L12: SDF1 Lecture 12: Stochastic Discount Factor and GMM Estimation The following topics will be...
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Transcript of L12: SDF1 Lecture 12: Stochastic Discount Factor and GMM Estimation The following topics will be...
L12: SDF 1
Lecture 12: Stochastic Discount Factor and GMM Estimation
• The following topics will be covered:• SDF
– Basic expression
– Risk free rate
– Risk correction
– Mean-variance frontier
– Time-varying expected returns
• GMM– GMM overview
– Applying GMM
• Also, more on “Hypothesis Testing”
L12: SDF 2
Stochastic Discount Factor Presentation
])('
)('[ 1
1
tt
ttt x
cu
cuEp
xt+1 is the payoff in t+1. β captures impatience and is called the subjective discount factor. U is utility function, ct denotes consumption in date t. To see this:
)]([)(max 1 ttt cuEcu
, s.t.
ttt pec
111 ttt xec
The first order condition is ])('[)(' 11 ttttt xcuEcup , equivalent to (1).
Why? In CLM, we have )]1)(('[)(' 11 tttt rcuEcu , known as the Euler equation.
L12: SDF 3
Stochastic Discount FactorIt states: the loss in utility if the investor buys another unit of the asset equates the increase in utility he obtains from the extra payoff at t+1. Stochastic Discount Factor Presentation:
][ 11 tttt xmEp )('
)(' 11
t
tt cu
cum
or even more simply:
p=E(mx) The variable mt+1 (m) is known as the stochastic discount factor, or pricing kernel. It is also known as the intertemporal marginal rate of substitution. m is always positive.
L12: SDF 4
Examples
)(1 mRE
)(( 11 ttt dpmEp
bt
at
et
et RRRwheremRE 1111)(0
)(1 1f
tmRE
Asset price
Stock return
Excess stock return
Risk free rate
See page 9 – 10 of Cochrane.
L12: SDF 5
Relating to EGS• Recall the consumption and saving in Chapter 6, EGS:
– β is a discount factor for delaying consumption
• The first order condition of this problem is similar to the SDF presentation
• Also on page 176, EGS, we have
• Z is the aggregate wealth in state z, q(z) denotes the expected payoff of the firm conditional on z. In equilibrium, the marginal utility of the representative agent in a given state equals the equilibrium state price itself.
)]([)(max 1 ttt cuEcu
Subject to wealth constraint on page 91
)(')()()()( zvzEqzzEqqP
L12: SDF 6
Risk-free rate
)(1 1f
tmRE For power utility
1)(
1ccu , we have ccu )('
)(
1 1
t
tf
c
cR or
)(
11 1
t
tf
c
cr .
Real interest rates are high when people are impatient, when β is low; they are high when consumption
growth is high.
Real interest rates are more sensitive to consumption growth if the power parameter γ is high.
With lognormal consumption and power utility function, we have
1)ln()2/()ln( ][ 122
1 tttt ccEft eeR
L12: SDF 7
Risk Corrections
),cov()(
),cov()()(
][
xmR
xEp
xmxEmEp
mxEp
f
The first term is the present value of E(x) (expected payoff). The second is a risk adjustment. An asset whose payoff co-varies positively with the discount factor has its price raised and vice versa.
The key u’(c) is inversely related to c. If you buy an asset whose payoff covaries negatively with consumption (hence u’(c)), it helps to smooth consumption and so is more valuable than its expected payoff indicates.
L12: SDF 8
Risk Corrections – Return Expression
)]('[
)),('cov()(
),cov()(
),cov()()(1
][1
1
11
t
ittfi
iffi
ii
i
cuE
RcuRRE
RmRRRE
RmREmE
mRE
All assets have an expected return equal to the risk-free rate, plus a risk adjustment.
Assets whose returns covary positively with consumption make consumption more volatile, and so must promise higher expected returns to induce investor to hold them, and vice versa.
L12: SDF 9
Expected Return-Beta Representation
mmifi
ifi
RRE
mE
m
m
mRRRE
,)(
))(
)var()(
)var(
),cov(()(
Where βis the regression coefficient of the asset return on m.
It says each expcted return should be proportional to the regression coefficient in a regression of that return on the discount factor m.
λis interpreted as the price of risk and β is the quantity of risk in each asset.
