {A_{k,l-1} } \right) } \right) \left( {\psi _b \left( {A_{k,l} } \right) -\psi _b \left( {A_{k,l-1} } \right) } \right) }$ is a positive constant.

Noting that ${\psi }'_a \left( x \right) /{\psi }''_a \left( x \right)$ is decreasing in $a,\forall x\ge 0.5$, we have$p_{k,l} \left( a \right) >p_{k,l} \left( b \right)$, which implies that the upward decision weights increase in the degree of probability weighting at the nodes where the option is exercisable. $\square$

Appendix 4

Proof of proposition 4

In the same way, assuming $p_e =0$, for all loss states (k, l) leading to state (i, j), where $x_{k,l} \le 0$ with

$x_{k,l} =p_{k,l} v_{\theta _k } \left( {V_{k+1,l+1}^0 } \right) +\overline{{p}}_{k,l} v_{\theta _k } \left( {V_{k+1,l}^0 } \right) ,$ the state subjective value writes:

$Ce_{k,l} =\left( {\theta _k -\left( {-\frac{x_{k,l} }{\lambda }} \right) ^{\frac{1}{\alpha }}} \right) e^{-r\delta t},$ which is decreasing in $\lambda$.

It follows that the state exercise probabilities $\pi _{k,l}$ are increasing in $\lambda$, which implies that the exercise probability at (i, j) is also increasing in loss aversion. $\square$