Overview of F400 Asset Pricing
Shuyi Ge
Viewers’ discretion advised: this note is an incomplete account of how I think the
lecture notes are connected. It is unofficial, does not indicate what may or may not
appear in the exam and may contain errors.
1 Risk Attitudes
When we talk about pricing an asset, two concepts are fundamental: payoff and risk. In Chapter
1, we start by measuring the economic agents’ preference towards risks, or risk attitudes. One
way to do that is to think about the curvature of the utility function, u′′(w), which leads to the
concepts of absolute and relative risk aversion. If the agent has constant absolute risk aversion
(CARA) or constant relative risk aversion (CRRA), the utiltiy functions can be proved to have
specific forms, and it implies nice properties when we consider portfolio choices.
2 Equilibrium Pricing
investors’ portfolio choice −→ aggregation to get the market −→ Security Market Line (SML)
2.1 Classic CAPM
In Chapter 2, there is a classic version of CAPM that assumes normality and CARA utility
function. We can find the demand for an asset by calculating and maximizing the certainty
equivalence, which gives the demand.
g =
1
ρ
V −1(m− rf1)
Then the Security Market Line(SML) follows naturally which relates excess returns to the betas
and market excess return.
1
2.2 Stock Market Economy and CCAPM
A richer economic model where there are states, firms that pay dividends, investors with two period
utility function. We solve an explicit optimization problem of the agent and arrive at the FOC:
qj =
S∑
s=1
piis
δiv
′
i(xis)
v′i(xi0)
yjs = E[k˜iy˜j]
which gives a formula that quantifies risk and return of an asset, that is the CCAPM1 equation
Ei(r˜j)− rif = δi(1 + rif )covi[−
v′i(x˜i)
v′i(xi0)
, r˜j]
CCAPM1 leads to classical SML under less restrictive assumptions than CAPM
2.3 Dynamic Equilibrium Pricing
History is important in this type of question.
3 Arbitrage Pricing
Arbitrage pricing theory tries to price an asset using other traded assets and the assumptions are
less restrictive.
3.1 Static Arbitrage Pricing
Linear pricing formula for assets in the asset span. Upper and lower bounds for non-traded assets.
3.2 Dynamic Arbitrage Pricing
Simple example is the binomial model, which is the discrete version of Black-Scholes.
3.3 Derivative Pricing in Continuous Time
Derivative is an asset whose value derives from and is dependent on the value of an underlying
asset. To price a derivative, we do the followings:
• model the prices of underlying assets (this is why we introduce all sorts of Wiener Processes
at the first place. Geometric W.P is particularly important, prices of risky assets like stocks
usually grow exponentially thus are modelled using geometric W.P)
2
• we are interested in the value of a derivative at t, which could be written as f(t, St). To
handle df , we need Ito’s lemma.
• We use no arbitrage condition to price an asset using other traded assets (this involves
replication. In option pricing, we replicate the option using the underlying and the safe
asset. In bond pricing, we replicate the safe asset using two bonds with different maturity
times). At the end of this step, we find a p.d.e that the price of the derivative must satisfy.
• From p.d.e, we apply RN transformation to the underlying so that we are ready to apply
the Feynman-Kac formula. (Feynman-Kac is a very powerful tool, it tells you a special type
of p.d.e would have an unique and interpretable solution!) After this stage, the price of the
derivative will be a expected present discounted value. Then we just need to evaluate the
expectation. (This step can get very nasty, revise probability and statistics! In particular
you should be knowing key properties of normal, log-normal distributions.)
Remark: In some cases, directly evaluating this expectation is beyond our ability so we ’guess
and check’ the solution (standing on the shoulders of mathematicians). See term structure
model as a example.
3
4 Questions Preview
Q1. What is the difference between market, market portfolio, and market return?
Q2. How to find the certainty equivalent (CE) in CAPM?
Q3. How to interpret risk neutral(RN) pricing?
Q4. What’s the implication of single agent economy (representative agent economy)?
Q5. When we derive Ito’s lemma heuristically, how to determine the order of Taylor expansion?How
to get the order of the remainder term?