L12: SDF 10
Mean-Variance Frontier
)()(
)(|)(| have we1exceedcannοann As
)()(
)()(
)()()()()(1
),cov()()()(1
,
,
ifi
i
Rm
fi
i
Rm
ii
iii
RmE
mRRE
RmE
mRRE
mRREmEmRE
mRREmEmRE
i
i
Implications:
(1) Means and variances of asset returns lie within efficient frontier.
(2) On the efficient frontier, returns are perfectly correlated with the discount factor.
(3) The priced return is perfectly correlated with the discount factor and hence perfectly correlated with any frontier return. The residual generates no expected return.
L12: SDF 11
Sharpe Ratio and Equity Premium Puzzle
fi
fi
RmmE
m
R
RRE)(
)(
)(
)(
|)(|
Let Rmv denote the return of a portfolio on the mean-variance efficient frontier and consider power utility. The slope of the frontier (Sharpe ratio) is
)ln(])/[(
])/[(
)(
)(
)(
|)(|
1
1 cccE
cc
mE
m
R
RRE
tt
ttmv
fmv
Sharpe ratio is higher if consumption is more volatile or if investors are more risk averse.
Over the last 50 years, average real stock return is 9% with a standard deviation of 16%. The real risk free rate is 1%. This suggests a real Sharpe ratio of _____
Aggregate nondurable and services consumption growth has a standard deviation of 1%. So
L12: SDF 12
Time-varying Expected Returns
),()()()(
),()()(
)(
)(
),(cov)(
1111
1111
111
ttttttttf
ti
t
ttttttt
tt
t
tttft
it
RmRcRRE
RmRmE
m
mE
RmRRE
The relation above is conditional. Conditional mean or other moment of a random variable could be different from its unconditional moment. E.g,, knowing tonight’s weather forecast, you can better predict rain tomorrow than just knowing the average rain for that date.
It suggests a link between conditional mean of stock returns and conditional variance of stock returns.
Little empirical support.
L12: SDF 13
Estimating SDF -- GMM
ttt
ttt
ttt
pxbmb
conditionmomentpxbmEeI
xbmEpE
111t
11
11
)()(u
vectors. typicallyare p andx .0])([.,.
])([)(
.parameters for the solve tois do try to weAll utility.power say form, specific a
it takes Assume factor.discount stochastic theestimate tomove weNow
Define gT(b) as the sample mean of the ut error, when the parameter vector is b in a sample of size T:
])([)(1
)( 111
tttT
T
ttT pxbmEbu
Tbg
The first-stage estimate of b minimizes a quadratic form of the sample mean of the error.
)()'(minarg }{
^
1 bWgbgb TTb
For some arbitrary matrix W (often, W=I).
L12: SDF 14
Estimating SDF – Second StageDefine an estimate S of
)(#~])[var('
)1,0(~
)var(
|)(
)'(1
)var(
)()'(minarg
]')()([
21^^
^
^
11^
2
1^
2
^
binincludedbbb
N
b
b
b
bgdwheredSd
Tb
bgSbgb
bubuES
jjjj
ii
i
bb
T
TTb
jjtt
JT Test: )#(#~)](')([ 2^
1^
parametersmomentsbgSbgTTJ TTT
L12: SDF 15
Implementing GMM
Moment condition 01)( 11 tt RbmE (1)
Involving instrumental variable z, we have 1)()( 11 ttt RbmEzE
tttt zRbmEzE 11 )()( , i.e., ttt zRbmE )1)((0 11 (2)
We have two returns R=[Ra, Rb] and one instrument z. Combining (1) and (2), we have:
0
0
0
0
1
1
)(
)(
)(
)(
11
11
11
11
t
t
tbtt
tatt
btt
att
z
z
zRbm
zRbm
Rbm
Rbm
E (we have 2 securities and z has (1, zt)’, thus having 2*2 equations)
Using the Kronecker product, we denote the above as: 01)( 11 ttt zRbmE
or 01 tt zuE
L12: SDF 16
GMM Example
Adapt from Moore (1998, USC lecture note)
Starting from 0111 ttt RmE
Where tj
tjtjtj P
DPR
,
1,1,1,
. Assume a power utility function, we have
t
tt c
cm 1
1
1)/( 1,11,
tjtttj Rccu
In other words, 0)( 1, tjt uE
L12: SDF 17
GMM Example (2)N securities, 3 instruments
NTNN
T
T
uuu
uuu
uuu
u
....
