Equity premium puzzle or faulty economic modelling?

Abstract

In this paper we revisit the equity premium puzzle reported in 1985 by Mehra and Prescott. We show that the large equity premium that they report can be explained by choosing a more appropriate distribution for the return data. We demonstrate that the high-risk aversion value observed by Mehra and Prescott may be attributable to the problem of fitting a proper distribution to the historical returns and partly caused by poorly fitting the tail of the return distribution. We describe a new distribution that better fits the return distribution and when used to describe historical returns can explain the large equity risk premium and thereby explains the puzzle.

Appendix

In this appendix, we derive the Lévy exponent and MGF of a doubly subordinated IG process.

Corollary: Let T(t) be an IG subordinator with Lévy exponent

$$\begin{aligned} \psi _{T(1)}(u)=-\ln E\left[ \exp \left( iu\,T(1) \right) \right] =\,-\frac{\lambda _T}{\mu _{T}}\left( 1-\sqrt{1-\frac{2\mu _{T}^2 \,iu}{\lambda _T}} \right) , \end{aligned}$$

(21)

and U(t), independent of T(t), be an IG process with Laplace exponent given

$$\begin{aligned} \phi _{U(1)}(s)=-\ln E\left[ \exp \left( -s\,U(1) \right) \right] =\,-\frac{\lambda _U}{\mu _{U}}\left( 1-\sqrt{1+\frac{2\mu _{U}^2 \,s}{\lambda _U}} \right) , \end{aligned}$$

(22)

then $V(1)=T\left( U(1) \right)$ is a subordinator with Lévy exponent given

$$\begin{aligned} \psi _{V(1)}(u)=\phi _{U(1)}\left( \psi _{T}(u) \right) =-\frac{\lambda _U}{\mu _{U}} \left( 1-\sqrt{1-2\,\frac{\mu _{U}^2\,\lambda _T}{\lambda _U\,\mu _{T}} \left( 1-\sqrt{1-\frac{2\mu _{T}^2}{\lambda _T}iu}\right) }\right) , \end{aligned}$$

(23)

and the MGF given

$$\begin{aligned} M_{V(1)}(v)=\,\exp \left( \frac{\lambda _U}{\mu _{U}} \left( 1-\sqrt{1-2\,\frac{\mu _{U}^2\,\lambda _T}{\lambda _U\,\mu _{T}} \left( 1-\sqrt{1-\frac{2\mu _{T}^2}{\lambda _T}v}\right) }\right) \right) , \end{aligned}$$

(24)

where $v\in \,\left( 0,\frac{\lambda _T}{2\mu _{T}^2}\left[ 1-\left( \frac{\lambda _T\,\mu _{T}}{2\mu _{U}^2\,\lambda _U}\right) ^2 \right] \right)$.

Proof

If T(t) is a Lévy subordinator with Lévy exponent, $\psi _{T}(u)\,=\,-\ln \,E\left[ \exp \left( iu\,T(1) \right) \right]$ $u\in {\mathbb {R}}$, and U(t), independent of T(t), is a Lévy subordinator with Laplace exponent $\phi _{U}(s)\,=\,-\ln E\left[ \exp \left( -s\,U(1)\right) \right]$, $s>0$, then the subordinator process $Y(t)=T\left( U(t) \right)$ is again a Lévy subordinator with Lévy exponent and probability transition given⁹

$$\begin{aligned} \psi _{Y}(u)= & \, \phi _{T}\left( \psi _{U}(u)\right) , \end{aligned}$$

(25)

$$\begin{aligned} P_{Y}(t,A)= & \, \int _{0}^{\infty } P_{T}(t,A)\,P_{U}(u,dt), \end{aligned}$$

(26)

respectively. Using (25), and substituting (21) in (22), the Lévy exponent and consequently the MGF are obtained.$\square$