Capital income taxation and risk taking under prospect theory

Abstract

This research examines capital income taxation for a prospect theory investor under some acceptable in the literature reference levels relative to which are the changes in the level of wealth valued. Depending on the reference level, some results indicate that it is possible for a capital income tax increase not to stimulate risk taking even if the tax code provides attractive full loss offset provisions. However, risk taking can be stimulated when investors compare their reference level with others. Risk taking can increase also if the investor interprets part of the tax as a loss instead as a reduced gain. Then the investor becomes risk seeking and moves away from the discomfort zone of relative losses. This later response to taxation causes private risk taking to increase.

Appendices

Appendix 1: When Ω≥0

Proof of Proposition 1

Let us assume at first that Ω>0. Based on this and the domain of α, there are three cases that can occur

(P1)::

W _2b ≥Γ _2b , W _2g ≥Γ _2g for $-\frac{\varOmega}{(1-\tau)W_{1}(x_{2g}-r)}\leq\alpha\leq\frac{\varOmega }{(1-\tau )W_{1}(r-x_{2b})}$

(P2)::

W _2b <Γ _2b , W _2g ≥Γ _2g for $\frac{\varOmega}{(1-\tau)W_{1}(r-x_{2b})}<\alpha\leq+\infty$

(P3)::

W _2b ≥Γ _2b , W _2g <Γ _2g for $-\infty \leq\alpha<-\frac{\varOmega}{(1-\tau)W_{1}(x_{2g}-r)}$

The corresponding problems are

$$ \left. \begin{array}{l@{\quad}l} \mathrm{Max_{\alpha}{:}} &pU_{G}(W_{2g}-\varGamma_{2})+(1-p)U_{G}(W_{2b}-\varGamma_{2})\\[2mm]&\quad{}\displaystyle= \frac{1}{1-\gamma}\bigl\{p\bigl[W_{1}(1-\tau)(x_{2g}-r)\alpha+\varOmega \bigr]^{1-\gamma}\\[4mm]&\qquad{}+(1-p)\bigl[-W_{1}(1-\tau) (r-x_{2b})\alpha+\varOmega\bigr]^{1-\gamma }\bigr\}\\[2mm]\mathrm{such\ that} &\displaystyle-\frac{\varOmega}{W_{1}(1-\tau)(x_{2g}-r)}\leq \alpha\leq\frac{\varOmega}{W_{1}(1-\tau)(r-x_{2b})}\end{array} \right\} $$

(P1)

$$ \left.\begin{array}{@{}l@{\quad}l} \mathrm{Max_{\alpha}{:}} &pU_{G}(W_{2g}-\varGamma_{2})+(1-p)\lambda U_{L}(W_{2b}-\varGamma_{2})\\[2mm]&\quad{}\displaystyle=\frac{1}{1-\gamma}\bigl\{ p\bigl[W_{1}(1-\tau)(x_{2g}-r)\alpha+\varOmega \bigr]^{1-\gamma}\\[4mm]&\qquad{}-\lambda(1-p)\bigl[W_{1}(1-\tau) (r-x_{2b})\alpha-\varOmega \bigr]^{1-\gamma}\bigr\}\\[2mm]\mathrm{such\ that} & \displaystyle\frac{\varOmega}{W_{1}(1-\tau)(r-x_{2b})}<\alpha \leq+\infty \end{array} \right\} $$

(P2)

$$ \left.\begin{array}{@{}l@{\quad}l} \mathrm{Max_{\alpha}{:}} & p\lambda U_{L}(W_{2g}-\varGamma_{2})+(1-p)U_{G}(W_{2b}-\varGamma_{2})\\[2mm]&\quad{}\displaystyle= -\frac{\lambda p}{1-\gamma}\bigl[-W_{1}(1-\tau) (x_{2g}-r)\alpha-\varOmega \bigr]^{1-\gamma}\\[4mm]&\qquad{}\displaystyle+\frac{1-p}{1-\gamma}\bigl[-W_{1}(1-\tau) (r-x_{2b})\alpha+\varOmega \bigr]^{1-\gamma }\\[2mm]\mathrm{such\ that} & \displaystyle-\infty\leq\alpha<-\frac{\varOmega }{W_{1}(1-\tau )(x_{2g}-r)} \end{array}\right\} $$