Q6. In term-structure model, why we introduce another bond, the S-bond?
Q7. How to solve the term structure equation using guess and check? Handout seems to miss
many technical details.
5 Questions and Answers
Q1. In CAPM, what is the difference between market, market portfolio, and market return?
After deriving CE, an agent’s portfolio problem is transformed in to maximizing CE. From the
FOC, we can get an agent with CARA coefficient ρi and wealth wi0 has optimal asset allocation
vector gi (J × 1 weight vector on risky assets):
gi =
1
ρi
V −1(m− rf1)
In equilibrium, the total demand and total supply of risky assets are equal, which gives the market
asset
M =
∑
i
wi0gi =
∑
i
wi0
ρi
V −1(m− rf1)
where Mj is the market value of the j
th asset. The money value of this market asset is MT1 =
M1 + · · ·+MJ =
∑
iwi0. Market return r˜M on this composite asset is given by
1 + r˜M =
∑
i w˜i1∑
iwi0
If we transform this portfolio in terms of proportions of holdings, we will have the market portfolio
Mp as the allocation vector
Mp = M/(
∑
i
wi0) = (
∑
iwi0/ρi∑
iwi0
)V −1(m− rf1)
4
Market portfolio is the value weighted index of all tradable assets in a given time period.
Q3. How to interpret risk neutral (RN) pricing?
Asset pricing in a risk neutral world is easy. If agents are RN, we can price assets using standard
expected present value formula
pt =
1
1 + rf
E(xt+1) (1)
If agents are risk averse, price is lower than above to compensate for risk-bearing.
There are two ways to incorporate risk in the Equation 1, (1)adjust discount factor (discount at a
rate higher than risk free rate) (2) transform probabilities to RN probabilities (RN pricing).
RN probabilities is a set of distorted probabilities which reflect agents’ risk aversions. As a toy
example, think of a bet, where you win 10 with 50% (up) and 0 with 50% (down). There is no
discounting. As a risk-averse agent, you are willing to pay less than 5 (the expected value of the
bet) to take the bet. Say the market price for taking this bet is 3, then RN probabilities for up
and down states are (0.3,0.7). p = ERN(x).
To see how the two ways of incorporate risk are equivalent to each other, take lecture note 2,
section 4.6 RN pricing as an example. Recall Asset Pricing equation
qj = p
iyj
The set of RN probabilities for state 1, ..., S are
pˆi =
1
qif
pi
The price for asset j can be derived using two ways
qj = p
iyj = q
i
f pˆ
iyj
blue part corresponds to adjusting discount factor (for more information, see John Cochrane’s
book Asset Pricing, section 1.2 Marginal rate of substitution and stochastic discount factor)
qj = p
iyj =
∑
s
δiv
′(xis)
v′(xi0)
piisyjs = E(my) (2)
red part corresponds to RN pricing
qj = q
i
f pˆ
iyj =
∑
s
1
1 + rif
δiv
′(xis)
v′(xi0) ∗ qif
piisyjs =
1
1 + rif
[
∑
s
δiv
′(xis)
v′(xi0) ∗ qif
piisyjs] =
1
1 + rif
EiRN [y˜j]
5
Mechanisms of RN adjustment in continuous time models are beyond the scope of this course. But
the same intuition applies.
Q4. What’s the implication of single agent economy (representative agent economy)? In
PS2Q4, why the agent consumes all endowment (e0) in date 0 and all dividends (ys) in state s date
1?
In a single-agent economy, there is no trading. The agent owns 100% of the company in the
economy. In a representative agent economy, we again have this ’No trading equilibrium’. If there
is an asset, and there is a price that one agent wants to sell it. Since everyone is the same, everyone
wants to sell it, so there is no demand. Similarly, if there is a price where one wants to buy it
there is no supply. Financial market is in equilibrium, if and only if, supply equals demand and
both are simultaneously zero.
