Initial results examining the Equity Premium puzzle using Habit Formation
and non-Gaussian stock price movement
Oren Shmuel
The Graduate Center, City University of New York
Current version: October 2015
Abstract
Mehra and Prescott (1985) found that for reasonable values of a relative risk aversion coefficient
(=RRA) and given small variance of the growth rate in per capita consumption , an Equity
Premium puzzle exists. In other words, the difference between the expected rate of return on the
stock market and the riskless rate of interest (= 𝜇 − 𝑟 ) is too big. I will use a model with no
time separability of preferences and with habit persistence (= adjacent complementarity in
consumption) to try and explain the equity premium puzzle. In addition, and unique to this paper,
I represent the stock price movement using a right skewed non-Gaussian model (the Gumbel
distribution). My results suggest that habit persistence can generate the Mehra and Prescott’s
(1985) sample mean and variance of the consumption growth rate with low risk aversion, hence
giving a plausible explanation to the equity premium puzzle.
I) Introduction
The equity premium puzzle emanates from the inability of the theoretical models to explain the
empirically observed high equity premium (when the average stock returns so much higher than
the average bond returns). It is based on the fact that in order to reconcile the much higher return
on stock compared to government bonds in the United States, individuals must have very high
risk aversion according to standard economics models.
Mehra and Prescott (1985) reported the existence of an equity premium puzzle. They calibrated
model parameters to match the sample mean, variance, and first order autocorrelation of the
annual growth rate of per capita consumption in the years 1889-1978. They used a time- and
state-separable utility. They were unable to determine pairs of discount rate and relative risk
aversion (RRA) coefficients to match the sample mean of the annual real interest rate and of the
equity premium between 1889-1978, hence a puzzle. In other words, the consumption growth
rate was too smooth to justify the average equity premium.
One type of tests, to resolve the equity premium puzzle, examines the Euler equation restriction
on the product of asset returns with the marginal rate of substitution of the representative agent.
Habit persistence usually causes specific lag coefficients in the Euler equation to be negative.
Positive coefficients in the Euler equation are considered to be evidence of durability of goods,
rather than habit persistence. Ferson and Constantinides (1989) used quarterly and annual data
and found negative coefficients in the Euler equation which supports habit persistence. They also
rejected the time separable model and supported instead a model with habit persistence with
respect of explaining the equity premium puzzle (as in this paper).
Hansen and Jagannathan (1988), using monthly data, support habit persistence. Heaton (1988)
found evidence to support habit persistence by checking the autocorrelations in consumption,
while considering time aggregation.
However other papers, such as Dunn and Singleton (1986), Gallant and Tauchen (1989),
Eichenbaum Hansen and Singleton (1988) and Eichen and Hansen (1990), found that specific lag
coefficients in the Euler equation to be positive which support durability of goods rather than
habit persistence.
In addition, Hansen and Singleton (1982, 1983), Ferson (1983), Grosssman, Melino, and Shiller
(1987) rejected the Euler equation restriction on asset returns and the marginal rate of
substitution caused by the time- and state-separable preferences.
Shiller (1981) found that stock market returns are too volatile relative to the volatility of
dividends.
Duesenberry (1949) is the first comprehensive examination of the effects of habit persistence.
Duesenberry’s demonstration effect is a type of consumption externality where an individual’s
utility depends not only upon his consumption level (Ct ) but also upon the social average level
(or Habit formation) of consumption ( C t ). Duesenberry (1949) claims that the demonstration
effect causes unhappiness with current levels of consumption, which affects savings rates and
macroeconomic growth.
The Brock (1982) asset pricing model can estimate a significant equity premium. Using the
Brock’s (1982) asset pricing model, Akdeniz and Dechert (2007) show that there are
parameterizations of the Brock model that have equity premia that are more consistent with the
empirical evidence than the equity premia that were observed by Mehra and Prescott (1985).
Kocherlakota (1996) tries to resolve the equity premium puzzle and the risk free rate puzzle by
reviewing the literature. Kocherlakota report the papers that try to explain these two puzzles.
Campbell and Cochrane (1999) find the equity premium puzzle using a consumption-based
model with external habit formation. They use a power utility and the Sharpe ratio inequality to
explain the equity premium puzzle.
In this paper, I represent the stock price movement using a right skewed non-Gaussian model
(the Gumbel distribution) unlike other papers, such as Constantinides (1990), that use a Gaussian
model for this purpose. To better represent a risk averse investor, I use a right skew distribution
for the stock price since such a distribution gives higher weight to negative stock return than
positive stock return.
The objective of this paper is to explain the equity premium puzzle in a rational expectations
model with time nonseparability of preferences and with habit persistence (i.e. adjacent
complementarity in consumption) while representing the stock price movement using a right
skewed non-Gaussian model (the Gumbel distribution).
The paper is organized as follows: In section II, the theoretical model and its assumptions are
presented. In section III, I find an optimal Consumption policy and an optimal Investment policy.
In section IV, I find the relation between the RRA coefficient and the intertemporal Elasticity of
Substitution in Consumption. In section V, I attempt to explain the Equity Premium Puzzle. In
section VI, I examine the effect of time separability in utility preferences on the Equity Premium
puzzle. In section VII, I examine the effect of both time separability in utility preferences and
Habit persistence on the Equity Premium puzzle. In section VIII, I present my conclusions.
II) The Model and Assumptions
I assume the utility function for the representative agent in this economy to be of the
following power type:
1
𝑢(𝑐(𝑡), 𝑥(𝑡)) = 𝛾 [𝑐(𝑡) − 𝑥(𝑡)]𝛾
(1)
where 𝑥(𝑡) is the subsistence level of per capita consumption generated by Habit persistence (i.e.
the habit-forming state variable), 𝑐(𝑡) is the level of per capita consumption, 𝛾 is the utility
sensitivity to the difference between consumption and the subsistence level of consumption.
𝑥(𝑡) is defined by:
𝑡
𝑥(𝑡) = 𝑥0 𝑒 −𝑎𝑡 + 𝑏 ∫0 𝑒 𝑎(𝑠−𝑡) 𝑐(𝑠)𝑑𝑠
(2)
The special case of 𝑏 = 0 corresponds to a time-separable utility.
Thus, for a time-separable utility, the coefficient of absolute risk aversion is:
(𝛾 − 1) [𝑐(𝑡) − 𝑥(𝑡)]𝛾−2
𝑢′′ (𝑐𝑡 , 𝑥𝑡 )
1−𝛾
𝐴[𝑐(𝑡) − 𝑥(𝑡)] = − [ ′
] = −[
]
=
𝑢 (𝑐𝑡 , 𝑥𝑡 )
[𝑐(𝑡) − 𝑥(𝑡)]𝛾−1
𝑐(𝑡) − 𝑥(𝑡)
The coefficient of relative risk aversion is:
𝑅[𝑐(𝑡) − 𝑥(𝑡)] = 𝑅𝑅𝐴 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑐 − 𝑥 = [𝑐(𝑡) − 𝑥(𝑡)] 𝐴(𝑐𝑡 ) = 1 − 𝛾
Another RRA definition :
𝑢𝑥𝑥
−(𝛾 − 1) [𝑐(𝑡) − 𝑥(𝑡)]𝛾−2
𝑅𝑅𝐴 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑥 = 𝑥(𝑡) [−
] = 𝑥(𝑡) [
]=
(−1)[𝑐(𝑡) − 𝑥(𝑡)]𝛾−1
𝑢𝑥
= (𝛾 − 1)
𝑥(𝑡)
= (𝛾 − 1)𝑦(𝑡)
𝑐(𝑡) − 𝑥(𝑡)
For simplicity, I assume that there are only two assets (without loss of generality), one risky
stock and one risk-free bond, to invest in.
To better represent a risk averse investor, I use a right skew distribution for the stock price since
such a distribution gives higher weight to negative stock return than positive stock return.
To represent the stock price motion I use the Gumble probability distribution which is a nonGaussian distribution with an asymmetric right skew. Thus, the formulation of the stock price St
is:
𝑑𝑆𝑡 = [𝜇𝑡 + 𝜀𝜎𝑡 ]𝑆𝑡 𝑑𝑡 +
(3)
Where
𝜋
𝜎
√6 𝑡
𝜋
𝜎 𝑆 𝑑𝑤𝑡
√6 𝑡 𝑡
is the Gumble distribution standard deviation, 𝜇𝑡 + 𝜀𝜎𝑡 is the Gumble distribution
mean, 𝜀 = 0.577216 the Euler’s constant, 𝜎𝑡 is the distribution scale (i.e. the normal distribution
standard deviation) of the price of stock (which is a risky asset), 𝜇𝑡 is the distribution location
(i.e. the normal distribution mean) of the risky rate of return and 𝑤𝑡 is a Wiener process.
The percent change, i.e. the return, of the stock price is described as:
(4)
𝑑𝑆𝑡
𝑆𝑡
= [𝜇 + 𝜀𝜎]𝑑𝑡 +
𝜋
√6
𝜎𝑑𝑤𝑡
The infinitely lived representative consumer has wealth 𝑊𝑡 .
The change in the per capita wealth ( W t ) is determined by the return from the risky asset (the
stock) and the return from the riskless asset (the risk-free bond) less consumption. Hence:
(5)
𝑑𝑊𝑡 = (1 − 𝛼𝑡 )𝑊𝑡 𝑟𝑑𝑡 + 𝛼𝑡 𝑊𝑡
𝑑𝑆𝑡
𝑆𝑡
− 𝑐𝑡 𝑑𝑡
where r is risk-free rate of return, Ct is per capita consumption, and 𝛼𝑡 is the portion, i.e. weight,
of the per capita wealth ( W t ) that is invested in the risky asset.
Inserting equation (4) into equation (5) yields the equation governing the changes in the wealth
written as follows:
(6)
𝜋
√6
𝑑𝑊𝑡 = {[𝛼𝑡 (𝜇 − 𝑟 + 𝜀𝜎) + 𝑟]𝑊𝑡 −𝑐𝑡 }𝑑𝑡 + ( ) 𝛼𝑡 𝑊𝑡 𝜎𝑑𝑤𝑡
where 𝑊𝑡 is the per capita wealth, 𝑟 is risk-free rate of return, 𝜇 is risky rate of return, 𝛼𝑡 is the
proportion of wealth invested in the risky asset, and 𝑤𝑡 is a Wiener process.
My goal is not to study the most general utility function that exhibits habit persistence but rather
to use the most simple utility specification that may explain the equity premium puzzle.
If 𝛼𝑡 = 0, equation (6) becomes:
𝑑𝑊𝑡 = {𝑟𝑊𝑡 −𝑐𝑡 }𝑑𝑡
(7)
𝑑𝑊𝑡
𝑑𝑡
Thus
= 𝑟𝑊𝑡 −𝑐𝑡
Using a constant wealth path in equation (7), i.e.
𝑑𝑊𝑡
𝑑𝑡
= 0 , I derive
𝑐𝑡 = 𝑟𝑊𝑡
For a particular consumption policy (when 𝛼𝑡 = 0) where
𝑊𝑡 = 𝑊0 𝑒 (𝑏−𝑎)𝑡 > 0
and
𝑐𝑡 = (𝑟 + 𝑎 − 𝑏)𝑊𝑡 = (𝑟 + 𝑎 − 𝑏)𝑊0 𝑒 (𝑏−𝑎)𝑡 > 0
thus, using equation (2) I derive
(8)
𝑡
𝑐(𝑡) − 𝑥(𝑡) = (𝑟 + 𝑎 − 𝑏)𝑊0 𝑒
(𝑏−𝑎)𝑡
− [𝑥0
𝑒−𝑎𝑡
+ 𝑏 ∫ 𝑒𝑎(𝑠−𝑡) 𝑐(𝑠)𝑑𝑠]
0
𝑥(𝑡)
= (𝑟 + 𝑎 − 𝑏) [𝑊(𝑡) −
]
𝑟+𝑎−𝑏
For a time-separable utility (i.e. 𝑏 = 0 ) I derive
𝑊𝑡 = 𝑊0 𝑒 (𝑏−𝑎)𝑡 = 𝑊0 𝑒 (−𝑎)𝑡
𝑡
𝑥(𝑡) = 𝑥0 𝑒
−𝑎𝑡
+ 𝑏 ∫ 𝑒 𝑎(𝑠−𝑡) 𝑐(𝑠)𝑑𝑠 = 𝑥0 𝑒 −𝑎𝑡
0
Hence
𝑐(𝑡) − 𝑥(𝑡) = (𝑟 + 𝑎 − 𝑏)𝑊0 𝑒 (−𝑎)𝑡 − 𝑥0 𝑒−𝑎𝑡 = (𝑟 + 𝑎 − 𝑏)𝑒 (−𝑎)𝑡 [𝑊0 −
To require 𝑐(𝑡) − 𝑥(𝑡) > 0 I need to make the following conditions:
(9)
𝑊0 > 0
(10)
0
𝑊0 − 𝑟+𝑎−𝑏
>0
(11)
𝑟+𝑎−𝑏 >0
𝑥
𝑥0
𝑟+𝑎−𝑏
]
𝑟+𝑎 >𝑏 >0
Or
In addition, I can use a constant consumption path which can be described as
𝑐(𝑡) =
(12)
𝑎𝑥(𝑡)
𝑏
or
𝑏𝑐(𝑡) − 𝑎𝑥(𝑡) = 0
III) Finding an optimal Consumption policy (𝒄∗𝒕 ) and Investment policy (𝜶∗𝒕 )
Theorem 1 shows existence and uniqueness of an optimal policy which determine the utility of
capital, and the dynamics of capital and consumption.
From the proof of Theorem 1 in Appendix A, I find the following optimal Consumption policy
(𝑐𝑡∗ ) and Investment policy (𝛼𝑡∗ ):
𝛾
(13) (A17)
∗ (𝑡)
(14) (A5)
𝛼
(15) (A19)
𝑚=
(16) (A20)
𝑥(𝑡)
(𝑟+𝑎)
𝑐 ∗ (𝑡) = 𝑥(𝑡) − (ℎ)𝛾−1 [𝑊(𝑡) − 𝑟+𝑎−𝑏] [𝐻(𝑟+𝑎−𝑏)]
= 𝑚 [1 −
𝑥(𝑡)
𝑊(𝑡)
𝑟+𝑎−𝑏
(𝜇−𝑟+𝜀𝜎)
where 0 ≤ 𝑚 ≤ 1
𝜋2
( )𝜎2 (1−𝛾)
6
𝑟+𝑎−𝑏
]
ℎ = (𝑎+𝑟)(1−𝛾) {𝜌 − 𝛾𝑟 −
(𝜇 −𝑟+𝜀𝜎)2 (−1)
𝜋2 2
)𝜎 (1−𝛾)
6
4(
The utility of wealth (=capital):
(17) (A10)
(18) (A21)
The wealth (=capital) is
𝑉[𝑊(𝑡), 𝑥(𝑡)] =
1
(ℎ)𝛾
𝛾
𝑥(𝑡)
𝛾 −1
[𝑊(𝑡) − 𝑟+𝑎−𝑏] ( 𝐻 )
9𝛾(𝜇 −𝑟+𝜀𝜎)2
𝐻 = (1−𝛾) {𝜌 − 𝛾𝑟 − 2(𝜋2 )𝜎2 (1−𝛾)}
}>0
𝑊(𝑡) =
(19) (A24)
𝑥(𝑡)
𝑟+𝑎−𝑏
𝜋
+ [𝑊(0) −
𝑥(0)
[𝑛𝑡+( )𝜎𝑚𝑤(𝑡)]
√6
]𝑒
𝑟+𝑎−𝑏
where
𝑛=
(20) (A22)
(𝜇−𝑟+𝜀𝜎)2
𝜋2
( )𝜎2 (1−𝛾)
6
+𝑟−(
1
1−𝛾)
{𝜌 − 𝛾𝑟 +
(𝜇−𝑟+𝜀𝜎)2
𝜋2
4( 6 )𝜎2 (1−𝛾)
}
The above optimal policies (𝑐 ∗ , 𝛼 ∗ ) are unique as I show in appendix A.
The consumption growth rate is
(21) (A30)
𝑑𝑐(𝑡)
𝑐(𝑡)
= [𝑛 + 𝑏 −
(𝑛+𝑎)𝑥(𝑡)
𝑐(𝑡)
] 𝑑𝑡 + [1 −
𝑥(𝑡)
𝑐(𝑡)
𝜋
] ( ) 𝜎𝑚𝑑𝑤(𝑡)
√6
I use the optimal policies found in theorem 1 as the equilibrium paths in a representativeconsumer economy. Specifically the consumption growth rate (defined in equation 21) is
considered to be the per capita consumption growth rate.
I will now define 𝑧(𝑡) the subsistence rate of consumption generated by Habit Persistence.
(22)
𝑧(𝑡) =
𝑥(𝑡)
𝑐(𝑡)
Where 𝑥(𝑡) is the subsistence level of consumption generated by Habit Persistence.
Since
(23)
𝑥(𝑡)
𝑅𝑅𝐴 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑥 = (𝛾 − 1) 𝑐(𝑡)−𝑥(𝑡)
I will define
(24)
𝑥(𝑡)
𝑧(𝑡)
𝑦(𝑡) = 𝑐(𝑡)−𝑥(𝑡) = 1−𝑧(𝑡)
Theorem 2 derives the stationary distribution of the state variable (= z) and calculates the
unconditional mean and variance of the consumption growth rate.
From Theorem 2 proof in Appendix B, 𝑦(𝑡) has a stationary probability distribution with
density
𝜋2
−12[𝑛+𝑎−𝑏+(
{
(25) (B14)
where
𝑝𝑦 (𝑦𝑡 , 𝑡) = 𝑘𝑦
6
𝜋2 𝜎2 𝑚2
)𝜎2 𝑚2 ]
}
𝑒
{
−12𝑏
}
𝜋2 𝜎2 𝑚2 𝑦
𝜋2
12[𝑛+𝑎−𝑏+(
{1−
𝑘
(26) (B15)
−1
6
)𝜎2 𝑚2 ]
}
𝜋2 𝜎2 𝑚2
12𝑏
= (𝜋2 𝜎2 𝑚2 )
𝜋2
𝛤{
12[𝑛+𝑎−𝑏+( )𝜎2 𝑚2 ]
6
𝜋2 𝜎2 𝑚2
− 1}
And 𝛤( ) is the gamma function.