(P3)

The idea of the proof is to show that (P1) is a concave programming problem with the global maximum being α ^∗ as defined by (5) which is such that $-\frac{\varOmega}{(1-\tau)W_{1}(x_{2g}-r)}\leq \alpha ^{\ast}\leq\frac{\varOmega}{(1-\tau)W_{1}(r-x_{2b})}$, the utility of (P2) is decreasing for λ>1/K _γ and the utility of (P3) is increasing for λ>K _γ . First-order conditions (FOC) for (P1) are

it can be shown easily that the second-order conditions are satisfied which implies that (P1) is a concave problem. Finally, it can be seen that

$$\alpha^{\ast}=\frac{\varOmega( 1-K_{0}^{1/\gamma})}{(1-\tau )W_{1}[ r-x_{2b}+K_{0}^{1/\gamma}(x_{2g}-r)]}$$

satisfies the FOC and that $-\frac{\varOmega}{W_{1}(1-\tau )(x_{2g}-r)}\leq \alpha^{\ast}\leq\frac{\varOmega}{W_{1}(1-\tau)(r-x_{2b})}$. As ${\mathbb{E}}(x_{2}-r)>0$ implies K ₀<1 then α ^∗>0.

Regarding problem (P2), note that

(9)

Thus, based on (9) and λ>1/K _γ it follows that

$$\lambda>\frac{p[W_{1}(1-\tau)(r-x_{2b})\alpha-\varOmega]^{\gamma }(x_{2g}-r)}{(1-p)[W_{1}(1-\tau)(x_{2g}-r)\alpha+\varOmega]^{\gamma}(r-x_{2b})}$$

and consequently

$$0>\frac{d{\mathbb{E}}(U_{LA}(W_{2}-\varGamma_{2}))}{d\alpha}$$

implying that the utility of (P2) is a decreasing function in α. The property of the utility of (P3) being an increasing function in α for λ>K _γ can be shown in a similar way.

Note that for Ω=0 the set of feasible solutions of (P1) consists only from α=0. As the rest is the same, i.e., utility increases for α<0 and decreases for α≥0, then the maximum is reached at α ^∗=0. This concludes the proof. □

Appendix 2: When Ω<0

Proof of Propositions 2 and 3

(P4)::

W _2b <Γ _2b , W _2g <Γ _2g for $\frac {\varOmega}{(1-\tau)W_{1}(r-x_{2b})}<\alpha<\frac{-\varOmega}{(1-\tau)W_{1}(x_{2g}-r)}$

(P5)::

W _2b <Γ _2b , W _2g ≥Γ _2g for $\alpha \ge \frac{-\varOmega}{(1-\tau)W_{1}(x_{2g}-r)}$

(P6)::

W _2b ≥Γ _2b , W _2g <Γ _2g for $\alpha <\frac{\varOmega}{(1-\tau)W_{1}(r-x_{2b})}$

Thus, the corresponding problems are

$$ \left. \begin{array} {l@{\quad}l} \mathrm{Max_{\alpha}{:}}& p\lambda U_{L}(W_{2g})+(1-p)\lambda U_{L}(W_{2b})\\[2mm]&\quad{}\displaystyle= -\frac{1}{1-\gamma}\lambda p\bigl[-(x_{2g}-r) (1-\tau)W_{1}\alpha-\varOmega \bigr]^{1-\gamma}\\[4mm]&\qquad{}\displaystyle-\frac{1}{1-\gamma}\lambda(1-p)\bigl[(r-x_{2b}) (1-\tau)W_{1}\alpha -\varOmega \bigr]^{1-\gamma}\\[4mm]\mathrm{such\ that} & \displaystyle\frac{\varOmega}{(1-\tau)W_{1}(r-x_{2b})}<\alpha <\frac{-\varOmega}{(1-\tau)W_{1}(x_{2g}-r)} \end{array}\right\} $$