In a general heterogeneous agent model, like the stock market economy model, in equilibrium
there are tradings of assets. For agent i, her endowment wij of shares in firm j not necessarily
equals to θij, which is her optimal holdings of shares in j. Trading happens because agents have
different endowments and different risk attitudes.
Example of single agent economy. Exam 2016 Q1
6
In equilibrium, the consumption of the agent (c0, c1, c2) will be the same as the endowments
(e0, e1, e2). Using that, you can derive the stochastic discount factor (SDF). Recall Equation 2
qj = p
iyj =
∑
s
δiv
′(xis)
v′(xi0)
piisyjs = E(my)
m =
δiv
′(xis)
v′(xi0)
where xi0 = e0, xi1 = e1, and xi2 = e2. Now you can see how SDF generalizes standard discount
factor ideas by incorporating risk and state-dependency. You can use that to price a cash flow
stream like (d) where y = (1, 3, 5).
Q5. When we derive Ito’s lemma heuristically, how to determine the order of Taylor expansion?
How to get the order of the remainder term? To be precise, in handout 6 page 8, why we keep the
second order expansion terms here? and why the reminder is O(∆T 3/2)?
∆Y =
∂f
∂t
∆T +
∂f
∂x
∆X +
1
2
∂2f
∂x2
∆X2 +O(∆T 3/2)
This is an asymptotic analysis. Asymptotics is a method of describing limiting behavior. In this
exercise, we are approximating continuous time processes using discrete time ones by pushing
∆T → 0. A term is O(∆T p) means that it goes to zero with the same speed as ∆T p. As ∆T → 0,
O(∆T p) terms with p > 1 can be ignored because they goes to zero faster than ∆T . Similarly, we
can not ignore terms with p <= 1.
∆X = O(
√
∆T ), so that ∆X2 = O(∆T )
We apply Taylor expansion to Y = f(T,X),
∆Y = (
∂f
∂t
∆T+
∂f
∂x
∆X)+
1
2
(
∂2f
∂t2
∆T 2+2
∂2f
∂x∂t
∆T∆X+
∂2f
∂x2
∆X2)+
1
6
(
∂3f
∂t3
∆T 3+...
∂3f
∂X3
∆X3)+...
we only keep O(∆T p) terms with p <= 1. We have three terms that go to zero not faster than
∆T , including ∆T,∆X,∆X2. Others go to zero faster than ∆T . To see what is the order of the
remainder term,
∂f
∂t
∆T +
∂f
∂x
∆X +
1
2
∂2f
∂x2
∆X2 + [
1
2
(
∂2f
∂t2
∆T 2 + 2
∂2f
∂x∂t
∆T∆X +
1
6
(
∂3f
∂t3
∆T 3 + ...
∂3f
∂X3
∆X3) + ...]
Asymptotically, the remainder term goes to zero with the same speed as ∆T∆X = O(∆T 3/2)
7
Q6. In term-structure model, why we introduce another bond, the S-bond?
We consider the pricing of zero-coupon bonds as a derivative written upon the interest rate.
There is only one traded asset B which is the (locally) risk free asset.
dr(t) = µ(t, r(t))dt+ σ(t, r(t))dZ(t)
dB(t) = r(t)B(t)dt
Let pT (t, r) be the price of the zero-coupon bond with maturity date T . We aim to determine pT
with arbitrage pricing. The dynamics of pT will be:
dpT =
∂pT
∂t
dt+
∂pT
∂r
dr +
1
2
∂2pT
∂r2
σ2dt
=
[
∂pT
∂t
+
∂pT
∂r
µ+
1
2
∂2pT
∂r2
σ2
]
dt+
∂pT
∂r
σ dZ
Why do we introduce another asset pS?
Observe that we can’t replicate the zero-coupon bond with only the risk free asset: because
interest rate r is not a traded asset and there is no dZ term in dB, we can’t perfectly match the
dynamics of pT with B. There is more “randomness” in the price of the derivative than in the
only traded asset B .