Using the above probability distribution density I will calculate the following:
𝑦(𝑡) has a single mode (see appendix B)
𝑦̂ =
(27) (B12)
𝑏
𝜋2
𝑛+𝑎−𝑏+( )𝜎2 𝑚2
<∞
6
𝜋2
Thus requires the condition 𝑛 + 𝑎 − 𝑏 + ( 6 ) 𝜎 2 𝑚2 > 0
And a mean
𝑏
𝑦̅ = 𝑛+𝑎−𝑏 < ∞
(28) (B11)
Thus requires the condition 𝑛 + 𝑎 − 𝑏 > 0
𝑧(𝑡) =
𝑥(𝑡)
𝑐(𝑡)
has a stationary probability distribution with the density in equation (29)
𝜋2
12[𝑏−𝑎−𝑛−(
(29) (B17)
𝑝𝑧 (𝑧𝑡 , 𝑡) = 𝑘𝑒
12𝑏
}
𝜋2 𝜎2 𝑚2
{
(1 − 𝑧)
[
12(𝑏+𝑎−𝑏)
]
𝜋2 𝜎2 𝑚2
{
6
)𝜎2 𝑚2 ]
}
𝜋2 𝜎2 𝑚2
𝑧
𝑒
{
−12𝑏
}
𝜋2 𝜎2 𝑚2 𝑧
Where 0 ≤ 𝑧 < 1
For the stationary distribution, 𝑧(𝑡) has a single mode
2
2
6
6
2
2
𝜋
𝜋
𝜋
[𝑛+𝑎+( )𝜎2 𝑚2 ]−√[𝑛+𝑎+( )𝜎2 𝑚2 ] −4( )𝜎2 𝑚2 𝑏
𝑧̂ =
(30) (B19)
6
𝜋2
2( )𝜎2 𝑚2
6
𝐸[
Next, I will derive the unconditional mean of consumption growth
In appendix C, i derive
𝐸[
(31) (C1)
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
1
= 𝑛 + 𝑏 − (𝑛 + 𝑎)𝐸[𝑧(𝑡)] = 𝑛 + 𝑏 − (𝑛 + 𝑎) ∫0 𝑧(𝑡) 𝑝𝑧 (𝑧𝑡 , 𝑡)𝑑𝑧
Since a closed form expression for the integral is unavailable, the integration is done
numerically.
𝑣𝑎𝑟[
Next, I will derive the unconditional variance of consumption growth
In appendix C, i derive
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
(32) (C2)
𝑣𝑎𝑟[
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
𝜋2
𝜋2
1
= ( 6 ) 𝜎 2 𝑚2 𝐸{[1 − 𝑧(𝑡)]2 } = ( 6 ) 𝜎 2 𝑚2 ∫0 [1 − 𝑧(𝑡)]2 𝑝𝑧 (𝑧𝑡 , 𝑡)𝑑𝑧
Since a closed form expression for the integral is unavailable, the integration is done
numerically.
IV) The relation between the RRA coefficient and the intertemporal Elasticity of
Substitution in Consumption (=s)
Constantinides (1990) asserts that “Habit Persistence drives a wegde between the RRA
coefficient and the inverse of the intertemporal Elasticity of Substitution in Consumption (=s)”.
That means that the relation can be described as
1
𝑅𝑅𝐴 = 𝑤𝑒𝑑𝑔𝑒 ∗ 𝑠
(32.5)
Or
𝑅𝑅𝐴 ∗ 𝑠 = 𝑤𝑒𝑑𝑔𝑒
For a time-separable model (b=0), 𝑅𝑅𝐴 ∗ 𝑠 = 1 . I will show that for a nonseparable model and
for specific variable values that may explain the equity premium puzzle the wedge is extremely
below one i.e. 𝑅𝑅𝐴 ∗ 𝑠 << 1.
In appendix D, I derive the expressions for RRA coefficient and s (=intertemporal Elasticity of
Substitution in Consumption).
RRA coefficient defined as
(33) (D1)
𝑅𝑅𝐴 =
−𝑊𝑉𝑊𝑊
= (1 − 𝛾)
𝑉𝑊
1
1−
𝑥(𝑡)
𝑊(𝑟 + 𝑎 − 𝑏)
In the short run, a shock of a decrease in wealth does not change 𝑥(𝑡) but increases the RRA
coefficient. In the long run, the stationary distribution of the RRA coefficient decreases the RRA
coefficient back.
thus
(34) (D3)
ℎ
𝑅𝑅𝐴 = (1 − 𝛾) {1 + 𝑦(𝑡) [(𝑟+𝑎−𝑏)]}
Since 𝑦(𝑡) has a steady state distribution, so does the RRA coefficients. Thus,
(35) (D4)
ℎ𝑏
̅̅̅̅̅̅ = (1 − 𝛾) {1 + [
𝑅𝑅𝐴
]}
(𝑛+𝑎−𝑏)(𝑟+𝑎−𝑏)
When 𝑧(𝑡) = 𝑧̂ (=modal value of z)
(36) (D5)
𝑧̂
ℎ
𝑅𝑅𝐴(𝑧 = 𝑧̂ ) = (1 − 𝛾) {1 + [1−𝑧̂ ] [(𝑟+𝑎−𝑏)]}
𝑑𝑐
)
𝑐 ]
𝑑𝑡
𝐸(
𝜕[
(37) (D6)
𝑠=
𝜕𝑟
=
1−𝑧(𝑡)
1−𝛾
when
𝑧(𝑡), 𝜇 − 𝑟, 𝑎𝑛𝑑 𝜎 2
held
constant
(38) (D7)
ℎ
𝑠 ∗ 𝑅𝑅𝐴 = 1 − {1 − [(𝑟+𝑎−𝑏)]} 𝑧(𝑡)
The modal value of s*RRA is
(39) (D8)
ℎ
𝑚𝑜𝑑𝑒(𝑠 ∗ 𝑅𝑅𝐴) = 1 − {1 − [(𝑟+𝑎−𝑏)]} 𝑧̂
For the equilibrium of my specific model I can write the following equations
𝜕𝑐 ⁄𝑐
(40)
𝑠 ∗ 𝑅𝑅𝐴 = 𝜕𝑊⁄𝑊 when 𝑥(𝑡) is held constant
(41)
𝑠 ∗ 𝑅𝑅𝐴 = 𝑠𝑡𝑑(𝜕𝑊⁄𝑊)
𝑠𝑡𝑑(𝜕𝑐 ⁄𝑐 )
Which is the ratio of the standard deviation of the consumption growth rate and the standard
deviation of the capital (wealth) growth.
V) Examining the Equity Premium Puzzle
Mehra and Prescott (1985) estimated the mean of the annual growth rate of per capita real
consumption of nondurables and services in the years 1889-1978 to be 0.0183 per year. Thus I
will use his estimate and set
𝐸[
𝑑𝑐(𝑡)
]
𝑐(𝑡)
(42)
= 0.0183 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟.
𝑑𝑡
In addition, they also estimated the standard deviation of the growth rate in the years 1889-1978
to be 0.0357 per year. Thus I will use his estimate and set
𝑣𝑎𝑟[
(43)
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
= (0.0357)2 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟.
Mehra and Prescott (1985) also estimated the mean annual real rate of return on a relatively
riskless security to be 0.008. Thus I will set
(44)
r = 0.01 per year.
This paper uses an economy that allows for production, thus I define 𝐾(𝑡) as the capital of a
firm. The firm invest 𝛿1 𝐾(𝑡) capital in risky technology and (1 − 𝛿1 )𝐾(𝑡) capital in riskless
technology. Thus 0 < 𝛿1 ≤ 1. The firm can finance its activity with equity (= 𝑆(𝑡)) and with
𝑆(𝑡)
riskless debt (= 𝐵(𝑡)) . The firm ratio of equity value to entire capital is 𝑆(𝑡)+𝑩(𝑡) = 𝛿2 which is
kept constant and it’s domain is 0 < 𝛿2 ≤ 1. The firm’s capital (= 𝐾(𝑡)) is equal to the sum of
the value of equity and the value of riskless debt, i.e. [= 𝐾(𝑡) = 𝑆(𝑡) + 𝐵(𝑡)]. Since bonds are
riskless, the rate of return for riskless assets is estimated by the bonds return which is equal to
𝑑𝐵(𝑡)⁄𝐵(𝑡) = 𝑟𝑑𝑡 . Since equity is risky, the rate of return for risky assets is estimated by the
equity return which is equal to 𝑑𝑆(𝑡)⁄𝑆(𝑡) .
The total capital change of the firm should equal to the returns from its investments thus
(45)
𝑑𝑆(𝑡) + 𝐵(𝑡)𝑟𝑑𝑡 = 𝛿1 𝐾(𝑡)
𝑑𝑆(𝑡)
𝑆(𝑡)
+ (1 − 𝛿1 )𝐾(𝑡)𝑟𝑑𝑡
From equation (4)
𝑑𝑆(𝑡)
(4)
𝑆(𝑡)
= (𝜇 + 𝜀𝜎)𝑑𝑡 +
𝜋
√6
𝜎𝑑𝑤(𝑡)
Insert equation (4) into (45)
(45)
𝑑𝑆(𝑡) + 𝐵(𝑡)𝑟𝑑𝑡 = 𝛿1 𝐾(𝑡) [(𝜇 + 𝜀𝜎)𝑑𝑡 +
𝜋
√6
𝜎𝑑𝑤(𝑡)] + (1 − 𝛿1 )𝐾(𝑡)𝑟𝑑𝑡
Divide both sides by 𝑆(𝑡) to derive
(46)
𝑑𝑆(𝑡)
𝑆(𝑡)
𝐵(𝑡)
𝐾(𝑡)
+ 𝑆(𝑡) 𝑟𝑑𝑡 = 𝛿1 𝑆(𝑡) [(𝜇 + 𝜀𝜎)𝑑𝑡 +
𝜋
√6
𝐾(𝑡)
𝜎𝑑𝑤(𝑡)] + (1 − 𝛿1 ) 𝑆(𝑡) 𝑟𝑑𝑡
𝑑𝑆(𝑡)
1
1
𝜋
1
+ [ − 1] 𝑟𝑑𝑡 = 𝛿1 [(𝜇 + 𝜀𝜎)𝑑𝑡 +
𝜎𝑑𝑤(𝑡)] + (1 − 𝛿1 ) 𝑟𝑑𝑡
𝑆(𝑡)
𝛿2
𝛿2
𝛿2
√6
thus
(47)
𝑑𝑆(𝑡)
𝑆(𝑡)
𝛿
𝛿
𝜋
= [𝛿1 (𝜇 + 𝜀𝜎 − 𝑟) + 𝑟] 𝑑𝑡 + (𝛿1 ) ( ) 𝜎𝑑𝑤(𝑡)
2
2
√6
In appendix E I derive
𝐸[
(48) (E1)
𝑑𝑆(𝑡)
]
𝑆(𝑡)
𝑑𝑡
𝑣𝑎𝑟[
(49) (E2)
𝛿
= [𝛿1 (𝜇 + 𝜀𝜎 − 𝑟) + 𝑟]
2
𝑑𝑆(𝑡)
]
𝑆(𝑡)
𝑑𝑡
𝛿
2 𝜋2
= (𝛿1 ) ( 6 ) 𝜎 2
2
Mehra and Prescott (1985) estimated the annual real return on the Standard and Poor’s composite
stock price index in the 1889-1978 period to have a mean 0.0698 and a standard deviation
0.1654. Ibbotson and Sinquefield (1982) had generally similar estimates. Thus I set
(50)
𝐸[
𝑑𝑆(𝑡)
]
𝑆(𝑡)
𝑑𝑡
𝛿
= [𝛿1 (𝜇 + 𝜀𝜎 − 𝑟) + 𝑟] = 0.06 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟
2
Inserting equation (44) I derive
(51)
(52)
𝛿1
𝛿2
(𝜇 + 𝜀𝜎 − 𝑟) = 0.06 − 𝑟 = 0.06 − 0.01 = 0.05
𝑣𝑎𝑟[
𝑑𝑆(𝑡)
]
𝑆(𝑡)
𝑑𝑡
𝛿
2 𝜋2
= (𝛿1 ) ( 6 ) 𝜎 2 = (0.165)2 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟
2
From equation (15)
(15) (A19)
𝑚=
(𝜇−𝑟+𝜀𝜎)
𝜋2
( )𝜎2 (1−𝛾)
6
where 0 ≤ 𝑚 ≤ 1
I can set the condition
(53)
𝑚=
(𝜇−𝑟+𝜀𝜎)
𝜋2
( )𝜎2 (1−𝛾)
6
<1
Using equations (51) and (52) I derive
(1 − 𝛾) ≥
(𝜇 −𝑟+𝜀𝜎)
𝜋2
( )𝜎2
6
=
𝛿1 𝛿1
[ (𝜇 −𝑟+𝜀𝜎)]
𝛿2 𝛿2
𝛿 2 𝜋2
( 1 ) ( )𝜎2
𝛿2
6
=
𝛿
( 1 )0.05
𝛿2
(0.165)2
Thus
𝛿
(1 − 𝛾) ≥ ( 1 ) 1.836
𝛿
(54)
2
Using equation (32) and (15) I drive
𝑣𝑎𝑟[
(55)
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
𝜋2
𝜋2
= ( 6 ) 𝜎 2 𝑚2 = ( 6 ) 𝜎 2
(𝜇 −𝑟+𝜀𝜎)2
𝜋2
( )𝜎4 (1−𝛾)2
36
6(𝜇 −𝑟+𝜀𝜎)2
= 𝜋2 𝜎2 (1−𝛾)2
From equation (43)
𝑣𝑎𝑟[
(43)
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
= (0.0357)2
insert equations (43) into (55)
𝑣𝑎𝑟[
(56)
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
6(𝜇 −𝑟+𝜀𝜎)2
= 𝜋2 𝜎2 (1−𝛾)2 = (0.0357)2
𝛿
Multiply numerator and denominator by (𝛿1 )
2
2
𝛿1 2
( ) (𝜇 − 𝑟 + 𝜀𝜎)2
𝛿2
= (0.0357)2
𝜋 2 𝛿1 2 2
( 6 ) ( ) 𝜎 (1 − 𝛾)2
𝛿2
Insert equations (51) and (52) into the above expression
(0.05)2
= (0.0357)2
(0.165)2 (1 − 𝛾)2
Solving for 1 − 𝛾
(57)
1 − 𝛾 = 8.488
𝛿
= (𝛿1 ) 1.836
2
I now intend to show that habit persistence can generate the Mahra and Prescott’s sample mean
and variance of the consumption growth rate with a lower RRA coefficient i.e. with a lower
(1 − 𝛾) see equation (58).
To determine how low a (1 − 𝛾) I can choose I do the following.
Using equation (34) and (57)
ℎ
𝑅𝑅𝐴 = (1 − 𝛾) {1 + 𝑦(𝑡) [(𝑟+𝑎−𝑏)]} > (1 − 𝛾) = 8.488
(58)
Since 𝑦(𝑡) > 0, ℎ > 0, 𝑎𝑛𝑑 𝑟 + 𝑎 − 𝑏 > 0 according to conditions above.
From equation (54)
𝛿
(1 − 𝛾) ≥ ( 1 ) 1.836
𝛿
(54)
2
Inserting equation (57) into (58)
𝛿1
8.488 ≥ ( ) 1.836
𝛿2
Thus
𝛿
4.6233≥ (𝛿1 )
(59)
I thus set
2
𝛿1
𝛿2
= 1 which is consistent with equation (59) and will determine a lower value of
(1 − 𝛾) that I intend to use. Thus from equation (54) I derive
𝛿
(60)
(1 − 𝛾) ≥ ( 1 ) 1.836 = 1.836
𝛿
For simplicity I choose
(61)
(1 − 𝛾) = 1.836
2
Hence from equation (58) I can derive the following regarding the specific lower (1 − 𝛾) I use.
(62)
𝑅𝑅𝐴 > (1 − 𝛾) = 1.836
I set the rate of time preference in units (𝑦𝑒𝑎𝑟)−1 [i.e. the constant utility discount rate see
equation (A8)] to
(63)
𝜌 = 0.037.
I proceed with creating Table 1 using the equations found above and the following settings.
The annual rate of return of the riskless technology= r =0.01, the power in the utility function= =
−0.836 , the rate of time preference in units (𝑦𝑒𝑎𝑟)−1 = 𝜌 = 0.037 ,
𝐸[
𝑑𝑐(𝑡)
]
𝑐(𝑡)
the unconditional mean of the annual growth rate in consumption (line 4 table 1) = 𝑑𝑡 =
0.018 ,
the unconditional standard deviation of the annual growth rate in consumption (line 6 table 1) =
𝑣𝑎𝑟[
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
= 0.036 , and equations (51) and (52).
I have chosen pairs of parameters (a, b) that meet the conditions in equations (11) and (28) and
for which the mean and variance of the consumption growth rate [equations (31) and (32)] match
their sample estimates [equations (42) and (43)].