(P4)

$$ \left.\begin{array}{l@{\quad}l}\mathrm{Max_{\alpha}{:}} & pU_{G}(W_{2g})+(1-p)\lambda U_{L}(W_{2b})\\[3mm]&\quad{}\displaystyle=\frac{1}{1-\gamma}p[(x_{2g}-r)(1-\tau)W_{1}\alpha+\varOmega ]^{1-\gamma}\\[4mm]& \qquad{}\displaystyle-\frac{1}{1-\gamma}\lambda(1-p)[(r-x_{2b})(1-\tau)W_{1}\alpha -\varOmega ]^{1-\gamma} \\[4mm]& \qquad{}\displaystyle-\frac{1}{1-\gamma}\lambda p[-(x_{2g}-r)(1-\tau)W_{1}\alpha-\varOmega ]^{1-\gamma} \\[4mm]& \qquad{}\displaystyle+\frac{1}{1-\gamma}(1-p)[-(r-x_{2b})(1-\tau)W_{1}\alpha+\varOmega ]^{1-\gamma} \\[3mm]\mathrm{such\ that} &\displaystyle\alpha\geq\frac{-\varOmega}{(1-\tau )W_{1}(x_{2g}-r)}\end{array}\right\} $$

(P5)

$$ \left.\begin{array}{l@{\quad}l}\mathrm{Max_{\alpha}{:}} & p\lambda U_{L}(W_{2g})+(1-p)U_{G}(W_{2b})\\[3mm]&\quad{}=\displaystyle -\frac{1}{1-\gamma}\lambda p[-(x_{2g}-r)(1-\tau)W_{1}\alpha-\varOmega ]^{1-\gamma} \\[4mm]& \qquad{}\displaystyle+\frac{1}{1-\gamma}(1-p)[-(r-x_{2b})(1-\tau)W_{1}\alpha+\varOmega ]^{1-\gamma} \\[4mm]\mathrm{such\ that} &\displaystyle\alpha<\frac{\varOmega}{(1-\tau)W_{1}(r-x_{2b})}\end{array}\right\} $$

(P6)

Problem (P4) is a convex programming problem (in α), and thus its maximum will be reached at one of the end points. First-order conditions for (P5) are

and it can be easily verified that α ^∗,+ given by (7) satisfies them. The second-order conditions for (P5) indicate that if λ>1/K _γ then (P5) is concave for $\alpha<\hat{\alpha}_{U}$ and convex for $\alpha>\hat{\alpha}_{U}$, where

$$\hat{\alpha}_{U}=\frac{-\varOmega [ ( \frac{1}{\lambda K_{\gamma }})^{1/(1+\gamma)}+\frac{r-x_{2b}}{x_{2g}-r} ] }{(1-\tau )W_{1}(r-x_{2b}) [ 1- ( \frac{1}{\lambda K_{\gamma}} )^{1/(1+\gamma)} ] }$$

It can be easily seen from (7) that $\alpha^{*,+}>\frac {-\varOmega}{(1-\tau)W_{1}(x_{2g}-r)}$ and also that $\alpha^{*,+}< \hat{\alpha }_{U}$ as λ>1/K _γ and $\lim_{\alpha\rightarrow+\infty}{\mathbb {E}}(U_{LA}(W_{2}-\varGamma_{2}))=-\infty$ as λ>1/K _γ , and thus the maximum is reached at α ^∗,+. In can be also shown that (P5) is concave when

As $\lambda>\frac{1}{K_{\gamma}}$ then

First-order conditions for (P6) are

and it can be easily verified that α ^∗,− given by (8) satisfies them. The second-order conditions for (P3) indicates that if λ>K _γ then (P6) is concave for $\alpha>\hat{\alpha}_{L}$ and convex for $\alpha<\hat{\alpha}_{L}$, where

$$\hat{\alpha}_{L}=\frac{\varOmega [ ( \frac{K_{\gamma}}{\lambda})^{1/(1+\gamma)}+\frac{x_{2g}-r}{r-x_{2b}} ] }{(1-\tau )W_{1}(x_{2g}-r) [ 1- ( \frac{K_{\gamma}}{\lambda} )^{1/(1+\gamma)} ] }$$