Compared with Black-Scholes model, where we have two traded assets S and B and deter-
ministic r:
dS = µ dt+ σ dZ
dB = rB dt
and we can form a portfolio consisting of S and B that perfectly match the dynamics of the deriva-
tive f(t, St) written on St. The “randomness” in f(t, S(t)) comes from the “randomness” in S, so
we can find a way to perfectly match the dynamics.
df =
[
∂f
∂t
+ µS
∂f
∂S
+
1
2
σ2S2
∂2f
∂S2
]
dt+ σS
∂f
∂S
dZ
In the bond pricing case, if we introduce another pS, there will be dZ terms in both pT and pS,
enabling us to form a portfolio of pS and pT in which the dZ terms cancelled.
8
This is 2018 Q4 (it doesn’t explicitly tell you to introduce a new asset), and the idea also relates
to Exam 2018 Q3
Remark: the derivative has two underlying assets, so it is subject to two sources of risks (Z1
and Z2). To replicate the derivative, we need the two risky assets and the safe asset.
Sketch solution: recall the four steps of derivative pricing given in section 3.3. The first step
(model the prices of underlying assets) is given, we start from step 2.
Step2: we denote the price of the derivative at t as f(t, S1t, S2t). Using Ito’s lemma to get df .
df =
∂f
∂t
dt+
∂f
∂S1
dS1 +
∂f
∂S2
dS2 +
1
2
(
∂2f
∂S21
dS21 +
∂2f
∂S22
dS22 + 2
∂2f
∂S1∂S2
dS1dS2)
= (
∂f
∂t
+ µ1S1
∂f
∂S1
+ µ2S2
∂f
∂S2
+
1
2
σ21S
2
1
∂2f
∂S21
+
1
2
σ22S
2
2
∂2f
∂S22
+ σ1σ2S1S2
∂2f
∂S1∂S2
)dt
+ σ1S1
∂f
∂S1
dZ1 + σ2S2
∂f
∂S2
dZ2
Step3: use no arbitrage condition to price an asset using other traded assets (this step involves
replication). Consider a portfolio Π consists of α units of B, β1 units of S1 and β2 units of S2.
dΠ = (αrB + β1µ1S1 + β2µ2S2)dt+ β1σ1S1dZ1 + β2σ2S2dZ2
The ex-post value of the portfolio and the derivative is equal (i.e., we replicate the derivative using
the portfolio) iff
Π + dΠ = f + df
9
Matching terms:
β1σ1S1 = σ1S1
∂f
∂S1
β2σ2S2 = σ2S2
∂f
∂S2
αrB + β1µ1S1 + β2µ2S2 =
∂f
∂t
+ µ1S1
∂f
∂S1
+ µ2S2
∂f
∂S2
+
1
2
σ21S
2
1
∂2f
∂S21
+
1
2
σ22S
2
2
∂2f
∂S22
+ σ1σ2S1S2
∂2f
∂S1∂S2
αrB + β1µ1S1 + β2µ2S2 = f
get the p.d.e of the derivative
0 =
∂f
∂t
− rf + (rS1 ∂f
∂S1
+ rS2
∂f
∂S2
) +
1
2
(σ21S
2
1
∂2f
∂S21
+ σ22S
2
2
∂2f
∂S22
+ 2σ1σ2S1S2
∂2f
∂S1∂S2
)
with boundary solution
f(T, S1(T ), S2(T )) = Φ(S1(T ), S2(T )) = K + S1(T )− S2(T )
Under RN dynamics
dSˆ1 = rSˆ1dt+ σ1Sˆ1dZˆ1
dSˆ2 = rSˆ2dt+ σ2Sˆ2dZˆ2
Apply Feynman-Kac formula,
f(t, S1(t), S2(t)) = Et(e
−r(T−t)Φ(S1(t), S2(t))) = Et[e−r(T−t)(K + S1(T )− S2(T ))]
price at time 0, plug in t = 0.
Q7. How to solve the term structure equation using guess and check? Handout seems to miss
many technical details. Example, 2015Q4
10
Given that we know the affine term structure:
pT (t, r) = exp
{
AT (t)−BT (t)r(t)}
and the term structure equation:
0 =
∂p
∂t
− rp+ µˆ∂p
∂r
+
1
2
σ2
∂2p
∂r2
and we assume the Vasicek model:
drˆ = (b− arˆ)dt+ σdZˆ
Solve for AT (t) and BT (t).