Table1
Mean and Variance of the Consumption Growth Rate generated by the model with Habit resistence
Parameter a, per year
0.1
0.2
0.3
0.4
0.5
Parameter b
0.093
0.172
0.25
0.328
0.405
Mode(z)
0.68
0.73
0.75
0.75
0.76
Mean annual growth rate in
consumption:
unconditional mean
0.018
0.019
0.018
0.018
0.018
at z= mode(z)
0.034
0.034
0.034
0.034
0.034
Standard deviation of the
annual growth rate in
unconditional mean
0.036
0.036
0.036
0.036
0.036
at z= mode(z)
0.053
0.045
0.042
0.041
0.040
RRA coefficient:
unconditional mean
3.21
2.67
2.43
2.30
2.21
at z= mode(z)
2.94
2.58
2.38
2.27
2.19
Elasticity of substitution (s):
at z= mode(z)
0.18
0.15
0.14
0.13
0.13
s*RRA at z=mode(z):
0.52
0.38
0.33
0.30
0.29
The modal value of 𝑧(𝑡) (line 3 table 1) is calculated using equation (30)
𝜋2
𝜋2
6
6
2
𝜋2
[𝑛+𝑎+( )𝜎2 𝑚2 ]−√[𝑛+𝑎+( )𝜎2 𝑚2 ] −4( )𝜎2 𝑚2 𝑏
(30) (B19)
𝑧̂ =
𝜋2
2( )𝜎2 𝑚2
6
Where
(15) (A19)
𝑚=
(𝜇−𝑟+𝜀𝜎)
𝜋2
( )𝜎2 (1−𝛾)
6
6
0.6
0.492
0.78
0.018
0.033
0.034
0.037
2.18
2.16
0.12
0.26
𝑛=
(20) (A22)
(𝜇−𝑟+𝜀𝜎)2
𝜋2
( )𝜎2 (1−𝛾)
6
1
+ 𝑟 − (1−𝛾) {𝜌 − 𝛾𝑟 +
(𝜇−𝑟+𝜀𝜎)2
𝜋2
4( 6 )𝜎2 (1−𝛾)
}
The mean of the annual growth rate in consumption at 𝑧(𝑡) = 𝑧̂ (line 5 table 1) =
𝑛 + 𝑏 − (𝑛 + 𝑎)𝑧̂ [equation (30) (C1)]
𝐸[
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
at 𝑧̂ =
The standard deviation of the annual growth rate in consumption at 𝑧(𝑡) = 𝑧̂
𝑣𝑎𝑟[
(line 7 table 1) =
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
𝜋2
at 𝑧̂ = ( 6 ) 𝜎 2 𝑚2 [1 − 𝑧̂ ]2 [equation (32) (C2)]
The unconditional mean of the RRA coefficient (line 8 table 1) is calculated using equation (35)
(35) (D4)
ℎ𝑏
̅̅̅̅̅̅
𝑅𝑅𝐴 = (1 − 𝛾) {1 + [(𝑛+𝑎−𝑏)(𝑟+𝑎−𝑏)]}
Where
(16) (A20)
𝑟+𝑎−𝑏
ℎ = (𝑎+𝑟)(1−𝛾) {𝜌 − 𝛾𝑟 −
(𝜇 −𝑟+𝜀𝜎)2 (−1)
𝜋2 2
)𝜎 (1−𝛾)
6
4(
}>0
The RRA coefficient at 𝑧(𝑡) = 𝑧̂ (line 9 table 1) is calculated using equation (36)
(36) (D5)
𝑧̂
ℎ
𝑅𝑅𝐴(𝑧 = 𝑧̂ ) = (1 − 𝛾) {1 + [1−𝑧̂ ] [(𝑟+𝑎−𝑏)]}
The Elasticity of Substitution (=s) at 𝑧(𝑡) = 𝑧̂ (line 10 table 1) is calculated using equation (37)
(37) (D6)
1−𝑧̂
𝑠(𝑧 = 𝑧̂ ) = 1−𝛾
when
𝑧(𝑡) = 𝑧̂ , 𝜇 − 𝑟, 𝑎𝑛𝑑 𝜎 2
held
constant
The product of s*RRA at 𝑧(𝑡) = 𝑧̂ (line 11 table 1) is calculated using the product of equations
(36) and (37) or by using equation (39)
(39)(D8)
ℎ
𝑠 ∗ 𝑅𝑅𝐴 at [𝑧(𝑡) = 𝑧̂ ] = 𝑠(𝑧 = 𝑧̂ ) ∗ 𝑅𝑅𝐴(𝑧 = 𝑧̂ ) = 1 − {1 − [(𝑟+𝑎−𝑏)]} 𝑧̂
Table 1 results interpretations:
First, the mean RRA decreases as one moves to the right of the table (3.21 down to 2.18), and
approaches the value (1 − 𝛾) = 1.836 . The equity premium puzzle may be explained since the
model, where the stock price movement is represented using a right skewed non-Gaussian model
(the Gumbel distribution), generates the mean and variance of the consumption growth rate with
the mean RRA coefficient approaching the value (1 − 𝛾) = 1.836 .
Habit persistence reduces the product s*RRA extremely below one smoothing consumption
growth beyond the smoothing achieved by the life cycle permanent income hypothesis with time
separable utility [see equations (40) and (41)]. Equation (56)
𝑣𝑎𝑟[
(56)
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
6(𝜇 −𝑟+𝜀𝜎)2
= 𝜋2 𝜎2 (1−𝛾)2
represent the effect of the consumption smoothing on the equity premium.
Second, the modal value of the state variable (= 𝑧̂ ) is about 0.75 for all (a,b) pairs. The model
estimates that the subsistence level of consumption generated by habit persistence (= 𝑥(𝑡)) is
about 75% of the consumption level.
Third, the intertemporal elasticity of substitution in consumption (= s) is extremely below one
(around 0.15).
Forth, the product of the elasticity of substitution and the RRA coefficient (=s*RRA=wedge) is
about 0.26 for the (a,b) pairs that explain the equity premium puzzle with low RRA coefficient.
Hence, the habit persistence creates a wedge between the RRA coefficient and the invers of the
elasticity substitution (=1/s) [as it appears in equation (32.5)] and thus may explain the equity
premium puzzle.
In summary, Table 1 shows that habit persistence can generate the sample mean and variance of
the consumption growth rate with low risk aversion, hence giving a plausible explanation to the
equity premium puzzle.
VI) The effect of time separability in utility preferences on the Equity Premium puzzle
To show that time separability in utility preferences causethe equity premium puzzle observed in
Hehra and Prescott (1985) paper, I will follow the methodology used by Constantinides (1990).
I define 𝑚𝑡+1 as the marginal rate of substitution at t+1 [different than the constant m defined in
equation (15) and equation (A19)].
𝑅𝐹𝑡 = 1 + 𝑟𝑓𝑡 where 𝑟𝑓𝑡 is the riskless rate of interest between periods t and t+1.
𝑅𝑡 = 1 + 𝑟𝑡 where 𝑟𝑡 is the rate of return between periods t and t+1.
The Euler equation is
𝐸[ 𝑚𝑡+1 𝑅𝐹𝑡 ∣ 𝐼𝑡 ] = 1
(64)
Where 𝐼𝑡 is the public information in period t.
Since at time t 𝑅𝐹𝑡 is known (i.e. in the information set 𝐼𝑡 ) then
𝐸[ 𝑚𝑡+1 𝑅𝐹𝑡 ∣ 𝐼𝑡 ] = 𝑅𝐹𝑡 𝐸[ 𝑚𝑡+1 ∣ 𝐼𝑡 ] = 1
Hence
𝐸[ 𝑚𝑡+1 ∣ 𝐼𝑡 ] = (𝑅𝐹𝑡 )−1
(65)
Applying the expectation operator on both sides of equation (65) yields
𝐸(𝑚) = 𝐸[(𝑅𝐹 )−1 ]
(66)
Utilizing the Jensen’s inequality on equation (66) I get
𝐸(𝑚) = 𝐸[(𝑅𝐹 )−1 ] ≥ [𝐸(𝑅𝐹 )]−1
(67)
Apply the Euler equation using 𝑅𝑡+1 (i.e. one plus the rate of return) thus
𝐸[ 𝑚𝑡+1 𝑅𝑡+1 ∣ 𝐼𝑡 ] = 1
(68)
Applying the expectation operator on both sides of equation (68) yields
𝐸[𝑚𝑅] = 1
(69)
Following the methodology of Hansen and Jagannathan (1988) I get
1 = 𝐸[𝑚𝑅] = 𝐸(𝑚)𝐸(𝑅) + 𝑐𝑜𝑣(𝑚, 𝑅)
(70)
from covariance definition:
𝑐𝑜𝑣(𝑚, 𝑅) = 𝜌 𝑠𝑡𝑑(𝑚) 𝑠𝑡𝑑(𝑅) , −1 < 𝜌 < 1
using 𝜌 = −1 yields the minimum of the covariance expression, so when inserting this into
equation (70) I get
(70)
1 = 𝐸[𝑚𝑅] = 𝐸(𝑚)𝐸(𝑅) + 𝑐𝑜𝑣(𝑚, 𝑅) ≥ 𝐸(𝑚)𝐸(𝑅) − 𝑠𝑡𝑑(𝑚) 𝑠𝑡𝑑(𝑅)
Inserting equation (67) into (70) yields
(70)
1 = 𝐸[𝑚𝑅] = 𝐸(𝑚)𝐸(𝑅) + 𝑐𝑜𝑣(𝑚, 𝑅) ≥ 𝐸(𝑚)𝐸(𝑅) − 𝑠𝑡𝑑(𝑚) 𝑠𝑡𝑑(𝑅)
𝐸(𝑅)
𝐹)
≥ 𝐸(𝑅
𝐸(𝑅)
𝐹)
1 ≥ 𝐸(𝑅
Thus
− 𝑠𝑡𝑑(𝑚) 𝑠𝑡𝑑(𝑅)
− 𝑠𝑡𝑑(𝑚) 𝑠𝑡𝑑(𝑅)
Solving for 𝑠𝑡𝑑(𝑚)
[
𝑠𝑡𝑑(𝑚) ≥
(71)
𝐸(𝑅)
−1]
𝐸(𝑅𝐹 )
𝑠𝑡𝑑(𝑅)
Mehra and Prescott (1985) use the following discrete utility form
1
𝑡
𝛾
𝑢(𝑐𝑡 ) = ∑∞
𝑡=0 𝛽 (𝛾) (𝑐𝑡 )
(72)
The marginal rate of substitution is calculated
(73)
𝑚𝑡+1 =
𝜕𝑢
)
𝜕𝑐𝑡+1
𝜕𝑢
( )
𝜕𝑐𝑡
(
1
𝛾
1
𝛽 𝑡 ( )𝛾 (𝑐𝑡 )𝛾−1
𝛾
𝛽 𝑡+1 ( )𝛾 (𝑐𝑡+1 )𝛾−1
=
𝑐𝑡+1 𝛾−1
= 𝛽(
𝑐𝑡
)
Assuming the consumption growth rate is bounded such that
𝑔1 ≤
(74)
1
Or
𝑔2
𝑐𝑡+1
≤
𝑐𝑡
≤ 𝑔2
1
1
≤𝑔
𝑐𝑡+1
𝑐𝑡
1
Since both 𝛽 and (1 − 𝛾) are positive, applying on both sides of the inequality yields
1 1−𝛾
𝛽 (𝑔 )
2
1−𝛾
1
≤ 𝛽 ( 𝑐𝑡+1 )
𝑐𝑡
1 1−𝛾
≤ 𝛽 (𝑔 )
1
Inserting equation (73) above yields
(75)
𝛽(𝑔2 )𝛾−1 ≤ 𝑚𝑡+1 ≤ 𝛽(𝑔1 )𝛾−1
Since equation (75) defines the upper and lower bounds of 𝑚𝑡+1 the standard deviation is
(76)
𝑠𝑡𝑣(𝑚) ≤
𝛽(𝑔1 )𝛾−1 −𝛽(𝑔2 )𝛾−1
2
Using both equation (76) and (71) I get
[
(77)
𝐸(𝑅)
−1]
𝐸(𝑅𝐹 )
𝑠𝑡𝑑(𝑅)
≤ 𝑠𝑡𝑣(𝑚) ≤
𝛽(𝑔1 )𝛾−1 −𝛽(𝑔2 )𝛾−1
2
Hence
2[
(78)
𝐸(𝑅)
−1]
𝐸(𝑅𝐹 )
𝑠𝑡𝑑(𝑅)
≤ 𝛽(𝑔1 )𝛾−1 − 𝛽(𝑔2 )𝛾−1
Mehra and Prescott (1985) used their 1889-1978 data sample to make the following estimates:
𝐸(𝑅𝐹 ) = 1.01 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟,
𝑔2 = 1.054 ,
𝐸(𝑅) = 1.07 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟,
𝑠𝑡𝑑(𝑅) = 0.165 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟, 𝑔1 = 0.982,
Inserting the above estimates into equation (78) I get
(79)
(0.982)𝛾−1 − (1.054)𝛾−1 ≥
0.72
𝛽
Solving for (1 − 𝛾) and inserting specific 𝛽 values I can calculate the level of risk aversion in
each case.
From equation (58) I get 𝑅𝑅𝐴 > (1 − 𝛾) , hence calculating (1 − 𝛾) levels represent the lower
bound of RRA coefficient (= relative risk aversion coefficient).
From equation (79) I find: for 𝛽 = 0.8 the (1 − 𝛾) = 16 , for 𝛽 = 0.9 the (1 − 𝛾) = 14 , for
𝛽 = 1 the (1 − 𝛾) = 12 , for 𝛽 = 1.1 the (1 − 𝛾) = 11 , for 𝛽 = 1.2 the (1 − 𝛾) = 10 .
One can see that the levels of (1 − 𝛾) for a time separable (=discrete) model (10 ≤ 1 − 𝛾 ≤ 16 )
are much higher than the levels in a nonsaparable model as seen in table 1 (1 − 𝛾 = 1.836 ).
This demonstrates that time separability in utility preferences creates the equity premium puzzle
observed by Mehra and Prescott (1985).
The issue is that the lower bound on the consumption growth rate (=𝑔1 ) affects the upper bound
on the marginal rate of substitution [= 𝑚𝑡+1 ,see equation (75)]. Reitz (1988) also shows the
effect of lower bound on consumption growth on the equity premium puzzle.
VII) The effect of time separability in utility preferences and Habit persistence on the
Equity Premium puzzle
Taking into consideration both time separability in utility preferences and Habit persistence, I
will redo the calculations performed in part (VI).
Equation (71) is derived in the same manner as in part VI.
[
𝑠𝑡𝑑(𝑚) ≥
(71)
𝐸(𝑅)
−1]
𝐸(𝑅𝐹 )
𝑠𝑡𝑑(𝑅)
Will now use a discrete utility function that includes Habit persistence
1
𝑡
𝛾
𝑢(𝑐𝑡 , 𝑥𝑡 ) = ∑∞
𝑡=0 𝛽 (𝛾) (𝑐𝑡 − 𝑥𝑡 )
(80)
Where 𝑥𝑡 represent the subsistence level of consumption generated by Habit Persistence.
Inserting equation (22) into (80)
𝑧(𝑡) =
(22)
𝑥(𝑡)
i.e. 𝑥(𝑡) = 𝑧(𝑡)𝑐(𝑡) I derive
𝑐(𝑡)
1
𝑡
𝛾
𝑢(𝑐𝑡 , 𝑥𝑡 ) = ∑∞
𝑡=0 𝛽 (𝛾) {[1 − 𝑧𝑡 ]𝑐𝑡 }
(81)
Deriving the marginal rate of substitution for this utility
𝑚𝑡+1 =
(82)
𝜕𝑢
)
𝜕𝑐𝑡+1
𝜕𝑢
( )
𝜕𝑐𝑡
(
=
1
𝛾
𝛽 𝑡+1 ( )𝛾 (𝑐𝑡+1 )𝛾−1 (1−𝑧𝑡+1 )𝛾
1
𝛽 𝑡 ( )𝛾 (𝑐𝑡 )𝛾−1 (1−𝑧𝑡 )𝛾
𝛾
= 𝛽(
𝑐𝑡+1 𝛾−1 1−𝑧𝑡+1 𝛾
𝑐𝑡
)
(
1−𝑧𝑡
)
Assuming the consumption growth rate is bounded such that
𝑔1 ≤
(74)
1
Or
𝑔2
𝑐𝑡+1
𝑐𝑡
≤
≤ 𝑔2
1
1
≤𝑔
𝑐𝑡+1
𝑐𝑡
1
Since both 𝛽 ,(1 − 𝛾) and (1 − 𝑧𝑡 ) are positive, applying on both sides of the inequality yields
1−𝛾
1 1−𝛾 1−𝑧𝑡+1 𝛾
𝛽 (𝑔 )
2
(
1−𝑧𝑡
1
) ≤ 𝛽 ( 𝑐𝑡+1 )
𝑐𝑡
1 1−𝛾 1−𝑧𝑡+1 𝛾
𝛽 (𝑔 )
1
(
1−𝑧𝑡
1−𝑧𝑡+1 𝛾
(
1−𝑧𝑡
) ≤
)
Inserting equation (82) above yields
(83)
1−𝑧𝑡+1 𝛾
𝛽(𝑔2 )𝛾−1 (
1−𝑧𝑡
) ≤ 𝑚𝑡+1 ≤ 𝛽(𝑔1 )𝛾−1 (
1−𝑧𝑡+1 𝛾
1−𝑧𝑡
)
Since equation (83) defines the upper and lower bounds of 𝑚𝑡+1 the standard deviation is
𝑠𝑡𝑣(𝑚) ≤ [
(84)
𝛽(𝑔1 )𝛾−1 −𝛽(𝑔2 )𝛾−1
2
1−𝑧𝑡+1 𝛾
](
1−𝑧𝑡
)
Using both equation (84) and (71) I get
[
(85)
𝐸(𝑅)
−1]
𝐸(𝑅𝐹 )
𝑠𝑡𝑑(𝑅)
≤ 𝑠𝑡𝑣(𝑚) ≤ [
𝛽(𝑔1 )𝛾−1 −𝛽(𝑔2 )𝛾−1
2
](
1−𝑧𝑡+1 𝛾
1−𝑧𝑡
)
Hence
2[
(86)
𝐸(𝑅)
−1]
𝐸(𝑅𝐹 )
𝑠𝑡𝑑(𝑅)
1−𝑧𝑡+1 𝛾
≤ [𝛽(𝑔1 )𝛾−1 − 𝛽(𝑔2 )𝛾−1 ] (
1−𝑧𝑡
)
Mehra and Prescott (1985) used their 1889-1978 data sample to make the following estimates:
𝐸(𝑅𝐹 ) = 1.01 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟,
𝑔2 = 1.054 ,
𝐸(𝑅) = 1.07 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟,
𝑠𝑡𝑑(𝑅) = 0.165 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟, 𝑔1 = 0.982,
Inserting the above estimates into equation (86) I get
1−𝑧𝑡+1 𝛾
(87)
(
Or
(
1−𝑧𝑡
) [(0.982)𝛾−1 − (1.054)𝛾−1 ] ≥
1−𝑧𝑡+1 𝛾
1−𝑧𝑡
1
1
) [(0.982)1−𝛾 − (1.054)1−𝛾 ] ≥
0.72
𝛽
0.72
𝛽
Using the same values of 𝛽 used in part VI, the right hand side of equation (87) will have the
same values as in part VI.