(8) implies that $\alpha^{*,-}<\frac{\varOmega}{(1-\tau )W_{1}(r-x_{2b})}$ and also that $\alpha^{*,-}>\hat{\alpha}_{L}$ as λ>K _γ and $\lim_{\alpha\rightarrow-\infty}{\mathbb{E}}(U_{LA}(W_{2}-\varGamma_{2}))=-\infty$ when λ>K _γ . Thus, the maximum is reached at α ^∗,− which concludes the proof. □

Proposition 4

Let λ>max{K _γ ,1/K _γ }. Then the following holds:

• (i) ${\mathbb{E}}(U_{LA}(\alpha^{\ast ,+}))>{\mathbb{E}}(U_{LA}(\alpha^{\ast,-}))$ if K _γ <1,

• (ii)  ${\mathbb{E}}(U_{LA}(\alpha^{\ast,+}))<{\mathbb{E}}(U_{LA}(\alpha^{\ast,-}))$ if K _γ >1, and

• (iii)  ${\mathbb{E}}(U_{LA}(\alpha^{\ast ,+}))={\mathbb{E}}(U_{LA}(\alpha^{\ast,-}))$ if K _γ =1.

Proof

Case (i). It can be shown that

$${\mathbb{E}}\bigl(U_{LA}\bigl(\alpha^{\ast,+}\bigr)\bigr)=\frac{[(x_{2g}-x_{2b})(-\varOmega )]^{1-\gamma}}{1-\gamma}\frac{p ( \frac{1}{K_{0}} )^{\frac{1-\gamma}{\gamma}}-\lambda^{\frac{1}{\gamma}}(1-p)}{ [ \lambda^{\frac{1}{\gamma}}(x_{2g}-r)- ( \frac{1}{K_{0}} )^{\frac {1}{\gamma }}(r-x_{2b}) ]^{1-\gamma}}$$

and

$${\mathbb{E}}\bigl(U_{LA}\bigl(\alpha^{\ast,-}\bigr)\bigr)=\frac{[(x_{2g}-x_{2b})(-\varOmega )]^{1-\gamma}}{1-\gamma}\frac{-\lambda^{\frac{1}{\gamma}}p (\frac{1}{K_{0}} )^{\frac{1-\gamma}{\gamma}}+1-p}{ [ ( \frac{\lambda}{K_{0}} )^{\frac{1}{\gamma}}(r-x_{2b})-(x_{2g}-r) ]^{1-\gamma}}$$

Thus, showing ${\mathbb{E}}(U_{LA}(\alpha^{\ast,+}))>{\mathbb{E}}(U_{LA}(\alpha^{\ast,-}))$, in case (i), boils down to proving

$$\frac{\lambda^{\frac{1}{\gamma}}p ( \frac{1}{K_{0}} )^{\frac {1-\gamma}{\gamma}}-1+p}{\lambda^{\frac{1}{\gamma}}(1-p)-p (\frac{1}{K_{0}} )^{\frac{1-\gamma}{\gamma}}}\ >\ \biggl[ \frac{ (\frac{\lambda}{K_{0}} )^{\frac{1}{\gamma }}(r-x_{2b})-(x_{2g}-r)}{\lambda^{\frac{1}{\gamma}}(x_{2g}-r)- ( \frac{1}{K_{0}} )^{\frac {1}{\gamma }}(r-x_{2b})} \biggr]^{1-\gamma}$$

which follows from the fact that

$$\frac{\lambda^{\frac{1}{\gamma}}p ( \frac{1}{K_{0}} )^{\frac {1-\gamma}{\gamma}}-1+p}{\lambda^{\frac{1}{\gamma}}(1-p)-p (\frac{1}{K_{0}} )^{\frac{1-\gamma}{\gamma}}}\ =\ \frac{ ( \frac {\lambda}{K_{0}} )^{\frac{1}{\gamma}}(r-x_{2b})-(x_{2g}-r)}{\lambda^{\frac {1}{\gamma}}(x_{2g}-r)- ( \frac{1}{K_{0}} )^{\frac{1}{\gamma}}(r-x_{2b})}>1$$

where the last inequality is implied by K _γ <1. Cases (ii) and (iii) directly follow. This concludes the proof. □