First, because for all r, pT (T, r) = 1, we get boundary condition AT (T ) = BT (T ) = 0.
Second, using the drift coefficient and diffusion coefficient that Vasicek specified, we can find
the p.d.e that AT , BT must follow.
Using Ito’s Lemma:
dpT (t, r) =
∂pT
∂t
dt+
∂pT
∂r
dr +
1
2
σ2
∂2pT
∂r2
dt
get partial derivatives from the affine term structure equation:
∂pT
∂t
= pT (t, r)
[
dAT
dt
− dB
T
dt
r(t)
]
∂pT
∂r
= pT (t, r)
(−BT )
∂2pT
∂r2
= pT (t, r)(BT )2
Plug into the term structure equation:
0 = p
[
dA
dt
− dB
dt
r
]
− rp+ (b− ar)p(−B) + 1
2
σ2pB2
I omit the T and t but the dependence should be clear. Divide both sides by p, and collect all the
coefficients of r:
0 = −
[
dB
dt
+ 1− aB
]
r +
(
dA
dt
− bB + 1
2
σ2B2
)
(3)
11
Equation 11 holds for all t and r, we have two (ordinary) differential equations:
dB
dt
+ 1− aB = 0 (4)
dA
dt
− bB + 1
2
σ2B2 = 0 (5)
Solution to Equation 4: Notice if that B′ = aB − 1 implies B is an exponential function. Let’s
define a very general exponential function B = b+ c ∗ edt and solve the coefficients b, c, d.
B′ = c ∗ d ∗ edt
From Equation 4
B′ − aB + 1 = (c ∗ d− a ∗ c)edt + 1− a ∗ b = 0
which implies
c ∗ d− a ∗ c = 0 =⇒ d = a (6)
1− a ∗ b = 0 =⇒ b = 1/a (7)
we determine c by the boundary condition B(T ) = 0:
1
a
+ ceaT = 0 =⇒ c = −1
a
e−aT
As a result, if we define H = e−aT eat:
B =
1
a
− 1
a
e−aT eat =
1
a
(1−H) (8)
Solution to Equation 5:
dA = (bB − 1
2
σ2B2)dt
A =
∫
(bB − 1
2
σ2B2) dt
Plug in B and
A =
∫ [
b
a
[1−H]− 1
2
σ2
1
a2
(1 +H2 − 2H)
]
dt[
b
a
− σ
2
2a2
]
t−
[
b
a
− σ
2
a2
] ∫
H dt− σ
2
2a2
∫
H2 dt
12
We only need to deal with
∫
H dt and
∫
H2 dt:∫
H dt =
∫
e−aT eat dt
=
1
a
e−aT eat + C1
=
1
a
H + C1∫
H2 dt =
∫
e−2aT e2at dt
=
1
2a
e−2aT e2at + C2
=
1
2a
H2 + C2
Hence:
A =
[
b
a
− σ
2
2a2
]
t−
[
b
a
− σ
2
a2
] ∫
H dt− σ
2
2a2
∫
H2 dt
=
[
b
a
− σ
2
2a2
]
t−
[
b
a2
− σ
2
a3
]
H − σ
2
4a3
H2 + C3 (9)
where C1, C2, C3 are arbitrary constants. We can determine C3 by the boundary condition that
AT = 0, notice that HT = 1:
C3 = −
{[
b
a
− σ
2
2a2
]
T −
[
b
a2
− σ
2
a3
]
− σ
2
4a3
}
= −
{[
b
a
− σ
2
2a2
]
T − 4ba− 3σ
2
4a3
}
Put C3 back to Equation 9, we have the result:
A =
4ba− 3σ2
4a3
−
[
b
a
− σ
2
2a2
]
(T − t)− H
4a3
(
4ba− 4σ2 + σ2H) (10)
Remark: nothing too technical, just calculus. Below are some commonly used tricks in ordinary
calculus.