Assuming increase in consumption over time (i.e. 𝑐𝑡+1 > 𝑐𝑡 ), I can assume a decrease in
subsistence rate of consumption generated by Habit persistence (i.e. 𝑧𝑡+1 < 𝑧𝑡 ) in order to keep
the subsistence level of consumption generated by Habit persistence fixed (i.e. 𝑥𝑡+1 ≅ 𝑥𝑡 ).
1−𝑧𝑡+1
Thus,
> 1 . Hence, for the same values of 𝛽 I need lower levels of (1 − 𝛾) to meet
1−𝑧𝑡
equation (87).
This shows that adding Habit Persistence to a utility function will reduce levels of risk aversion
(i.e. RRA) since (1 − 𝛾) is the lower bound of the RRA.
VIII) Conclusion
My results suggest that habit persistence can generate the Mehra and Prescott’s (1985) sample
mean and variance of the consumption growth rate with low risk aversion, hence giving a
plausible explanation to the equity premium puzzle.
The habit persistence creates a wedge between the RRA coefficient and the invers of the
elasticity substitution (=1/s) [as it appears in equation (32.5)] and thus may explain the equity
premium puzzle.
Habit persistence reduces the product s*RRA extremely below one, thus smoothing consumption
growth (see equations 40 and 41) beyond the smoothing achieved by the life cycle permanent
income hypothesis with time separable utility. From equation (56), such smoothing of
𝑣𝑎𝑟[
𝑑𝑐(𝑡)
]
𝑐(𝑡)
consumption growth (i.e. reduction in
) will reduce the equity premium (by reducing
𝑑𝑡
𝜇 − 𝑟 + 𝜀𝜎 ) ,hence may explain the equity premium puzzle. Furthermore, I show that time
separability in utility preferences ( i.e. b=0) creates the equity premium puzzle (i.e. very high
equity premium) presented in the Mehra and Prescott (1985) paper. Remember, Mehra and
Prescott use a discrete-time economy in which the state variable is determined by a Markov
process with two realization. Thus, their model present high time separability that generates the
equity premium puzzle. My model represent a continuous time economy in which the forcing
process is a diffusion, thus present a nonseparable utility (b>0 in table 1) which reduces the
equity premium thus explain the equity premium puzzle.
The model in this paper incorporates habit persistence in utility which causes low risk aversion (
minimal unconditional mean of the RRA coefficient = 2.18, since low 1 − 𝛾 means low RRA)
which generates low variability in consumption growth rate [mean value around (0.036)2 and
see equations 40 and 41] and low variability in the marginal rate of substitution in consumption
(𝑚𝑡 see equations 73 and 82). Since the modal value of the subsistence rate of consumption
generated by Habit Persistence (= z) is around 0.75 in my model, a high value, the subsistence
level of consumption (= x) is about 75% of the normal consumption rate. Thus, a small decrease
in consumption causes a big decrease in consumption net of the subsistence level (= c - x) thus
causing a big decrease in the marginal rate of substitution (= 𝑚𝑡 see equation 73 and 82). Such
low risk aversion can explain the observed equity premium.
The results displayed in the paper show that a rational expectations model with time
nonseparability of preferences, with habit persistence and with stock price movement represented
by a right skewed non-Gaussian model (the Gumbel distribution) can explain the equity premium
puzzle. The model is able to find acceptable relative risk aversion (RRA) of the representative
agent to match the sample estimates.
Habit persistence differs from the known state and time-separable preferences. Habit persistence
should be embedded in new models of the business cycle, labor behavior, public finance and
dynamics.
Appendices:
Appendix A: Proof of Theorem 1
I use a technique used by Davis and Norman (1987). If 𝑐(𝑡) and 𝛼(𝑡) are not optimal at t, I
assume an optimal policy 𝑐 ∗ (𝑠) and 𝛼 ∗ (𝑠) for 𝑠 ≥ 𝑡.
Equation (6)
(6)
𝜋
𝑑𝑊𝑡 = {[(𝜇 − 𝑟 + 𝜀𝜎)𝛼𝑡 + 𝑟]𝑊𝑡 −𝑐𝑡 }𝑑𝑡 + ( 6) 𝛼𝑡 𝑊𝑡 𝜎𝑑𝑤𝑡
√
Becomes
(A1)
𝜋
√6
𝑑𝑊(𝑠) = {[(𝜇 − 𝑟 + 𝜀𝜎)𝛼∗ (𝑠) + 𝑟]𝑊(𝑠) − 𝑐∗ (𝑠)}𝑑𝑠 + ( ) 𝛼∗ (𝑠)𝑊(𝑠)𝜎𝑑𝑤(𝑠)
Equation (11) above
(11)
𝑏𝑐(𝑡) − 𝑎𝑥(𝑡) = 0
Becomes
(A2)
𝑏𝑐 ∗ (𝑠) − 𝑎𝑥(𝑠) =
𝑑𝑥(𝑠)
𝑑𝑠
or
𝑑𝑥(𝑠) = [𝑏𝑐 ∗ (𝑠) − 𝑎𝑥(𝑠)]𝑑𝑠
Equation (8)
𝑥(𝑡)
𝑐(𝑡) − 𝑥(𝑡) = (𝑟 + 𝑎 − 𝑏) [𝑊(𝑡) − 𝑟+𝑎−𝑏]
(8)
Becomes
𝑥(𝑠)
𝑐 ∗ (𝑠) − 𝑥(𝑠) = ℎ [𝑊(𝑠) − 𝑟+𝑎−𝑏]
(A3)
Next I will derive
𝑑 [𝑊(𝑠) −
𝑥(𝑠)
𝑑𝑥(𝑠)
] = 𝑑𝑊(𝑠) −
𝑟+𝑎−𝑏
𝑟+𝑎−𝑏
I will insert equation (A1) and (A2) into the above expression to derive (A4)
(A4)
𝑥(𝑠)
𝜋
𝑑 [𝑊(𝑠) − 𝑟+𝑎−𝑏] = {[(𝜇 − 𝑟 + 𝜀𝜎)𝛼∗ (𝑠) + 𝑟]𝑊(𝑠) − 𝑐∗ (𝑠)}𝑑𝑠 + ( 6) 𝛼∗ (𝑠)𝑊(𝑠)𝜎𝑑𝑤(𝑠) −
√
−
[𝑏𝑐∗ (𝑠) − 𝑎𝑥(𝑠)]𝑑𝑠
𝑟+𝑎−𝑏
= {[(𝜇 − 𝑟 + 𝜀𝜎)𝛼∗ (𝑠) + 𝑟]𝑊(𝑠) − 𝑐∗ (𝑠) −
=
[𝑏𝑐∗ (𝑠) − 𝑎𝑥(𝑠)]
𝑟+𝑎−𝑏
𝜋
} 𝑑𝑠 + ( ) 𝜎𝛼∗ (𝑠)𝑊(𝑠)𝑑𝑤(𝑠) =
√6
I will define m such that:
𝛼 ∗ (𝑠)𝑊(𝑠) = 𝑚 [𝑊(𝑠) −
𝑥( 𝑠 )
𝑟+𝑎−𝑏
]
Hence
𝛼
(A5)
∗ (𝑠)
= 𝑚 [1 −
𝑥(𝑠)
𝑊(𝑠)
𝑟+𝑎−𝑏
]
Inserting 𝛼 ∗ (𝑠) into (A4) yield
(A4) 𝑑 [𝑊(𝑠) −
𝑥(𝑠)
]
𝑟+𝑎−𝑏
= {[(𝜇 − 𝑟 + 𝜀𝜎)𝑚 [1 −
𝑥(𝑠)
𝑊(𝑠)
] + 𝑟] 𝑊(𝑠) −
𝑟+𝑎−𝑏
𝑎[𝑐∗ (𝑠)−𝑥(𝑠)]
𝜋
𝑥(𝑠)
+ ( ) 𝜎𝑚 [𝑊(𝑠) −
] 𝑑𝑤(𝑠) =
𝑟+𝑎−𝑏
√6
𝑟+𝑎−𝑏
−
[𝑐∗ (𝑠)𝑟]
𝑟+𝑎−𝑏
} 𝑑𝑠 +
Inserting equation (A3) into equation (A4) yields
(A4) = {[(𝜇 − 𝑟 + 𝜀𝜎)𝑚 [1 −
𝑥(𝑠)
𝑊(𝑠)
𝑥(𝑠)
] + 𝑟] 𝑊(𝑠) −
𝑟+𝑎−𝑏
𝑎ℎ[𝑊(𝑠)−𝑟+𝑎−𝑏]
𝑟+𝑎−𝑏
𝑥(𝑠)
−
[𝑥(𝑠)+ℎ[𝑊(𝑠)−𝑟+𝑎−𝑏]]
𝑟+𝑎−𝑏
} 𝑑𝑠 +
𝜋
𝑥(𝑠)
+ ( ) 𝜎𝑚 [𝑊(𝑠) −
] 𝑑𝑤(𝑠) =
𝑟+𝑎−𝑏
√6
Arriving at
(A4)
𝑥(𝑠)
]
𝑟+𝑎−𝑏
𝑥(𝑠)
[𝑊(𝑠)−
]
𝑟+𝑎−𝑏
𝑑[𝑊(𝑠)−
= {(𝜇 − 𝑟 + 𝜀𝜎)𝑚 + 𝑟 −
ℎ(𝑎+𝑟)
𝜋
𝜋
𝑟+𝑎−𝑏
√
√
} 𝑑𝑠 + ( 6) 𝜎𝑚𝑑𝑤(𝑠) = 𝑛𝑑𝑠 + ( 6) 𝜎𝑚𝑑𝑤(𝑠)
Where i define
𝑛 = (𝜇 − 𝑟 + 𝜀𝜎)𝑚 + 𝑟 −
ℎ(𝑎 + 𝑟)
𝑟+𝑎−𝑏
Since
𝑥(𝑠)
𝑑 [𝑊(𝑠) −
]
𝑥(𝑠)
𝑟+𝑎−𝑏
𝑑𝐿𝑛 [𝑊(𝑠) −
]=
𝑥(𝑠)
𝑟+𝑎−𝑏
[𝑊(𝑠) −
]
𝑟+𝑎−𝑏
Equation (A4) becomes
(A6)
𝑥(𝑠)
𝜋
𝑑𝐿𝑛 [𝑊(𝑠) − 𝑟+𝑎−𝑏] = 𝑛𝑑𝑠 + ( 6) 𝜎𝑚𝑑𝑤(𝑠)
√
Using exponent on sides yield
𝑒
{𝑑𝐿𝑛[𝑊(𝑠)−
𝑥(𝑠)
]}
𝑟+𝑎−𝑏
= 𝑒 {𝑛𝑑𝑠} 𝑒
𝜋
{( )𝜎𝑚𝑑𝑤(𝑠)}
√6
Hence
𝑥(𝑠)
𝑑 [𝑊(𝑠) − 𝑟+𝑎−𝑏] = 𝑒 {𝑛𝑑𝑠} 𝑒
{(
𝜋
)𝜎𝑚𝑑𝑤(𝑠)}
√6
Solving the above differential equation yields the following solution
(A7)
𝑥(𝑠)
𝑥(𝑡)
𝑊(𝑠) − 𝑟+𝑎−𝑏 = [𝑊(𝑡) − 𝑟+𝑎−𝑏] 𝑒 {𝑛(𝑠−𝑡)} 𝑒
{(
𝜋
)𝜎𝑚[𝑤(𝑠)−𝑤(𝑡)]}
√6
,𝑠 ≥ 𝑡
I define the state valuation function, V (.,.), as a function of the two variables, 𝑊(𝑡) and 𝑥(𝑡).
Thus, the maximization problem is:
(A8)
∞
𝑉[𝑊(𝑡), 𝑥(𝑡)] = max 𝐸𝑡 ∫ 𝑒
∞
−𝜌(𝑠−𝑡)
𝑢[𝑐(𝑠), 𝑥(𝑠)]𝑑𝑠 = max 𝐸𝑡 ∫ 𝑒 −𝜌(𝑠−𝑡)
𝑡
𝑡
1
[𝑐(𝑡) − 𝑥(𝑡)]𝛾 𝑑𝑠
𝛾
where  represents the constant utility discount rate, and Et represents the operator for
expectations conditional on the information set at time t.
