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Licensed Unlicensed Requires Authentication Published by De Gruyter August 31, 2020

Maximum Entropy Analysis of Consumption-based Capital Asset Pricing Model and Volatility

Tae-Hwy Lee, Millie Yi Mao and Aman Ullah

Abstract

Based on the maximum entropy (ME) method, we introduce an information theoretic approach to estimating conditional moment functions with incorporating a theoretical constraint implied from the consumption-based capital asset pricing model (CCAPM). Using the ME conditional mean/variance functions obtained from the ME density, we analyze the relationship between asset returns and consumption growth under the theoretical constraint of the CCAPM. We evaluate the predictability of asset return using consumption growth through in-sample estimation and out-of-sample prediction in the ME mean regression function. We also examine the ME variance regression function for the asset return volatility as a function of the consumption growth. Our findings suggest that incorporating the CCAPM constraint can capture the nonlinear predictability of asset returns in mean especially in tails, and that the consumption growth has an effect on reducing stock return volatility, indicating the counter-cyclical variation of stock market volatility.

JEL Classification: C1; C5; G1

Corresponding author: Tae-Hwy Lee, Department of Economics, University of California, Riverside, CA, 92521, USA, E-mail:

A Appendix

A.1 4th-Order Taylor Expansion of Theoretical Constraint

In this subsection, we show the mathematical derivation of the 4th-order Taylor expansion of consumption Euler equation in (10) to obtain (11). After we drop the subscripts of X and Y, we define G(Y,X)βXαY1. According to the Taylor Theorem, G (Y, X) is approximated as

G(Y,X)G(Y0,X0)+GY(YY0)+GX(XX0)+2GY2(YY0)22+2GX2(XX0)22+2GYX(YY0)(XX0)+3GY3(YY0)33!+3GY2X(YY0)22(XX0)+3GYX2(YY0)(XX0)22+3GX3(XX0)33!+4GY4(YY0)44!+4GY3X(YY0)33!(XX0)+4GY2X2(YY0)22(XX0)22+4GYX3(YY0)(XX0)33!+4GX4(XX0)44!,

under the 4th-order expansion. The partial derivatives are computed as

GY=βXα,GX=βαXα1Y,2GY2=0,2GX2=βα(α+1)Xα2Y,2GYX=βαXα1,3GY3=3GY2X=0,3GYX2=βα(α+1)Xα2,3GX3=βα(α+1)(α+2)Xα3Y,4GY4=4GY3X=4GY2X2=0,4GYX3=βα(α+1)(α+2)Xα3,4GX4=βα(α+1)(α+2)(α+3)Xα4Y.

All the above partial derivatives are evaluated at (X0, Y0). After rearranging all the terms, (11) is obtained.

A.2 Recursive Integration

In this subsection, we explain the mathematical details of the recursive integration method in subsection 2.3 with indefinite range. When the integral range of y is definite from a to b, the procedure to compute m (x) and β(x) is similar.

When the range for y is from to +, define the following integrals as functions of x.

Fr(x)+yrexp[λ20y2+λ10(x)y]dy,

where r = 0, 1, 2, ….

F0(x)+exp[λ20y2+λ10(x)y]dy,F1(x)+yexp[λ20y2+λ10(x)y]dy,F2(x)+y2exp[λ20y2+λ10(x)y]dy.

Firstly, assuming λ20>0,

0=+d exp[λ20y2+λ10(x)y]=+(2λ20yλ10(x))e[λ20y2+λ10(x)y]dy=2λ20F1(x)λ10(x)F0(x).

Secondly,

F0(x)=+exp[λ20y2+λ10(x)y]dy=ye[λ20y2+λ10(x)y]|++yde[λ20y2+λ10(x)y]=+(2λ20y2+λ10(x)y)e[λ20y2+λ10(x)y]dy=2λ20F2(x)+λ10(x)F1(x).

Thirdly,

F1(x)=+yexp[λ20y2+λ10(x)y]dy=+exp[λ20y2+λ10(x)y]d(12y2)=12y2e[λ20y2+λ10(x)y]|++12y2de[λ20y2+λ10(x)y]=+12y2(2λ20y+λ10(x))e[λ20y2+λ10(x)y]dy=λ20F3(x)+12λ10(x)F2(x).

Thus,

(33)F3(x)=1λ20F1(x)λ10(x)2λ20F2(x).

Define

F0(x)dF0(x)dx,F1(x)dF1(x)dx,F2(x)dF2(x)dx,andλ10(x)dλ10(x)dx.

Firstly, solve for F0(x)

F0(x)ddx+exp[λ20y2+λ10(x)y]dy=+ddxexp[λ20y2+λ10(x)y]dy=+λ10(x)ye[λ20y2+λ10(x)y]dy=λ10(x)F1(x).

Secondly, solve for F1(x)

F1(x)ddx+yexp[λ20y2+λ10(x)y]dy=+ddxyexp[λ20y2+λ10(x)y]dy=+λ10(x)y2e[λ20y2+λ10(x)y]dy=λ10(x)F2(x).

Thirdly, solve for F2(x)

F2(x)ddx+y2exp[λ20y2+λ10(x)y]dy=+ddxy2exp[λ20y2+λ10(x)y]dy=+λ10(x)y3e[λ20y2+λ10(x)y]dy=λ10(x)F3(x).

Substitute F3 (x) for (33) to obtain

F2(x)=λ10(x)λ10(x)2λ20F2(x)λ10(x)λ20F1(x).

The expressions for F0(x), F1(x), and F2(x) can be written in a linear system,

F0(x)=Λ01(x)F1(x),F1(x)=Λ12(x)F2(x),F2(x)=Λ21(x)F1(x)+Λ22(x)F2(x),

where Λ’s denote the corresponding coefficients.

Since

F0(x0+h)F0(x0)+F0(x)h,F1(x0+h)F1(x0)+F1(x)h,F2(x0+h)F2(x0)+F2(x)h,

for a given initial value x0 and a small increment h, the functions of x, F0 (x), F1 (x) and F2 (x) can be traced out. Given that

m(x)=F1(x)F0(x),
β(x)=F1(x)F0(x)F1(x)F0(x)F02(x),

the ME conditional mean function and its response function can be traced out as well.

Acknowledgments

The authors are thankful to Raffaella Giacomini (the editor) and two anonymous referees for many valuable comments, to Amos Golan, Claudio Morana, and Liangjun Su for discussions on the subject matter of this paper, and to the seminar participants at the World Finance & Banking Symposium, New Delhi.

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Received: 2019-09-22
Accepted: 2020-07-16
Published Online: 2020-08-31

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