• Integral as antiderivative ∫
f(x)dx = F (x) + C
• Integral of Exponential and Logarithmic Functions
– Integral of the exponential function
∫
exdx = ex + C
13
– Integral of the exponential function with base a
∫
axdx = a
x
lna
+ C, a > 0
–
∫
eaxdx = e
ax
a
+ C, a 6= 0
– Integral of the natural logarithm
∫
lnxdx = xlnx− x+ C
• Integral of Power functions ∫
xndx =
xn+1
n+ 1
+ C
• Integral of Reciprocal ∫
1
x
dx = ln | x | +C
• Integration by Parts ∫
udv = uv −
∫
vdu
• Integration by Substitution (reverse chain rule)∫
f(g(x))g′(x)dx
Notice that we have g(x) and its derivative. Let’s reverse the chain rule by letting g(x) = u
and write the integral as ∫
f(u)u′dx =
∫
f(u)du
Then we can integrate f(u), and finish by putting g(x) back as u.
Example: ∫
cos(x2)2xdx
where g(x) = x2 and g′(x) = 2x. We let g(x) = u = x2. The above integral is equivalent to∫
cos(u)du = sin(u)
finally putting u = x2 back, we have
∫
cos(x2)2xdx = sin(x2)
Q2. How to find the certainty equivalent (CE) in CAPM? (completing the square in the exponent)?
We are interested inE
[
u
(
(1 + rf )w0 +
∑J
j=1(r˜j − rf )gj
)]
, where r˜ ∼ N(m,V ), a J−dimensional
multivariate normal distribution, i.e., its probability density is
f(x) = (2pi)−
J
2 det(V )−
1
2 exp
{
−1
2
(x−m)TV −1(x−m)
}
= C exp
{
−1
2
(x−m)TV −1(x−m)
}
14
The utility function is of CARA, so from Lecture Notes 1 we can write it as
u(w) =
1− e−ρw
ρ
Then we handle the expectation by the following calculations:
J∑
j=1
(r˜j − rf )gj = (r˜ − rf1)Tg (11)
where xT is the transpose of x.
The (stochastic) utility is
u
(
(1 + rf )w0 + (r˜ − rf1)Tg
)
= 1−
=
1
ρ
− 1
ρ
exp{[−ρ(1 + rf )w0]} exp [ρ(rf1)Tg] exp [−ρr˜Tg]
= A−B
So the expectaion is:
E
[
u
(
(1 + rf )w0 +
J∑
j=1
(r˜j − rf )gj
)]
= E[A−B = A−BE[ = A−B
∫
= A−B
∫
C = A−B
∫
C
Now we handle D to make it look like a normal density:
D = −ρxTg − 1
2
(x−m)TV −1(x−m)
= −1
2
[(x−m)TV −1(x−m) + 2ρxTV −1V g]
= −1
2
[(x−m+ ρV g)TV −1(x−m+ ρV g)−ρ2gTV g + 2ρmTg]
Plug in:
E[u] = A−B
∫
C exp
{
−1
2
[(x−m+ ρV g)TV −1(x−m+ ρV g)−ρ2gTV g + 2ρmTg]
}
dx
= A−B exp
[
1
2
ρ2gTV g − ρmTg
]∫
C exp
[
−1
2
(x−m+ ρV g)TV −1(x−m+ ρV g)
]
dx
= A−B exp
[
1
2
ρ2gTV g − ρmTg
]
× 1
Because it’s the integral of normal density and total probability is 1.
15
E(u) = A−B exp
[
1
2
ρ2gTV g − ρmTg
]
=
1
ρ
− 1
ρ
exp[−ρ(1 + rf )w0] exp
[
ρ(rf1)
Tg
]
exp
[
1
2
ρ2gTV g − ρmTg
]
=
1
ρ
− 1
ρ
exp{−ρ[(1 + rf )w0−(rf1)Tg+mTg−1
2
ρgTV g]}
from our utility function, we can see CE is the red part
u(w) =
1
ρ
− 1
ρ
exp{−ρw}
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