By inserting the optimal values 𝑐 ∗ (𝑠) and 𝛼 ∗ (𝑠) ,for 𝑠 ≥ 𝑡 , the maximum achieved, thus
(A9)
∞
𝑉[𝑊(𝑠), 𝑥(𝑠)] = 𝐸𝑡 ∫ 𝑒 −𝜌(𝑠−𝑡)
𝑠
1 ∗
[𝑐 (𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠 =
𝛾
By inserting equation (A3) into equation (A9) I derive
∞
1
𝑥(𝑠) 𝛾
𝑉[𝑊(𝑠), 𝑥(𝑠)] = 𝐸𝑡 ∫ 𝑒 −𝜌(𝑠−𝑡) (ℎ)𝛾 [𝑊(𝑠) −
] 𝑑𝑠 =
𝛾
𝑟+𝑎−𝑏
𝑠
∞
1
𝑥(𝑠) 𝛾
= ∫ 𝑒 −𝜌(𝑠−𝑡) 𝐸𝑡 { (ℎ)𝛾 [𝑊(𝑠) −
] } 𝑑𝑠
𝛾
𝑟+𝑎−𝑏
𝑠
Concentrating on developing the expression within the integral and inserting equation (A7)
yields
𝛾
𝑒
−𝜌(𝑠−𝑡)
1
𝑥( 𝑠 )
𝐸𝑡 { (ℎ)𝛾 [𝑊(𝑠) −
] }=
𝛾
𝑟+𝑎−𝑏
𝛾
𝜋
1
𝑥(𝑡)
{𝑛(𝑠−𝑡)} {(√6)𝜎𝑚[𝑤(𝑠)−𝑤(𝑡)]}
−𝜌(𝑠−𝑡)
𝛾
=𝑒
𝐸𝑡 { (ℎ) [[𝑊(𝑡) −
]𝑒
𝑒
] }=
𝛾
𝑟+𝑎−𝑏
𝛾
𝛾
𝛾
2
𝜋
(ℎ)𝛾
𝑥(𝑡)
{( )𝜎𝑚[𝑤(𝑠)−𝑤(𝑡)]}
−𝜌(𝑠−𝑡) [𝑛(𝑠−𝑡)] 𝛾
√
(
)
=
[𝑊 𝑡 −
] 𝑒
[𝑒
] 𝐸𝑡 { [𝑒 6
] }=
𝛾
𝑟+𝑎−𝑏
𝛾 𝜋
(ℎ)𝛾
𝑥(𝑡)
{ ( )𝜎𝑚[𝑤(𝑠)−𝑤(𝑡)]}
𝛾
=
[𝑊(𝑡) −
] 𝑒 −𝜌(𝑠−𝑡) [𝑒[𝑛(𝑠−𝑡)] ] 𝐸𝑡 { [𝑒 2 √6
] }=
𝛾
𝑟+𝑎−𝑏
𝛾 𝜋
𝛾
2
(ℎ)𝛾
𝑥(𝑡)
𝛾 {𝐸𝑡 [2(√6)𝜎𝑚[𝑤(𝑠)−𝑤(𝑡)]] }
=
[𝑊(𝑡) −
] 𝑒 −𝜌(𝑠−𝑡) [𝑒[𝑛(𝑠−𝑡)] ] 𝑒
=
𝛾
𝑟+𝑎−𝑏
Since with a Wiener process 𝐸𝑡 [𝑤(𝑠)] = 𝐸𝑡 [𝑤(𝑡)] = 0 thus
𝛾
𝛾 𝜋
{𝑣𝑎𝑟 [ ( )𝜎𝑚[𝑤(𝑠)−𝑤(𝑡)]]}
(ℎ)𝛾
𝑥(𝑡)
2 √6
[𝑛(𝑠−𝑡)] 𝛾
−𝜌(𝑠−𝑡)
=
[𝑊(𝑡) −
] 𝑒
[𝑒
] 𝑒
=
𝛾
𝑟+𝑎−𝑏
Since with a Wiener process 𝑣𝑎𝑟[𝑤(𝑠)] = 𝑠 and 𝑣𝑎𝑟[𝑤(𝑡)] = 𝑡 thus
𝛾
𝛾 2 𝜋2
(ℎ)𝛾
𝑥(𝑡)
𝛾 {( )( )(𝑚2 )(𝜎2 )(𝑠−𝑡)}
=
[𝑊(𝑡) −
] 𝑒 −𝜌(𝑠−𝑡) [𝑒[𝑛(𝑠−𝑡)] ] 𝑒 4 6
=
𝛾
𝑟+𝑎−𝑏
𝛾
𝜋2 𝛾 2 𝑚 2 𝜎 2
(ℎ)𝛾
𝑥(𝑡)
)](𝑠−𝑡)}
{[−𝜌+𝛾𝑛+( )(
6
4
=
[𝑊(𝑡) −
] 𝑒
=
𝛾
𝑟+𝑎−𝑏
𝛾
𝜋2 𝛾 2 𝑚 2 𝜎 2
(ℎ)𝛾
𝑥(𝑡)
)](𝑡−𝑠)}
{[𝜌−𝛾𝑛−( )(
6
4
=
[𝑊(𝑡) −
] 𝑒
=
𝛾
𝑟+𝑎−𝑏
I define
𝜋2
𝛾 2 𝑚2 𝜎 2
𝐻 = 𝜌 − 𝛾𝑛 − ( ) (
)
6
4
Thus equation (A9) becomes
∞
𝛾
(ℎ)𝛾
𝑥(𝑡)
𝑉[𝑊(𝑠), 𝑥(𝑠)] = ∫
[𝑊(𝑡) −
] 𝑒 {𝐻(𝑡−𝑠)} 𝑑𝑠 =
𝛾
𝑟+𝑎−𝑏
𝑠
𝛾 ∞
(ℎ)𝛾
𝑥(𝑡)
=
[𝑊(𝑡) −
] ∫ 𝑒 {𝐻(𝑡−𝑠)} 𝑑𝑠 =
𝛾
𝑟+𝑎−𝑏
𝑠
Solving the integral yields equation (A10)
(A10)
Where
𝑉[𝑊(𝑠), 𝑥(𝑠)] =
(ℎ)𝛾
𝛾
𝑥(𝑡)
𝛾 −1
[𝑊(𝑡) − 𝑟+𝑎−𝑏] ( 𝐻 )
𝜋2
𝛾 2 𝑚2 𝜎 2
𝐻 = 𝜌 − 𝛾𝑛 − ( ) (
)
6
4
ℎ(𝑎+𝑟)
𝑛 = (𝜇 − 𝑟 + 𝜀𝜎)𝑚 + 𝑟 − 𝑟+𝑎−𝑏
(A11)
I now define M(t) as follows
𝑡
∞
1
1
𝑀(𝑡) = ∫ 𝑒 −𝜌(𝑠−0) [𝑐(𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠 + ∫ 𝑒 −𝜌(𝑠−0) [𝑐(𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠 =
𝛾
𝛾
0
𝑡
𝑡
∞
= ∫𝑒
−𝜌(𝑠−0)
0
1
1
[𝑐(𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠 + 𝑒 −𝜌𝑡 ∫ 𝑒 −𝜌(𝑠−𝑡) [𝑐(𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠 =
𝛾
𝛾
𝑡
Inserting equation (A9) into the above equation to derive equation (A12)
(A12)
𝑡
1
𝑀(𝑡) = ∫0 𝑒 −𝜌(𝑠−0) 𝛾 [𝑐(𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠 + 𝑒 −𝜌𝑡 𝑉[𝑊(𝑡), 𝑥(𝑡)]
I now define the derivative
(A13) 𝑁(𝑡) =
𝑑𝑀(𝑡)
𝑑𝑡
1
= 𝑒 −𝜌𝑡 (𝛾) [𝑐(𝑡) − 𝑥(𝑡)]𝛾 + (−𝜌)𝑒 −𝜌𝑡 𝑉[𝑊(𝑡), 𝑥(𝑡)] + 𝑒 −𝜌𝑡 [
I use Ito’s Lemma to calculate
𝑑𝑉[𝑊(𝑡),𝑥(𝑡)]
𝑑𝑉[𝑊(𝑡),𝑥(𝑡)]
𝑑𝑡
The closed form problem is defined below:
(A10)
∞
1
max 𝐸𝑡 ∫𝑡 𝑒 −𝜌(𝑠−𝑡) 𝛾
𝑉[𝑊(𝑡), 𝑥(𝑡)] =
(ℎ)𝛾
𝛾
𝑥(𝑡)
𝛾 −1
[𝑊(𝑡) − 𝑟+𝑎−𝑏] ( 𝐻 ) =
𝛾
[𝑐(𝑡) − 𝑥 (𝑡)] 𝑑𝑠
Where
ℎ(𝑎+𝑟)
𝜋2
𝛾 2 𝑚2 𝜎 2
(A14)
𝐻 = 𝜌 − 𝛾 [(𝜇 − 𝑟 + 𝜀𝜎)𝑚 + 𝑟 − 𝑟+𝑎−𝑏 ] − ( 6 ) (
(6)
𝑑𝑊(𝑡) = {[𝛼𝑡 (𝜇 − 𝑟 + 𝜀𝜎) + 𝑟]𝑊(𝑡)−𝑐𝑡 }𝑑𝑡 + ( 6) 𝛼𝑡 𝑊(𝑡)𝜎𝑑𝑤𝑡
(A2)
𝑑𝑥(𝑡) = [𝑏𝑐 ∗ (𝑡) − 𝑎𝑥(𝑡)]𝑑𝑡 + 0𝑑𝑤𝑡
𝜋
√
4
)
𝑑𝑡
]
Applying Ito’s Lemma formulation:
dV (Wt , xt ) 
  2V
V
V
1   2V 
1   2V 
2
2








dWt 
dxt  
dW

dx

t
t
Wt
xt
2  Wt 2 
2  xt 2 
 Wt xt

dWt dxt 

By dividing both side of the equation by dt I get:
( A15)
dV  V

dt  Wt
 dWt

 dt
  V
  
  xt
2
2
2
2
2
 1   V  dWt   1   V  dxt     V





2
2

 2  Wt  dt  2  xt  dt   Wt xt
 dxt

 dt
 dWt dxt 

dt

Solving the expressions using equations (6), (A2) and (A10):
∞
𝑑𝑉
1
𝛾
= 𝜌 max 𝐸𝑡 ∫ 𝑒−𝜌(𝑠−𝑡) [𝑐(𝑡) − 𝑥(𝑡)] 𝑑𝑠 = 𝜌𝑉
𝑑𝑡
𝛾
𝑡
𝑑𝑊(𝑡)𝑑𝑥(𝑡) = 0 [Since dt   dtdwt  0 and dwt dwt  dt ]
2
𝜋2
(𝑑𝑊𝑡 )2 = ( ) 𝛼 2 𝜎 2 (𝑊𝑡 )2 𝑑𝑡
6
(𝑑𝑥𝑡 )2 = 0
Thus equation (A15) is
(A15)
𝑑𝑉
𝑑𝑡
𝜋2
1
∗
= 𝑉𝑊 {[𝛼𝑡 (𝜇 − 𝑟 + 𝜀𝜎) + 𝑟]𝑊(𝑡)−𝑐𝑡 } + 𝑉𝑥 {𝑏𝑐 (𝑡) − 𝑎𝑥(𝑡)} + 2 𝑉𝑊𝑊 {( 6 ) 𝛼2 𝜎2 (𝑊𝑡 )2 }
I will, next, perform first order conditions on 𝑁(𝑡) [equation (A13)]:
1
(A13) 𝑁(𝑡) = 𝑒 −𝜌𝑡 (𝛾) [𝑐(𝑡) − 𝑥(𝑡)]𝛾 + (−𝜌)𝑒 −𝜌𝑡 𝑉[𝑊(𝑡), 𝑥(𝑡)] + 𝑒 −𝜌𝑡 [
𝑑𝑉[𝑊(𝑡),𝑥(𝑡)]
𝑑𝑡
]
Where
(A10)
𝑉[𝑊(𝑡), 𝑥(𝑡)] =
(ℎ)𝛾
𝛾
𝑥(𝑡)
𝛾 −1
[𝑊(𝑡) − 𝑟+𝑎−𝑏] ( 𝐻 )
The derivatives of equation (A10)
𝑉𝑊 =
(ℎ)𝛾
𝛾
𝑥(𝑡)
𝛾 [𝑊(𝑡) − 𝑟+𝑎−𝑏]
𝛾−1 −1
𝑥(𝑡)
( 𝐻 ) = (ℎ)𝛾 [𝑊(𝑡) − 𝑟+𝑎−𝑏]
𝛾−1 −1
(𝐻)
𝑉𝑊𝑊 =
(ℎ)𝛾 (𝛾
− 1) [𝑊(𝑡) −
𝑟+𝑎−𝑏
𝛾−1
(ℎ)𝛾
𝑥(𝑡)
𝑉𝑥 =
𝛾 [𝑊(𝑡) −
]
𝛾
𝑟+𝑎−𝑏
=
(ℎ)𝛾
[𝑊(𝑡) −
𝑥(𝑡)
𝑟+𝑎−𝑏
(
]
−1
( )
𝐻
−1
)( ) =
𝑟+𝑎−𝑏
𝐻
−1
𝛾−1
]
𝛾−2
𝑥(𝑡)
−1
(
𝑟+𝑎−𝑏
)(
−1
)
𝐻
FOC1: with respect to 𝑐(𝑡)
0=
𝑑𝑁(𝑡)
1
𝛾−1
= 𝑒−𝜌𝑡 ( ) 𝛾 [𝑐(𝑡) − 𝑥(𝑡)]
+ 0 + 𝑒−𝜌𝑡 {𝑉𝑊 (−1) + 𝑉𝑥 𝑏 + 0}
𝑑𝑐(𝑡)
𝛾
Thus
𝛾−1
0 = [𝑐(𝑡) − 𝑥(𝑡)]
(A16)
− 𝑉𝑊 + 𝑉𝑥 𝑏
Insert the derivatives into equation (A16):
0 = [𝑐(𝑡) − 𝑥(𝑡)]
𝛾−1
𝑥(𝑡)
− (ℎ)𝛾 [𝑊(𝑡) − 𝑟+𝑎−𝑏]
𝛾−1 −1
𝑥(𝑡) 𝛾−1
−1
−1
( ) + 𝑏(ℎ)𝛾 [𝑊(𝑡) −
(
)( )
]
𝐻
𝑟+𝑎−𝑏
𝑟+𝑎−𝑏
𝐻
Solving for 𝑐(𝑡)
𝛾
𝑥(𝑡)
(𝑟+𝑎)
𝑐 ∗ (𝑡) = 𝑥(𝑡) − (ℎ)𝛾−1 [𝑊(𝑡) − 𝑟+𝑎−𝑏] [𝐻(𝑟+𝑎−𝑏)]
(A17)
FOC2: with respect to 𝛼(𝑡)
0=
𝑑𝑁(𝑡)
1
𝜋2
= 0 + 0 + 𝑒−𝜌𝑡 {𝑉𝑊 (𝜇 − 𝑟 + 𝜀𝜎)𝑊(𝑡) + 𝑉𝑊𝑊 [( ) 2𝛼𝜎2 (𝑊𝑡 )2 ]}
𝑑𝛼(𝑡)
2
6
Thus
(A18)
𝜋2
6
0 = 𝑉𝑊 (𝜇 − 𝑟 + 𝜀𝜎)𝑊(𝑡) + 𝑉𝑊𝑊 [( ) 𝛼𝜎2 (𝑊𝑡 )2 ]
Insert the derivatives into equation (A18):
𝛾−1
0=
+(ℎ)
(ℎ)𝛾 [𝑊(𝑡) −
𝛾 (𝛾
𝑥(𝑡)
]
𝑟+𝑎−𝑏
− 1) [𝑊(𝑡) −
𝑥(𝑡)
𝑟+𝑎−𝑏
(
−1
) (𝜇 − 𝑟 + 𝜀𝜎)𝑊(𝑡) +
𝐻
𝛾−2
]
−1
𝜋2
( ) [( ) 𝛼𝜎 2 (𝑊𝑡 )2 ]
𝐻
6
Solve for 𝛼(𝑡)
(𝜇 − 𝑟 + 𝜀𝜎)
𝑥(𝑡)
𝛼 ∗ (𝑡) = [ 2
] [𝑊(𝑡) −
]
𝜋
𝑟+𝑎−𝑏
( 6 ) 𝜎 2 𝑊(𝑡)(1 − 𝛾)
Insert equation (A5) into the above equation
𝛼
(A5)
∗ (𝑡)
= 𝑚 [1 −
𝑥(𝑡)
𝑊(𝑡)
𝑟+𝑎−𝑏
]
𝑥(𝑡)
(𝜇 − 𝑟 + 𝜀𝜎)
𝑥(𝑡)
𝑊(𝑡)
𝑚 [1 −
]
]=[ 2
] [𝑊(𝑡) −
𝑟+𝑎−𝑏
𝑟+𝑎−𝑏
𝜋
2
( ) 𝜎 𝑊(𝑡)(1 − 𝛾)
6
Solve for m
𝑚=
(A19)
(𝜇−𝑟+𝜀𝜎)
𝜋2
( )𝜎2 (1−𝛾)
6
Where 0 ≤ 𝑚 ≤ 1
Insert equation (A19) into equation (A14)
(A14)
ℎ(𝑎+𝑟)
𝜋2
𝛾 2 𝑚2 𝜎 2
𝐻 = 𝜌 − 𝛾 [(𝜇 − 𝑟 + 𝜀𝜎)𝑚 + 𝑟 − 𝑟+𝑎−𝑏 ] − ( 6 ) (
4
)=
(𝜇 − 𝑟 + 𝜀𝜎)
ℎ(𝑎 + 𝑟)
= 𝜌 − 𝛾 {(𝜇 − 𝑟 + 𝜀𝜎) [ 2
]+𝑟−
}−
𝜋
𝑟+𝑎−𝑏
( 6 ) 𝜎 2 (1 − 𝛾)
2
−(
𝜋2
6
(𝜇 − 𝑟 + 𝜀𝜎)
𝛾2 [ 2
] 𝜎2
𝜋
2
( 6 ) 𝜎 (1 − 𝛾)
)
=
4
(
)
𝛾(𝜇 − 𝑟 + 𝜀𝜎)2 (4 − 3𝛾) 𝛾(𝑎 + 𝑟)ℎ
𝐻 = 𝜌 − 𝛾𝑟 −
+
=
𝜋2
𝑟+𝑎−𝑏
4 ( 6 ) 𝜎 2 (1 − 𝛾)2
I define
𝑟+𝑎−𝑏
ℎ = (𝑎+𝑟)(1−𝛾) {𝜌 − 𝛾𝑟 −
(A20)
(𝜇 −𝑟+𝜀𝜎)2 (−1)
𝜋2 2
)𝜎 (1−𝛾)
6
4(
}
thus
𝛾(𝜇 − 𝑟 + 𝜀𝜎)2 (4 − 3𝛾)
𝐻 = 𝜌 − 𝛾𝑟 −
+
𝜋2
4 ( 6 ) 𝜎 2 (1 − 𝛾)2
+[
𝛾(𝑎 + 𝑟)
𝑟+𝑎−𝑏
][
] {𝜌 − 𝛾𝑟 −
𝑟 + 𝑎 − 𝑏 (𝑎 + 𝑟)(1 − 𝛾)
(𝜇 − 𝑟 + 𝜀𝜎)2 (−1)
}=
𝜋2 2
4 ( 6 ) 𝜎 (1 − 𝛾)
Hence
1
9𝛾(𝜇 −𝑟+𝜀𝜎)2
𝐻 = (1−𝛾) {𝜌 − 𝛾𝑟 − 2(𝜋2 )𝜎2 (1−𝛾)}
(A21)
Using equations (A11), (A20) and (A19) I will derive 𝑛
ℎ(𝑎+𝑟)
𝑛 = (𝜇 − 𝑟 + 𝜀𝜎)𝑚 + 𝑟 − 𝑟+𝑎−𝑏 =
(A11)
= (𝜇 − 𝑟 + 𝜀𝜎) [
(𝜇 − 𝑟 + 𝜀𝜎)2 (−1) (𝑎 + 𝑟)
𝑟+𝑎−𝑏
+
𝑟
−
𝜌
−
𝛾𝑟
−
=
]
{
}
(𝑎 + 𝑟)(1 − 𝛾)
𝑟+𝑎−𝑏
𝜋2
𝜋2
( ) 𝜎2 (1 − 𝛾)
4 ( ) 𝜎2 (1 − 𝛾)
(𝜇 − 𝑟 + 𝜀𝜎)
6
6
thus
𝑛=
(A22)
(𝜇−𝑟+𝜀𝜎)2
𝜋2
( )𝜎2 (1−𝛾)
6
(𝜇−𝑟+𝜀𝜎)2
1
+ 𝑟 − (1−𝛾) {𝜌 − 𝛾𝑟 +
𝜋2
4( 6 )𝜎2 (1−𝛾)
}
Using equations (A10), (A20) and (A21)
𝑉[𝑊(𝑡), 𝑥(𝑡)] =
(A10)
(ℎ)𝛾
𝛾
𝛾 −1
𝑥(𝑡)
[𝑊(𝑡) − 𝑟+𝑎−𝑏] ( 𝐻 ) =
thus
𝛾
(A23)
𝑉[𝑊(𝑡), 𝑥(𝑡)] =
∗{
(𝜇−𝑟+𝜀𝜎)2 (−1)
𝑟+𝑎−𝑏
{( )(1−𝛾)[𝜌−𝛾𝑟−
]}
𝑎+𝑟
𝜋2
4( 6 )𝜎2 (1−𝛾)
𝛾
𝑥(𝑡)
𝛾
[𝑊(𝑡) − 𝑟+𝑎−𝑏] ∗
−1
}=
1
9𝛾(𝜇 − 𝑟 + 𝜀𝜎)2
(1 − 𝛾 ) [𝜌 − 𝛾𝑟 −
]
2(𝜋 2 )𝜎 2 (1 − 𝛾)
From equation (A7)
(A7)
𝑊(𝑠) −
𝑥(𝑠)
𝑟+𝑎−𝑏
𝜋
= [𝑊(𝑡) −
𝑥(𝑡)
{( )𝜎𝑚[𝑤(𝑠)−𝑤(𝑡)]}
] 𝑒 {𝑛(𝑠−𝑡)} 𝑒 √6