.
...
...
.
.
.
....
...
21
22221
11211
2,1,
12
1
1
...
...
1
1
1
TmTm
mm
m
RR
RR
R
Z
N=1
]/,/,/[
1
...
...
1
1
1
],...,)[/1(/
2,1,
2,1,
12
1
11
TRuTRuTu
RR
RR
R
uuuTTuG
tmttmtt
TmTm
mm
m
TZ
L12: SDF 18
GMM Example (3)
If N=2, we have
2,2
1,2
2
2,1
1,1
1
/1
tmt
tmt
t
tmt
tmt
t
Ru
Ru
u
Ru
Ru
u
Tg to solve this problem, we do Wgg 'min),(
Hensen(1982) proves that T times the minimized value of g’Wg is asymptotically
distributed as 2 random variable with degrees of freedom 4 for the case of N=2 with 3 instrumental variables.
L12: SDF 19
Program GMM using SAS /* N=5, 7 instruments */proc model data=gmm;
parms beta 1.0 gamma 1.0;endogenous cons0 cons1;exogenous r1 r2 r3 r4 r5;instruments lrm1 lrm2 lrm3 lrm4 lrm5 lrm6;eq.m1=1-(1+r0)*(beta*(cons0/cons1)**(-gamma);eq.m1=1-(1+r0)*(beta*(cons0/cons1)**(-gamma);eq.m1=1-(1+r0)*(beta*(cons0/cons1)**(-gamma);eq.m1=1-(1+r0)*(beta*(cons0/cons1)**(-gamma);eq.m1=1-(1+r0)*(beta*(cons0/cons1)**(-gamma);fit m1-m6/gmm kernel=(parzen, 1,0);ods output EstSummaryStats=parms;
run;
L12: SDF 20
More on Hypothesis Testing
• Testing J linear Restrictions
• We can base a test of H0 on the Wald criterion:
• The chi-squared statistic is not usable when σ2 is unknown. As an alternative, we have the following F statistic
R'X)R(X'R'bRqRbm
qRbm
qRβ
1
2
0
]}[{][)(
bestimator squaresleast Given the
:
VarVarVar
let
H
mmm' 12 ]}[{][ VarJW
J
sF
)(][)'( 12 qRbR'X)R(X'qRb 1
L12: SDF 21
Examples of J Restrictions
.:0 qRH
Each row of R is a single linear restriction on the coefficient vector.
One of the coefficients is zero, :0j
00100 R and q=0.
Two of the coefficients are zero, :jk
01100 R and q=0.
A set of the coefficients sum to one, :1432
01110R and q=1.
A subset of the coefficients are all zero, ,01 ,02 and :03
00100
00010
00001
R and
0
0
0
q
or, equivalently, 00: .
L12: SDF 22
More Examples
Several constraints hold simultaneously: ,132 ,064 and ,065
0
0
1
110000
101000
000110
6
5
4
3
2
1
L12: SDF 23
Example: Test of Structural Change
Denote the first 14 years of the data in y and X as 1y and 1X and the remaining years
as 2y and 2X . An unrestricted regression that allows the coefficients to be different in
two periods is
2
1
2
1
2
1
2
1
0
0
X
X
y
y. We have 2211 eeeeee .
Under a restricted model, the two coefficient vectors are the same, then (7-19) may be
written
2
1
2
1
2
1
X
X
y
y. The residual sum of squares from this restricted
regression, ee then forms the basis for the test. The test statistic is then given in (7-14),
where J, the number of restrictions, is the number of columns in 2X and the denominator
degrees of freedom is .221 knn
is qR , where :R and q=0. See page 289, Greene (2000)
L12: SDF 24
Test Based on Loss of Fit
• Least squares vector b is chosen to maximize R2.
• The overall fitness of a regression:
• To see if the coefficient of a particular variable is a given value, we can also apply the F-stat, where F[1,n-K]=t2[n-K]
• To see if the constraints on a set of variables hold, we use
• or
)/()1(
)1/(),1(
2
2
KnR
KRKnKF
)/()1(
/)(),1(
2
2*
2
KnR
JRRKnKF
)/()1(
/)(),1(
2 KnR
JKnKF
ee'ee *
'*
L12: SDF 25
Exercises
• CR 1.7
• Read through Chapter 11, CR