𝑟+𝑎−𝑏
For inserting 𝑠 = 𝑡 and 𝑡 = 0 in equation (A7) I derive
𝑥(𝑡)
𝑥(0)
𝑊(𝑡) − 𝑟+𝑎−𝑏 = [𝑊(0) − 𝑟+𝑎−𝑏] 𝑒 {𝑛(𝑡−0)} 𝑒
{(
𝜋
)𝜎𝑚[𝑤(𝑡)−𝑤(0)]}
√6
Since for a Wiener process 𝑤(0) = 0 , I derive the wealth (=capital)
(A24)
𝑥(𝑡)
𝑥(0)
𝑊(𝑡) = 𝑟+𝑎−𝑏 + [𝑊(0) − 𝑟+𝑎−𝑏] 𝑒
[𝑛𝑡+(
𝜋
)𝜎𝑚𝑤(𝑡)]
√6
where
(A22)
𝑛=
(𝜇−𝑟+𝜀𝜎)2
𝜋2
( )𝜎2 (1−𝛾)
6
1
+ 𝑟 − (1−𝛾) {𝜌 − 𝛾𝑟 +
(𝜇−𝑟+𝜀𝜎)2
𝜋2
4( 6 )𝜎2 (1−𝛾)
}
,𝑠 ≥ 𝑡
Next, I will derive the consumption growth rate
𝑑𝑐(𝑡)
𝑐(𝑡)
Insert equation (A24) into (A3)
(A25)
𝑥(𝑡)
𝑥(0)
𝑐(𝑡) − 𝑥(𝑡) = ℎ [𝑊(𝑡) − 𝑟+𝑎−𝑏] = ℎ {[𝑊(0) − 𝑟+𝑎−𝑏] 𝑒
[𝑛𝑡+(
𝜋
)𝜎𝑚𝑤(𝑡)]
√6
}
Apply natural logarithm (Ln) on both sides of equation (A25)
(A26)
𝑥(0)
𝜋
𝐿𝑛[𝑐(𝑡) − 𝑥(𝑡)] = 𝐿𝑛(ℎ) + 𝐿𝑛 [𝑊(0) − 𝑟+𝑎−𝑏] + 𝑛𝑡 + ( 6) 𝜎𝑚𝑤(𝑡)
√
Derive equation (A26) by t
𝑑𝐿𝑛[𝑐(𝑡)−𝑥(𝑡)]
𝑑𝑡
𝜋
= 0 + 0 + 𝑛 + ( 6) 𝜎 𝑚 [
√
𝑑𝑤(𝑡)
]
𝑑𝑡
Hence
(A27)
𝜋
𝑑𝐿𝑛[𝑐(𝑡) − 𝑥(𝑡)] = 𝑛𝑑𝑡 + ( 6) 𝜎𝑚𝑑𝑤(𝑡)
√
Derivative rules determine
(A28)
𝑑𝐿𝑛[𝑐(𝑡) − 𝑥(𝑡)] =
𝑑[𝑐(𝑡)−𝑥(𝑡)]
𝑐(𝑡)−𝑥(𝑡)
=
𝑑𝑐(𝑡)−𝑑𝑥(𝑡)
𝑐(𝑡)−𝑥(𝑡)
insert equation (A27) into (A28) and derive
𝑑𝑐(𝑡)−𝑑𝑥(𝑡)
𝑐(𝑡)−𝑥(𝑡)
(A29)
𝜋
= 𝑛𝑑𝑡 + ( 6) 𝜎𝑚𝑑𝑤(𝑡)
√
𝜋
𝑑𝑐(𝑡) − 𝑑𝑥(𝑡) = [𝑐(𝑡) − 𝑥(𝑡)]𝑛𝑑𝑡 + [𝑐(𝑡) − 𝑥(𝑡)] ( 6) 𝜎𝑚𝑑𝑤(𝑡)
√
Insert equation (A2) into (A29)
(A2)
𝑑𝑥(𝑡) = [𝑏𝑐(𝑡) − 𝑎𝑥(𝑡)]𝑑𝑡
thus
(A29)
𝜋
𝑑𝑐(𝑡) = [𝑏𝑐(𝑡) − 𝑎𝑥(𝑡)]𝑑𝑡 + [𝑐(𝑡) − 𝑥(𝑡)]𝑛𝑑𝑡 + [𝑐(𝑡) − 𝑥(𝑡)] ( 6) 𝜎𝑚𝑑𝑤(𝑡)
Divide both sides of equation (A29) by 𝑐(𝑡), and solve for
𝑑𝑐(𝑡)
𝑐(𝑡)
√
(A30)
𝑑𝑐(𝑡)
𝑐(𝑡)
= [𝑛 + 𝑏 −
(𝑛+𝑎)𝑥(𝑡)
𝑐(𝑡)
𝑥(𝑡)
𝜋
] 𝑑𝑡 + [1 − 𝑐(𝑡) ] ( 6) 𝜎𝑚𝑑𝑤(𝑡)
√
where
𝑛=
(A22)
(𝜇−𝑟+𝜀𝜎)2
(𝜇−𝑟+𝜀𝜎)2
1
𝜋2
( )𝜎2 (1−𝛾)
6
+ 𝑟 − (1−𝛾) {𝜌 − 𝛾𝑟 +
𝜋2
4( 6 )𝜎2 (1−𝛾)
}
I now show that the optimal policies 𝑐 ∗ (𝑡) and 𝛼 ∗ (𝑡) found above are unique:
From equation (A12)
(A12)
𝑡
1
𝑀(𝑡) = ∫0 𝑒 −𝜌(𝑠−0) 𝛾 [𝑐(𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠 + 𝑒 −𝜌𝑡 𝑉[𝑊(𝑡), 𝑥(𝑡)]
I define the differential
(A31)
𝑑𝑀(𝑡) =
𝑑𝑀(𝑡)
𝑑𝑡
From equation (A13)
𝑑𝑀(𝑡)
𝑑𝑡 + 𝑑𝑊(𝑡) 𝑑𝑊(𝑡)
𝑁(𝑡) =
𝑑𝑀(𝑡)
(A32)
𝑑𝑊(𝑡)
𝑑𝑀(𝑡)
𝑑𝑡
and
= 𝑒 −𝜌𝑡 𝑉𝑊
Insert equations (A13) and (A32) into (A31) to derive
𝑑𝑀(𝑡) = 𝑁(𝑡)𝑑𝑡 + 𝑒 −𝜌𝑡 𝑉𝑊 𝑑𝑊(𝑡)
(A33)
Thus, for an arbitrary (𝑐, 𝛼):
(A34)
For the optimal policies (𝑐 ∗ , 𝛼 ∗ ) when (𝑡) =
𝑑𝑀(𝑡)
𝑑𝑡
= 0 : (A35)
𝑑𝑀(𝑡) ≤ 𝑒 −𝜌𝑡 𝑉𝑊 𝑑𝑊(𝑡)
𝑑𝑀(𝑡) = 𝑒 −𝜌𝑡 𝑉𝑊 𝑑𝑊(𝑡)
Using equation (A12):
(A36)
𝑡=∞ −𝜌(𝑠−0) 1
𝑒
𝛾
𝑀(𝑡 = ∞) = ∫0
[𝑐(𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠 + 𝑒 −𝜌∞ 𝑉[𝑊(𝑡 = ∞), 𝑥(𝑡 = ∞)]
𝑡=∞
𝑡=∞
0
0
1
1
= ∫ 𝑒 −𝜌(𝑠−0) [𝑐(𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠 + 0 = ∫ 𝑒 −𝜌𝑠 [𝑐(𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠
𝛾
𝛾
Applying the expectation operator on both sides of equation (A36) I get
(A37)
𝑡=∞
𝐸𝑡=0 [𝑀(𝑡 = ∞)] = 𝐸𝑡=0 { ∫ 𝑒 −𝜌𝑠
0
1
[𝑐(𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠}
𝛾
For t=0
(A38)
𝑡=0
𝑀(𝑡 = 0) = ∫ 𝑒 −𝜌(𝑠−0)
0
1
[𝑐(𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠 + 𝑒 −𝜌0 𝑉[𝑊(𝑡 = 0), 𝑥(𝑡 = 0)] =
𝛾
= 0 + 𝑉[𝑊(𝑡 = 0), 𝑥(𝑡 = 0)] =
Inserting t=0 in equation (A9) I get
(A39)
∞
𝑉[𝑊(𝑡 = 0), 𝑥(𝑡 = 0)] = 𝐸𝑡=0 ∫ 𝑒 −𝜌(𝑠−0)
𝑡=0
1 ∗
[𝑐 (𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠
𝛾
Hence
(A40)
∞
𝑀(𝑡 = 0) = 𝑉[𝑊(𝑡 = 0), 𝑥(𝑡 = 0)] = 𝐸𝑡=0 ∫ 𝑒 −𝜌(𝑠−0)
𝑡=0
1 ∗
[𝑐 (𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠
𝛾
Applying the expectation operator on both sides of equation (A40) I get
(A41)
∞
𝐸𝑡=0 [𝑀(𝑡 = 0)] = 𝐸𝑡=0 {𝐸𝑡=0 ∫ 𝑒 −𝜌(𝑠−0)
𝑡=0
∞
= 𝐸𝑡=0 { ∫ 𝑒 −𝜌(𝑠−0)
𝑡=0
1 ∗
[𝑐 (𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠} =
𝛾
1 ∗
[𝑐 (𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠}
𝛾
Comparing the right hand sides of equations (A37) and (A41) the optimal effect 𝑐(𝑠) ≤ 𝑐 ∗ (𝑠)
I get
(A42)
𝑡=∞
𝐸𝑡=0 [𝑀(𝑡 = ∞)] = 𝐸𝑡=0 { ∫ 𝑒 −𝜌𝑠
0
1
[𝑐(𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠} ≤
𝛾
∞
≤ 𝐸𝑡=0 { ∫ 𝑒 −𝜌(𝑠−0)
𝑡=0
1 ∗
[𝑐 (𝑠) − 𝑥(𝑠)]𝛾 𝑑𝑠} = 𝐸𝑡=0 [𝑀(𝑡 = 0)]
𝛾
Thus
(A43)
𝐸𝑡=0 [𝑀(𝑡 = ∞)] ≤ 𝐸𝑡=0 [𝑀(𝑡 = 0)]
Which proves that 𝑀(𝑡) is a supermartingale.
Hence equation (A43) is an equality iff 𝑐(𝑠) = 𝑐 ∗ (𝑠) and 𝛼(𝑠) = 𝛼 ∗ (𝑠) for all s , 𝑠 ≥ 0 .
Thus, for 𝑡 > 𝑠 ≥ 0 the optimal policies found (𝑐 ∗ , 𝛼 ∗ ) are unique.
Appendix B: Proof of Theorem 2
I defined above 𝑧(𝑡) the subsistence rate of consumption generated by Habit Persistence.
(22)
𝑧(𝑡) =
𝑥(𝑡)
𝑐(𝑡)
Where 𝑥(𝑡) is the subsistence level of consumption generated by Habit Persistence.
I also defined
(24)
𝑥(𝑡)
𝑧(𝑡)
𝑦(𝑡) = 𝑐(𝑡)−𝑥(𝑡) = 1−𝑧(𝑡)
From equation (22) I can find
𝑐(𝑡) =
𝑥(𝑡)
𝑧(𝑡)
Derive both sides by t such that
𝑑𝑥(𝑡)
𝑑𝑧(𝑡)
𝑧(𝑡) − 𝑥(𝑡)
𝑑𝑐(𝑡)
𝑑𝑡
𝑑𝑡
=
[𝑧(𝑡)]2
𝑑𝑡
𝑑𝑡
Multiple both sides by 𝑐(𝑡) to derive
𝑑𝑐(𝑡)
(B1)
𝑐(𝑡)
=
𝑑𝑥(𝑡)
𝑥(𝑡)
−
𝑑𝑧(𝑡)
𝑧(𝑡)
From equation (A2)
𝑑𝑥(𝑠)
(A2)
𝑑𝑠
= 𝑏𝑐 ∗ (𝑠) − 𝑎𝑥(𝑠)
I can derive
𝑑𝑥(𝑡)
= 𝑏𝑐(𝑡) − 𝑎𝑥(𝑡)
𝑑𝑡
Hence
𝑑𝑥(𝑡)
(B2)
𝑥(𝑡)
𝑐(𝑡)
1
= [𝑏 𝑥(𝑡) − 𝑎] 𝑑𝑡 = [𝑏 𝑧(𝑡) − 𝑎] 𝑑𝑡
From equation (21), (22) and (B1)
𝑑𝑐(𝑡)
(21) (A30)
𝑐(𝑡)
= [𝑛 + 𝑏 −
(𝑛+𝑎)𝑥(𝑡)
𝑐(𝑡)
𝑥(𝑡)
𝜋
] 𝑑𝑡 + [1 − 𝑐(𝑡) ] ( 6) 𝜎𝑚𝑑𝑤(𝑡)
√
𝑑𝑥(𝑡) 𝑑𝑧(𝑡)
𝜋
−
= [𝑛 + 𝑏 − (𝑛 + 𝑎)𝑧(𝑡)]𝑑𝑡 + [1 − 𝑧(𝑡)] ( ) 𝜎𝑚𝑑𝑤(𝑡)
𝑥(𝑡)
𝑧(𝑡)
√6
Insert equation (B2) into the above
[𝑏
1
𝑑𝑧(𝑡)
𝜋
− 𝑎] 𝑑𝑡 −
= [𝑛 + 𝑏 − (𝑛 + 𝑎)𝑧(𝑡)]𝑑𝑡 + [1 − 𝑧(𝑡)] ( ) 𝜎𝑚𝑑𝑤(𝑡)
𝑧(𝑡)
𝑧(𝑡)
√6
Solve for 𝑑𝑧(𝑡)
(B3)
𝜋
𝑑𝑧(𝑡) = [𝑏 − 𝑎𝑧(𝑡) − 𝑛𝑧(𝑡)][1 − 𝑧(𝑡)]𝑑𝑡 − 𝑧(𝑡)[1 − 𝑧(𝑡)] ( 6) 𝜎𝑚𝑑𝑤(𝑡)
√
From equation (24)
(24)
Derive both sides by t
𝑧(𝑡)
𝑦(𝑡) = 1−𝑧(𝑡)
𝑑𝑧(𝑡)
𝑑[1 − 𝑧(𝑡)] 𝑑𝑧(𝑡) [1 − 𝑧(𝑡)] − 𝑧(𝑡) [−𝑑𝑧(𝑡)]
[1 − 𝑧(𝑡)] − 𝑧(𝑡)
𝑑𝑦(𝑡)
𝑑𝑡
𝑑𝑡
𝑑𝑡
= 𝑑𝑡
=
[1 − 𝑧(𝑡)]2
[1 − 𝑧(𝑡)]2
𝑑𝑡
Hence
𝑑𝑧(𝑡) = 𝑑𝑦(𝑡)[1 − 𝑧(𝑡)]2
(B4)
Insert equation (B4) into (B3)
𝜋
𝑑𝑦(𝑡)[1 − 𝑧(𝑡)]2 = [𝑏 − 𝑎𝑧(𝑡) − 𝑛𝑧(𝑡)][1 − 𝑧(𝑡)]𝑑𝑡 − 𝑧(𝑡)[1 − 𝑧(𝑡)] ( ) 𝜎𝑚𝑑𝑤(𝑡)
√6
I Divide both sides by [1 − 𝑧(𝑡)]2
𝑑𝑦(𝑡) = [
𝑏
𝑧(𝑡)
𝑧(𝑡)
𝜋
− (𝑛 + 𝑎)
] 𝑑𝑡 −
( ) 𝜎𝑚𝑑𝑤(𝑡)
1 − 𝑧(𝑡)
1 − 𝑧(𝑡)
1 − 𝑧(𝑡) √6
Inserting equation (24) into the above yields
𝜋
𝑑𝑦(𝑡) = {𝑏[1 − 𝑦(𝑡)] − (𝑛 + 𝑎)𝑦(𝑡)}𝑑𝑡 − 𝑦(𝑡) ( ) 𝜎𝑚𝑑𝑤(𝑡)
√6
(B5)
𝜋
𝑑𝑦(𝑡) = {𝑏 − (𝑛 + 𝑎 − 𝑏)𝑦(𝑡)}𝑑𝑡 − 𝑦(𝑡) ( 6) 𝜎𝑚𝑑𝑤(𝑡)
√
I continue with deriving 𝑦̅ and 𝑦̂ (the modal value of y)
The general case of the one dimension (only variable t) Fokker-Plank equation states
Given
𝑑𝑥𝑡 = 𝜇(𝑥𝑡 , 𝑡)𝑑𝑡 + 𝜎(𝑥𝑡 , 𝑡)𝑑𝑤𝑡
𝜎2 (𝑥 ,𝑡)
Where 𝜇(𝑥𝑡 , 𝑡) is the drift coefficient and 𝐷(𝑥𝑡 , 𝑡) = 2 𝑡 is the diffusion coefficient.
For 𝑝(𝑥𝑡 , 𝑡), the probability density of random variable 𝑥𝑡 , the Fokker-Plank equation is
𝜕𝑝(𝑥𝑡 , 𝑡)
𝜕[𝜇(𝑥𝑡 , 𝑡)𝑝(𝑥𝑡 , 𝑡)] 𝜕 2 [𝐷(𝑥𝑡 , 𝑡)𝑝(𝑥𝑡 , 𝑡)]
=−
+
𝜕𝑡
𝜕𝑥
𝜕𝑥 2
In my specific case: if the density of 𝑦(𝑡) is 𝑝𝑦 (𝑦𝑡 , 𝑡), given
𝜋
𝑑𝑦(𝑡) = [𝑏 − (𝑛 + 𝑎 − 𝑏)𝑦(𝑡)]𝑑𝑡 − 𝑦(𝑡) ( 6) 𝜎𝑚𝑑𝑤(𝑡)
(B5)
√
The Fokker-Plank equation is:
𝜕2
𝜕{[𝑏 − (𝑛 + 𝑎 − 𝑏)𝑦(𝑡)] 𝑝𝑦 (𝑦𝑡 , 𝑡)}
𝜕𝑝(𝑦𝑡 , 𝑡)
=−
+
𝜕𝑡
𝜕𝑦
𝜋
[−𝑦(𝑡) ( ) 𝜎𝑚]
√6
2
2
𝑝𝑦 (𝑦𝑡 , 𝑡)
{
}
𝜕𝑦 2
Hence
𝜕𝑝(𝑦𝑡 ,𝑡)
(B6)
𝜕𝑡
=−
𝜕{[𝑏−(𝑛+𝑎−𝑏)𝑦(𝑡)] 𝑝𝑦 (𝑦𝑡 ,𝑡)}
𝜕𝑦
𝜋2
+ (2∗6)
𝜕2 {[𝑦(𝑡)]2 𝜎2 𝑚2 𝑝𝑦 (𝑦𝑡 ,𝑡)}
𝜕𝑦 2
𝜕𝑝𝑦 (𝑦𝑡 ,𝑡)
Since 𝑝(𝑦𝑡 , 𝑡) represent a stationary distribution it satisfies 𝜕𝑡 = 0
Which make equation (B6) to become the Pearson equation as in equation (B7).
0=−
(B7)
𝜕{[𝑏−(𝑛+𝑎−𝑏)𝑦(𝑡)] 𝑝𝑦 (𝑦𝑡 ,𝑡)}
𝜕𝑦
𝜋2
+ (12)
𝜕2 {[𝑦(𝑡)]2 𝜎2 𝑚2 𝑝𝑦 (𝑦𝑡 ,𝑡)}
𝜕𝑦 2
I integrate both sides of equation (B7) by 𝑦(𝑡) to get
𝜋2
0 = (12)
(B8)
𝜋2
𝜕{[𝑦(𝑡)]2 𝜎2 𝑚2 𝑝𝑦 (𝑦𝑡 ,𝑡)}
𝜕𝑦
0 = ( ) 𝜎 2 𝑚2 {[𝑦(𝑡)]2
12
(B9)
− [𝑏 − (𝑛 + 𝑎 − 𝑏)𝑦(𝑡)] 𝑝𝑦 (𝑦𝑡 , 𝑡)
𝜕𝑝𝑦 (𝑦𝑡 , 𝑡)
+ 2𝑦(𝑡)𝑝𝑦 (𝑦𝑡 , 𝑡)} − [𝑏 − (𝑛 + 𝑎 − 𝑏)𝑦(𝑡)] 𝑝𝑦 (𝑦𝑡 , 𝑡)
𝜕𝑦
𝜋2
0 = (12) 𝜎 2 𝑚2 [𝑦(𝑡)]2
𝜕𝑝𝑦 (𝑦𝑡 ,𝑡)
𝜕𝑦
𝜋2
− {𝑏 − [(𝑛 + 𝑎 − 𝑏 + ( 6 ) 𝜎 2 𝑚2 )] 𝑦(𝑡)} 𝑝𝑦 (𝑦𝑡 , 𝑡)
I integrate both sides of equation (B9) by 𝑦(𝑡) to get
(B10)
𝜋2
∞
0 = (12) 𝜎 2 𝑚2 ∫0 [𝑦(𝑡)]2
𝜋2
𝜕𝑝𝑦 (𝑦𝑡 ,𝑡)
𝜕𝑦
∞
𝑑𝑦 − 𝑏 ∫0 𝑝𝑦 (𝑦𝑡 , 𝑡)𝑑𝑦 +
∞
+ [(𝑛 + 𝑎 − 𝑏 + ( ) 𝜎 2 𝑚2 )] ∫ 𝑦(𝑡) 𝑝𝑦 (𝑦𝑡 , 𝑡)𝑑𝑦
6
0
under normalization requirement
∞
∫ 𝑝𝑦 (𝑦𝑡 , 𝑡)𝑑𝑦 = 1
0
And that the average definition is
∞
∫ 𝑦(𝑡) 𝑝𝑦 (𝑦𝑡 , 𝑡)𝑑𝑦 = 𝑦̅
0
Equation (B10) becomes
∞
𝜋2
0 = ( ) 𝜎 2 𝑚2 ∫ [𝑦(𝑡)]2
12
0
𝜕𝑝𝑦 (𝑦𝑡 , 𝑡)
𝜋2
𝑑𝑦 − 𝑏 + [(𝑛 + 𝑎 − 𝑏 + ( ) 𝜎 2 𝑚2 )] 𝑦̅
𝜕𝑦
6
The remaining integral will be solved using integration by parts thus
0=(
∞
𝜋2
𝜋2
) 𝜎 2 𝑚2 {0 − 2 ∫ 𝑦(𝑡)𝑝𝑦 (𝑦𝑡 , 𝑡)𝑑𝑦} − 𝑏 + [(𝑛 + 𝑎 − 𝑏 + ( ) 𝜎 2 𝑚2 )] 𝑦̅
12
6
0
Hence
𝜋2
2
2 {0
0 = ( )𝜎 𝑚
12
𝜋2
− 2𝑦̅} − 𝑏 + [(𝑛 + 𝑎 − 𝑏 + ( ) 𝜎 2 𝑚2 )] 𝑦̅
6
Solving for 𝑦̅ I derive
𝑏
𝑦̅ = 𝑛+𝑎−𝑏 < ∞
(B11)
the condition 𝑛 + 𝑎 − 𝑏 > 0 assures 𝑦̅ < ∞ , i.e. 𝑦̅ is finite.
The mode of a variable, 𝑦(𝑡), that is stationary distributed satisfies
𝜕𝑝𝑦 (𝑦𝑡 ,𝑡)
𝜕𝑦𝑡
= 0 so equation
(B9) is
(B9)
𝜋2
𝜋2
0 = (12) 𝜎 2 𝑚2 [𝑦̂]2 0 − {𝑏 − [(𝑛 + 𝑎 − 𝑏 + ( 6 ) 𝜎 2 𝑚2 )] 𝑦̂} 𝑝𝑦 (𝑦̂, 𝑡)
Solving for 𝑦̂ I derive
𝑦̂ =
(B12)
𝑏
<∞
𝜋2
𝑛+𝑎−𝑏+( )𝜎2 𝑚2
6
𝜋2
the condition 𝑛 + 𝑎 − 𝑏 + ( 6 ) 𝜎 2 𝑚2 > 0 assures 𝑦̂ < ∞ , i.e. 𝑦̂ is finite.
Solving the Pearson equation (B9)
(B9)
𝜋2
𝜕𝑝𝑦 (𝑦𝑡 ,𝑡)
12
𝜕𝑦
0 = ( ) 𝜎 2 𝑚2 [𝑦(𝑡)]2
𝜋2
− {𝑏 − [(𝑛 + 𝑎 − 𝑏 + ( ) 𝜎 2 𝑚2 )] 𝑦(𝑡)} 𝑝𝑦 (𝑦𝑡 , 𝑡)
6
𝜕𝑝𝑦 (𝑦𝑡 , 𝑡)
𝑏
={ 2
−
𝜋
𝜕𝑦
2
2
2
(12) 𝜎 𝑚 [𝑦(𝑡)]
𝜋2
[𝑛 + 𝑎 − 𝑏 + ( 6 ) 𝜎 2 𝑚2 ]
𝜋2
(12
) 𝜎 2 𝑚2 𝑦(𝑡)
Hence
(B13)
𝜕𝑝𝑦 (𝑦𝑡 ,𝑡)
𝑝𝑦 (𝑦𝑡 ,𝑡)
𝜋2
={
𝑏
(
𝜋2
12
)𝜎2 𝑚2 [𝑦(𝑡)]2
−
[(𝑛+𝑎−𝑏+( )𝜎2 𝑚2 )]
6
(
𝜋2
12
)𝜎2 𝑚2 𝑦(𝑡)
Solving the equation (B13) yields
𝜋2
−12[𝑛+𝑎−𝑏+(
{
(B14)
6
)𝜎2 𝑚2 ]
𝜋2 𝜎2 𝑚2
𝑝𝑦 (𝑦𝑡 , 𝑡) = 𝑘𝑦
}
𝑒
{
Where 𝑘 is the integration coefficient and 0 ≤ 𝑦 < ∞
To find 𝑘 I use the normalization requirement
∞
∫ 𝑝𝑦 (𝑦𝑡 , 𝑡)𝑑𝑦 = 1
0
Thus
−12𝑏
}
𝜋2 𝜎2 𝑚2 𝑦
} 𝜕𝑦
} 𝑝𝑦 (𝑦𝑡 , 𝑡)
𝜋2
−12[𝑛+𝑎−𝑏+(
{
∞
6
)𝜎2 𝑚2 ]
}
𝜋2 𝜎2 𝑚2
∫ 𝑘𝑦
𝑒
{
−12𝑏
}
𝜋2 𝜎2 𝑚2 𝑦
𝑑𝑦 = 1
0
Solving the integral yields
1 𝑄+1 ∞ 𝑄 −𝑇
1 𝑄+1
𝑘 −1 = ( )
∫ 𝑇 𝑒 𝑑𝑇 = ( )
𝛤(𝑄 + 1)
𝑞
𝑞
0
Where 𝛤 is the gamma function and
𝑞=
𝑇=
12𝑏
𝜋 2 𝜎 2 𝑚2
12𝑏
𝜋 2 𝜎 2 𝑚2 𝑦
𝜋2
𝑄={
12 [𝑛 + 𝑎 − 𝑏 + ( 6 ) 𝜎 2 𝑚2 ]
𝜋2 𝜎 2 𝑚2
}−2
Hence
𝜋2
12[𝑛+𝑎−𝑏+(
{1−
(B15)
𝑘
−1
6
)𝜎2 𝑚2 ]
}
𝜋2 𝜎2 𝑚2
12𝑏
= (𝜋2 𝜎2 𝑚2 )
𝜋2
𝛤{
12[𝑛+𝑎−𝑏+( )𝜎2 𝑚2 ]
6
𝜋2 𝜎2 𝑚2
To summaries, the overall solution of the Pearson equation (B9) is
𝜋2
−12[𝑛+𝑎−𝑏+(
{
(B14)
where
𝑝𝑦 (𝑦𝑡 , 𝑡) = 𝑘𝑦
6
𝜋2 𝜎2 𝑚2
)𝜎2 𝑚2 ]
}
𝑒
{
−12𝑏
}
𝜋2 𝜎2 𝑚2 𝑦
− 1}
𝜋2
12[𝑛+𝑎−𝑏+(
{1−
(B15)
𝑘
−1
12𝑏
6
)𝜎2 𝑚2 ]
}
𝜋2 𝜎2 𝑚2
= (𝜋2 𝜎2 𝑚2 )
𝜋2
𝛤{
12[𝑛+𝑎−𝑏+( )𝜎2 𝑚2 ]
6
− 1}
𝜋2 𝜎2 𝑚2
I have defined above
𝑧(𝑡)
𝑦(𝑡) = 1−𝑧(𝑡)
(24)
And found
𝑑𝑧(𝑡) = 𝑑𝑦(𝑡)[1 − 𝑧(𝑡)]2
(B4)
Hence
𝑑𝑦(𝑡)
1
=
𝑑𝑧(𝑡) [1 − 𝑧(𝑡)]2
Since 𝑦(𝑡) is monotone increasing function with 0 ≤ 𝑦 < ∞, thus 𝑧(𝑡) is also monotone
increasing function in 𝑦(𝑡) in the domain of 𝑦(𝑡) and 0 ≤ 𝑧 < 1.
Since 𝑦(𝑡) has a stationary distribution 𝑝𝑦 (𝑦𝑡 , 𝑡), 𝑧(𝑡) also have a stationary distribution
𝑝𝑧 (𝑧𝑡 , 𝑡) such that
𝑑𝑦(𝑡)
𝑧
1
𝑝𝑧 (𝑧𝑡 , 𝑡) = 𝑝𝑦 (𝑦𝑡 , 𝑡) [ 𝑑𝑧(𝑡) ] = 𝑝𝑦 (𝑦𝑡 = 1−𝑧𝑡 ) {[1−𝑧(𝑡)]2 }
(B16)
𝑡
Insert equation (B14) and equation (24) into (B16) to find
𝜋2
12[𝑏−𝑎−𝑛−(
(B17)
𝑝𝑧 (𝑧𝑡 , 𝑡) = 𝑘𝑒
12𝑏
}
𝜋2 𝜎2 𝑚2
{
(1 − 𝑧)
[
12(𝑏+𝑎−𝑏)
]
𝜋2 𝜎2 𝑚2
{
6
)𝜎2 𝑚2 ]
}
𝜋2 𝜎2 𝑚2
𝑧
𝑒
{
−12𝑏
}
𝜋2 𝜎2 𝑚2 𝑧
Where 0 ≤ 𝑧 < 1
𝑧̂ is the mode of a variable with stationary distribution thus satisfy
Insert equation (B17) into the above condition
𝜕𝑝𝑧 (𝑧̂ ,𝑡)
𝜕𝑧̂
=0
0=
𝜕𝑝𝑧 (𝑧̂ , 𝑡)
𝜕𝑧̂
𝜋2
12[𝑏−𝑎−𝑛−(
= 𝑘𝑒
{
12(𝑏+𝑎−𝑏)
−12(𝑏 + 𝑎 − 𝑏)
[ 2 2 2 ]−1
𝜋 𝜎 𝑚
(1
[
]
−
𝑧)
𝑧
𝜋2 𝜎 2 𝑚2
12𝑏
}
𝜋2 𝜎2 𝑚2
{
+ (1 − 𝑧)
[
12 [𝑏
12(𝑏+𝑎−𝑏)
]
𝜋2 𝜎2 𝑚2
{
𝜋
𝜋2
2
12[𝑏−𝑎−𝑛−(
2
− 𝑎 − 𝑛 − ( 6 )𝜎 𝑚 ]
𝜋2 𝜎 2 𝑚2
)𝜎2 𝑚2 ]
}
𝜋2 𝜎2 𝑚2
{{
2
6
{
6
)𝜎2 𝑚2 ]
}−1
𝜋2 𝜎2 𝑚2
}𝑧
𝑒
{
−12𝑏
}
𝜋2 𝜎2 𝑚2 𝑧
}
𝜋2
12[𝑏−𝑎−𝑛−(
+
−12𝑏
{
}
𝑒 𝜋2 𝜎2 𝑚2 𝑧 [
−12𝑏
] (−1)𝑧 −2 (1 −
𝜋2 𝜎 2 𝑚2
12(𝑏+𝑎−𝑏)
[
]
𝑧) 𝜋2 𝜎2 𝑚2 𝑧
{
6
)𝜎2 𝑚2 ]
𝜋2 𝜎2 𝑚2
}
}
Solving for 𝑧̂ gives a quadratic equation
𝜋2
𝜋2
( 6 ) 𝜎 2 𝑚2 𝑧̂ 2 − [𝑛 + 𝑎 + ( 6 ) 𝜎 2 𝑚2 ] 𝑧̂ + 𝑏 = 0
(B18)
Usually a quadratic equation has two solution. However, since 0 ≤ 𝑧̂ < 1
𝜋2
𝜋2
𝜋2
𝜋2
For 𝑧̂ = 0 : ( 6 ) 𝜎 2 𝑚2 02 − [𝑛 + 𝑎 + ( 6 ) 𝜎 2 𝑚2 ] 0 + 𝑏 > 0
For 𝑧̂ = 1 : ( 6 ) 𝜎 2 𝑚2 12 − [𝑛 + 𝑎 + ( 6 ) 𝜎 2 𝑚2 ] 1 + 𝑏 = 𝑏 − 𝑎 − 𝑛 < 0
Hence in the domain 0 ≤ 𝑧̂ < 1 there is only one solution (cross the zero line) for equation (B18)
which is equal to
𝜋2
𝜋2
6
6
2
𝜋2
[𝑛+𝑎+( )𝜎2 𝑚2 ]−√[𝑛+𝑎+( )𝜎2 𝑚2 ] −4( )𝜎2 𝑚2 𝑏
(B19)
𝑧̂ =
6
𝜋2
2( )𝜎2 𝑚2
6
Appendix C: deriving the unconditional mean and variance of the consumption growth
𝐸[
I will derive the unconditional mean of consumption growth
From appendix A
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
𝑑𝑐(𝑡)
(A30)
𝑐(𝑡)
= [𝑛 + 𝑏 −
(𝑛+𝑎)𝑥(𝑡)
𝑐(𝑡)
𝑥(𝑡)
𝜋
] 𝑑𝑡 + [1 − 𝑐(𝑡) ] ( 6) 𝜎𝑚𝑑𝑤(𝑡)
√
where
𝑛=
(A22)
(𝜇−𝑟+𝜀𝜎)2
𝜋2
( )𝜎2 (1−𝛾)
6
1
+ 𝑟 − (1−𝛾) {𝜌 − 𝛾𝑟 +
(𝜇−𝑟+𝜀𝜎)2
𝜋2
4( 6 )𝜎2 (1−𝛾)
}
I will calculate the average on both sides of equation (A30) and insert equation (22)
𝐸[
𝑑𝑐(𝑡)
𝜋
] = [𝑛 + 𝑏 − (𝑛 + 𝑎)𝐸[𝑧(𝑡)]]𝑑𝑡 + {1 − 𝐸[𝑧(𝑡)]} ( ) 𝜎𝑚 𝐸[𝑑𝑤(𝑡)]
𝑐(𝑡)
√6
Since for a wiener process 𝐸[𝑤(𝑡)] = 0 thus
𝐸[𝑑𝑤(𝑡)] = 𝑑{𝐸[𝑤(𝑡)]} = 0
Hence
𝐸[
𝑑𝑐(𝑡)
] = {𝑛 + 𝑏 − (𝑛 + 𝑎)𝐸[𝑧(𝑡)]}𝑑𝑡 + 0
𝑐(𝑡)
So
𝐸[
(C1)
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
1
= 𝑛 + 𝑏 − (𝑛 + 𝑎)𝐸[𝑧(𝑡)] = 𝑛 + 𝑏 − (𝑛 + 𝑎) ∫0 𝑧(𝑡) 𝑝𝑧 (𝑧𝑡 , 𝑡)𝑑𝑧
Since a closed form expression for the integral is unavailable, the integration is done
numerically.
I will calculate the variance on both sides of equation (A30) and insert equation (22)
𝑣𝑎𝑟 [
𝑑𝑐(𝑡)
𝜋
] = 𝑣𝑎𝑟{[𝑛 + 𝑏 − (𝑛 + 𝑎)𝑧(𝑡)]𝑑𝑡} + 𝑣𝑎𝑟 {[1 − 𝑧(𝑡)] ( ) 𝜎𝑚 𝑑𝑤(𝑡)} +
𝑐(𝑡)
√6
+2𝑐𝑜𝑣 {[𝑛 + 𝑏 − (𝑛 + 𝑎)𝑧(𝑡)]𝑑𝑡, [1 − 𝑧(𝑡)] (
𝜋
√6
) 𝜎𝑚 𝑑𝑤(𝑡)} =
Since 𝑣𝑎𝑟(𝑑𝑡) = 0,
And for a wiener process 𝑣𝑎𝑟[𝑑𝑤(𝑡)] = 𝑑𝑣𝑎𝑟[𝑤(𝑡)] = 𝑑𝑡 I derive
𝑣𝑎𝑟 [
𝑑𝑐(𝑡)
𝜋2
] = ( ) 𝜎 2 𝑚2 𝐸{[1 − 𝑧(𝑡)]2 }𝑑𝑡
𝑐(𝑡)
6
hence
𝑣𝑎𝑟[
(C2)
𝑑𝑐(𝑡)
]
𝑐(𝑡)
𝑑𝑡
𝜋2
𝜋2
1
= ( 6 ) 𝜎 2 𝑚2 𝐸{[1 − 𝑧(𝑡)]2 } = ( 6 ) 𝜎 2 𝑚2 ∫0 [1 − 𝑧(𝑡)]2 𝑝𝑧 (𝑧𝑡 , 𝑡)𝑑𝑧
Since a closed form expression for the integral is unavailable, the integration is done
numerically.
Appendix D: deriving RRA coefficient and the intertemporal Elasticity of Substitution in
Consumption (=s)
In appendix A I derived
(ℎ)𝛾
𝑉[𝑊(𝑡), 𝑥(𝑡)] =
(A10)
𝛾
𝑥(𝑡)
𝛾 −1
[𝑊(𝑡) − 𝑟+𝑎−𝑏] ( 𝐻 )
The derivatives of equation (A10)
𝑉𝑊 =
(ℎ)𝛾
𝛾
𝑉𝑊𝑊 =
𝑥(𝑡)
𝛾 [𝑊(𝑡) − 𝑟+𝑎−𝑏]
(ℎ)𝛾 (𝛾
𝛾−1 −1
𝑥(𝑡)
( 𝐻 ) = (ℎ)𝛾 [𝑊(𝑡) − 𝑟+𝑎−𝑏]
− 1) [𝑊(𝑡) −
𝑥(𝑡)
𝑟+𝑎−𝑏
𝛾−2
]
𝛾−1 −1
(𝐻)
−1
( )
𝐻
I define RRA coefficient and insert the above derivatives
(D1)
𝑅𝑅𝐴 =
−𝑊𝑉𝑊𝑊
=
𝑉𝑊
𝛾−2
𝑥(𝑡)
−1
−𝑊(ℎ)𝛾 (𝛾 − 1) [𝑊(𝑡) − 𝑟 + 𝑎 − 𝑏]
(𝐻)
(ℎ)𝛾 [𝑊(𝑡) −
𝑥(𝑡)
]
𝑟+𝑎−𝑏
𝛾−1
−1
(𝐻)
Fron equation (A3)
(A3)
𝑥(𝑡)
𝑐(𝑡) − 𝑥(𝑡) = ℎ [𝑊(𝑡) − 𝑟+𝑎−𝑏]
Solving for 𝑊(𝑡)
(D2)
Insert equation (D2) into (D1)
𝑊(𝑡) = [
𝑐(𝑡)−𝑥(𝑡)
𝑥(𝑡)
] + 𝑟+𝑎−𝑏
ℎ
1
= (1 − 𝛾)
1−
𝑥(𝑡)
𝑊(𝑟 + 𝑎 − 𝑏)
𝑅𝑅𝐴 = (1 − 𝛾)
1
𝑥(𝑡)
1−
𝑐(𝑡) − 𝑥(𝑡)
𝑥(𝑡)
{[
] + 𝑟 + 𝑎 − 𝑏} (𝑟 + 𝑎 − 𝑏)}
{
ℎ
𝑥(𝑡)
=
ℎ
]}
𝑐(𝑡) − 𝑥(𝑡) (𝑟 + 𝑎 − 𝑏)
= (1 − 𝛾) {1 + [
][
Inserting equation (24) into the above expression
𝑥(𝑡)
𝑧(𝑡)
𝑦(𝑡) = 𝑐(𝑡)−𝑥(𝑡) = 1−𝑧(𝑡)
(24)
thus
ℎ
𝑅𝑅𝐴 = (1 − 𝛾) {1 + 𝑦(𝑡) [(𝑟+𝑎−𝑏)]}
(D3)
From equation (D3) and since 𝑦(𝑡) has a steady state distribution (with 𝑦̅ and 𝑦̂ ), RRA
coefficient also has a steady state distribution. Thus RRA average is
̅̅̅̅̅̅
𝑅𝑅𝐴 = (1 − 𝛾) {1 + ̅̅̅̅̅̅
𝑦(𝑡) [
ℎ
]}
(𝑟 + 𝑎 − 𝑏)
Insert equation (28) into the above equation
𝑏
𝑦̅ = 𝑛+𝑎−𝑏
(28) (B11)
Thus
(D4)
̅̅̅̅̅̅ = (1 − 𝛾) {1 + [
𝑅𝑅𝐴
𝑏
𝑛+𝑎−𝑏
ℎ
ℎ𝑏
] [(𝑟+𝑎−𝑏)]} = (1 − 𝛾) {1 + [(𝑛+𝑎−𝑏)(𝑟+𝑎−𝑏)]}
When 𝑧(𝑡) = 𝑧̂
(D5)
𝑧̂
ℎ
𝑅𝑅𝐴(𝑧 = 𝑧̂ ) = (1 − 𝛾) {1 + [1−𝑧̂ ] [(𝑟+𝑎−𝑏)]}
The elasticity of substitution in consumption (=s) is defined as the derivative of the expected
growth rate in consumption with respect to 𝑟 while 𝑧(𝑡), 𝜇 − 𝑟, 𝑎𝑛𝑑 𝜎 2 held constant thus
𝑑𝑐
)
𝑐
]
𝑑𝑡
𝐸(
𝜕[
𝑠=
while 𝑧(𝑡), 𝜇 − 𝑟, 𝑎𝑛𝑑 𝜎 2 held constant
𝜕𝑟
From equation (C1) when 𝑧(𝑡), 𝜇 − 𝑟, 𝑎𝑛𝑑 𝜎 2 held constant
(C1)
𝐸[
𝑑𝑐(𝑡)
]
𝑐(𝑡)
= 𝑛 + 𝑏 − (𝑛 + 𝑎)𝐸[𝑧(𝑡)] = 𝑛 + 𝑏 − (𝑛 + 𝑎)𝑧(𝑡) = 𝑛[1 − 𝑧(𝑡)] + 𝑏 − 𝑎𝑧(𝑡)
𝑑𝑡
And equation (20)
𝑛=
(20) (A22)
= 𝑟(
1
1−𝛾
)+
(𝜇−𝑟+𝜀𝜎)2
1
𝜋2
( )𝜎2 (1−𝛾)
6
+ 𝑟 − (1−𝛾) {𝜌 − 𝛾𝑟 +
(𝜇−𝑟+𝜀𝜎)2
𝜋2
4( 6 )𝜎2 (1−𝛾)
}=
(𝜇 − 𝑟 + 𝜀𝜎)2
(𝜇 − 𝑟 + 𝜀𝜎)2
1
+
𝑟
−
{𝜌
+
}
𝜋2 2
𝜋2 2
(1 − 𝛾)
( 6 ) 𝜎 (1 − 𝛾)
4 ( 6 ) 𝜎 (1 − 𝛾)
Insert equation (20) into (C1) and derive s when 𝑧(𝑡), 𝜇 − 𝑟, 𝑎𝑛𝑑 𝜎 2 held constant
𝑑𝑐
)
𝑐 ]
𝑑𝑡
𝐸(
𝜕[
(D6)
𝑠=
𝜕𝑟
1
= [1 (1−𝛾) + 0 − 0] [1 − 𝑧(𝑡)] + 0 =
1−𝑧(𝑡)
1−𝛾
To find s*RRA use equations (D6), (24) and (D3)
(D7)
𝑠 ∗ 𝑅𝑅𝐴 = [
1−𝑧(𝑡)
1−𝛾
𝑧(𝑡)
ℎ
] (1 − 𝛾) {1 + 𝑦(𝑡) [(𝑟+𝑎−𝑏)]} =
ℎ
ℎ
= [1 − 𝑧(𝑡)] {1 + [1−𝑧(𝑡)] [(𝑟+𝑎−𝑏)]} = 1 − {1 − [(𝑟+𝑎−𝑏)]} 𝑧(𝑡)
Hence
ℎ
𝑠 ∗ 𝑅𝑅𝐴 = 1 − {1 − [(𝑟+𝑎−𝑏)]} 𝑧(𝑡)
(D7)
The modal value of s*RRA is
ℎ
𝑚𝑜𝑑𝑒(𝑠 ∗ 𝑅𝑅𝐴) = 1 − {1 − [(𝑟+𝑎−𝑏)]} 𝑧̂
(D8)
Thus for a time separable utility (i.e. b=0) the following values found
Using equations (B19) and (D8)
𝜋2
𝜋2
6
6
2
𝜋2
[𝑛+𝑎+( )𝜎2 𝑚2 ]−√[𝑛+𝑎+( )𝜎2 𝑚2 ] −4( )𝜎2 𝑚2 0
𝑧̂ =
(B19)
6
=
𝜋2
2( )𝜎2 𝑚2
6
𝜋2
𝜋2
[𝑛+𝑎+( )𝜎2 𝑚2]−[𝑛+𝑎+( )𝜎2 𝑚2 ]
6
6
𝜋2
2(
(D8)
6
)𝜎2 𝑚2
=0
ℎ
ℎ
𝑚𝑜𝑑𝑒(𝑠 ∗ 𝑅𝑅𝐴) = 1 − {1 − [(𝑟+𝑎−𝑏)]} 𝑧̂ = 1 − {1 − [(𝑟+𝑎−𝑏)]} 0 = 1
Appendix E: derive mean and variance of risky asset return
Using equation (47)
(47)
𝑑𝑆(𝑡)
𝑆(𝑡)
𝛿
𝛿
𝜋
= [𝛿1 (𝜇 + 𝜀𝜎 − 𝑟) + 𝑟] 𝑑𝑡 + (𝛿1 ) ( ) 𝜎𝑑𝑤(𝑡)
2
2
√6
Apply expectation operator on both sides
𝑑𝑆(𝑡)
𝛿1
𝛿1
𝜋
𝐸[
] = [ (𝜇 + 𝜀𝜎 − 𝑟) + 𝑟] 𝐸[𝑑𝑡] + ( ) ( ) 𝜎𝐸[𝑑𝑤(𝑡)]
𝑆(𝑡)
𝛿2
𝛿2 √6
Since 𝐸[𝑑𝑡] = 𝑑𝐸[𝑡] = 𝑑𝑡 and
𝐸[𝑑𝑤(𝑡)] = 𝑑𝐸[𝑤(𝑡)] = 0 since for a wiener process 𝐸[𝑤(𝑡)] = 0 I derive
𝑑𝑆(𝑡)
𝛿1
𝐸[
] = [ (𝜇 + 𝜀𝜎 − 𝑟) + 𝑟] 𝑑𝑡
𝑆(𝑡)
𝛿2
Hence
(E1)
𝐸[
𝑑𝑆(𝑡)
]
𝑆(𝑡)
𝑑𝑡
𝛿
= [𝛿1 (𝜇 + 𝜀𝜎 − 𝑟) + 𝑟]
2
Apply variance on both sides of equation (44) above
2
2
𝑑𝑆(𝑡)
𝛿1
𝛿1 𝜋
𝑣𝑎𝑟 [
] = [ (𝜇 + 𝜀𝜎 − 𝑟) + 𝑟] 𝑣𝑎𝑟[𝑑𝑡] + [( ) ( ) 𝜎] 𝑣𝑎𝑟[𝑑𝑤(𝑡)] +
𝑆(𝑡)
𝛿2
𝛿2 √6
+2𝑐𝑜𝑣 {[
𝛿1
𝛿1 𝜋
(𝜇 + 𝜀𝜎 − 𝑟) + 𝑟] ( ) ( ) 𝜎(𝑑𝑡)𝑑𝑤 (𝑡)}
𝛿2
𝛿2 √6
since
𝑣𝑎𝑟[𝑑𝑡] = 𝑑𝑣𝑎𝑟(𝑡) = 0 and for a wiener process (𝑑𝑡)𝑑𝑤(𝑡) = 0, 𝑣𝑎𝑟[𝑤(𝑡)] = 𝑡 thus
𝑣𝑎𝑟[𝑑𝑤(𝑡)] = 𝑑𝑣𝑎𝑟[𝑤(𝑡)] = 𝑑𝑡 hence
2
𝑑𝑆(𝑡)
𝛿1
𝜋
𝑣𝑎𝑟 [
] = [( ) ( ) 𝜎] 𝑑𝑡
𝑆(𝑡)
𝛿2 √6
so
(E2)
𝑣𝑎𝑟[
𝑑𝑆(𝑡)
]
𝑆(𝑡)
𝑑𝑡
𝛿
𝜋
2
𝛿
2 𝜋2
= [(𝛿1 ) ( ) 𝜎] = (𝛿1 ) ( 6 ) 𝜎 2
2
√6
2
References
AKDENIZ, L. and W.D. DECHERT (2007), “The Equity Premium in Brock’s Asset Pricing
Model,” Journal of Economic Dynamics and Control, vol. 31, pp. 2263-2292
BROCK, W. A. (1982), “Asset Prices in a Production Economy,” in The Economics of
Information and Uncertainty, ed. by J. J. McCall, pp.1-46. The University of Chicago Press,
Chicago.
CAMPBELL, J.Y. and COCHRANE, J.H. (1999), “By Force of Habit: A ConsumptionBased Explanation of Aggregate Stock Market Behavior”, The Journal of Political Economy,
Volume 107, Issue 2, 205-251.
CONSTANTINIDES, G. M. (1990), “Habit Formation: A resolution of the Equity Premium
Puzzle,” Journal of Political Economy 98 (3): 519-543.
DAVIS, M. H. A. and NORMAN, A. R. (1987), “Portfolio selection with Transaction
Costs.” Manuscript. London: Imperial Coll.
DUESENBERRY, J.S. (1949), “Income, Saving and the Theory of Consumer Behavior”,
Harvard University Press, Cambridge.
DUNN, K. B., and SINGLETON, K. J. (1986), “Modeling the Term Structure of Interest
Rates under Non-separable Utility and Durability of Goods.” J. Financial Econ. 17
(September 1986): 27-55.
EICHENBAUM, M. S., and HANSEN, L. P. (1990). “Estimating Models with Intertemporal
Substitution Using Aggregate Time Series Data.” J. Bus. and Econ. Statis.
EICHENBAUM, M. S.; HANSEN, L. P. and SINGLETON, K. J. (1988), “A Time Series
Analysis of Representatives Agent Models of Consumption and Leisure Choice under
Uncertainty.” Q.J.E. 103 (February 1988): 51-78.
FERSON, W. E. (1983), “Expectations of Real Interest Rates and Aggregate Consumption:
Empirical Tests.” J. Financial and Quantitative Analysis 18 (December 1983): 477-97.
FERSON, W. E. and CONSTANTINIDES, G. M. (1989), “Habit Persistence and Durability
in Aggregate Consumption: Empirical Tests.” Manuscript. Chicago: Univ. Chicago.
FRIEND, I. and BLUME, M.E. (1975), “The demand for risky assets”, American Economic
Review 65, 900-922.
GALLANT, A. R. and TAUCHEN, G. (1989), “Seminonparametric Estimation of
Conditionally Constrained Heterogeneous Processes: Asset Pricing Applications.”
Manuscript. Durham, N.C.: Duke Univ.
GROSSMAN, S. J.; MELINO, A.; and SHILLER, R. J. (1987), “Estimating the ContinuousTime Consumption-based Asset-Pricing Model.” J. Bus. And Econ. Statis. 5 (July 1987):
315-27.
HANSEN, L. P. and JAGANNATHAN, R. (1988), “Restrictions on Intertemporal Marginal
Rates of Substitution Implied by Asset Prices.” Manuscript. Chicago: Univ. Chicago, 1988.
HANSEN, L. P. and SINGLETON, K. J. (1982), “Generalized Instrumental Variables
Estimation of Nonlinear Rational Expectations Models.” Econometrica 50 (September 1982):
1269-86.
HANSEN, L. P. and SINGLETON, K. J. (1983), “Dtochastic Consumption, Risk Aversion,
and the Temporal Behavior of Asset Returns.” J. P. E. 91 (April 1983): 249-65.
HEATON, J. (1988), “The Interaction between Time-Nonseparable Preferences and Time
Aggregation.” Manuscript. Chicago: Univ. Chicago.
IBBOTSON, R. G. and SINQUEFIELD, R. A. (1982), “Stocks, Bonds, Bills and Inflation:
The Past and the Future. Charlottesville, Va.: Financial Analysts Res. Found.
KOCHERLAKOTA, N. R. (1996), “The Equity Premium: It's Still a Puzzle,” Journal of
Economic Literature 34 (1): 42-71.
LUCAS, R. E. (1978), “Asset Prices in an Exchange Economy,” Econometrica, 46, 14291445.
MEHRA, R., and E.C. PRESCOTT (1985), “The Equity Premium: A Puzzle”. Journal of
Monetary Economics 15: 145-161.
MCGRATTAN, E. R., and E.C. PRESCOTT (2001), “Taxes, Regulations, and asset Prices”.
Working paper 610, Federal Reserve Bank of Minneapolis.
RIETZ, T. A. (1988), “The Equity Risk Premium: A Solution.” J.Monetary Econ. 22(July
1988): 117-31.
SHILLER, R. J, (1981), “Do Stock Prices Move Too Much to be Justified by Subsequent
Changes in Dividends? “American Economic Review, 71 (3): 421-436.