# 35. Elementary Asset Pricing Theory¶

## 35.1. Overview¶

This lecture summarizes the heart of applied asset-pricing theory.

From a single equation, we’ll derive

• a mean-variance frontier

• a single-factor model of excess returns on each member of a collection of assets

To do this, we use two ideas:

• an asset pricing equation

• a Cauchy-Schwartz inequality

In this way, we shall derive the basic capital asset pricing model, the celebrated CAPM.

We’ll describe the basic ways that practitioners have implemented the model using

• cross sections of returns on many assets

• time series of returns on various assets

For background and basic concepts, see our lecture orthogonal projections and their applications.

As a sequel to the material here, please see our lecture two modifications of mean-variance portfolio theory.

## 35.2. Key Equation¶

We begin with a key asset pricing equation:

(35.1)$E m R^i = 1$

for $$i=1, \ldots, I$$ and where

\begin{split} \begin{aligned} m &=\text { stochastic discount factor } \\ R^{i} &= \text {random gross return on asset } i \\ E &\sim \text { mathematical expectation } \end{aligned} \end{split}

The random gross returns $$R^i$$ and the scalar stochastic discount factor $$m$$ live in a common probability space.

[HR87] and [HJ91] explain how existence of a scalar stochastic discount factor that verifies equation (35.1) is implied by a law of one price that requires that all portfolios of assets that end up having the same payouts must have the same price.

They also explain how the absence of an arbitrage opportunity implies that the stochastic discount factor $$m \geq 0$$.

To say something about the uniqueness of a stochastic discount factor would require that we impose more theoretical structure than we do in this lecture.

In complete markets models like those illustrated in this lecture equilibrium capital structures with incomplete markets, the stochastic discount factor is unique.

In incomplete markets models like those illustrated in this lecture the Aiyagari model, the stochastic discount factor is not unique.

## 35.3. Implications of Key Equation¶

We combine key equation (35.1) with a remark of Lars Peter Hansen that “asset pricing theory is all about covariances”.

Note

Lars Hansen’s remark is a concise summary of ideas in [HR87] and [HJ91]. For other important foundations of these ideas, see [Ros76], [Ros78], [HK79], [Kre81], and [CR83].

This remark of Lars Hansen refers to the fact that interesting restrictions can be deduced by recognizing that $$E m R^i$$ is a component of the covariance between $$m$$ and $$R^i$$ and then using that fact to rearrange key equation (35.1).

Let’s do this step by step.

First note that the definition of a covariance $$\operatorname{cov}\left(m, R^{i}\right) = E (m - E m)(R^i - E R^i)$$ implies that

$E m R^i = E m E R^{i}+\operatorname{cov}\left(m, R^{i}\right)$

Substituting this result into key equation (35.1) gives

(35.2)$1 = E m E R^{i}+\operatorname{cov}\left(m, R^{i}\right)$

Next note that for a risk-free asset with non-random gross return $$R^f$$, equation (35.1) becomes

$1 = E R^f m = R^f E m.$

This is true because we can pull the constant $$R^f$$ outside the mathematical expectation.

It follows that the gross return on a risk-free asset is

$R^{f} = 1 / E(m)$

Using this formula for $$R^f$$ in equation (35.2) and rearranging, it follows that

$R^{f} = E R^{i}+\operatorname{cov}\left(m, R^{i} \right) R^{f}$

which can be rearranged to become

$E R^i = R^{f}-\operatorname{cov}\left(m, R^{i}\right) R^{f} .$

It follows that we can express an excess return $$E R^{i}-R^{f}$$ on asset $$i$$ relative to the risk-free rate as

(35.3)$E R^{i}-R^{f} = -\operatorname{cov}\left(m, R^{i}\right) R^{f}$

Equation (35.3) can be rearranged to display important parts of asset pricing theory.

## 35.4. Expected Return - Beta Representation¶

We can obtain the celebrated expected-return-Beta -representation for gross return $$R^i$$ by simply rearranging excess return equation (35.3) to become

$E R^{i}=R^{f}+\left(\underbrace{\frac{\operatorname{cov}\left(R^{i}, m\right)}{\operatorname{var}(m)}}_{\quad\quad\beta_{i,m} = \text{regression coefficient}}\right)\left(\underbrace{-\frac{\operatorname{var}(m)}{E(m)}}_{\quad\lambda_{m} = \text{price of risk}}\right)$

or

(35.4)$E R^{i}=R^{f}+\beta_{i, m} \lambda_{m}$

Here

• $$\beta_{i,m}$$ is a (population) least squares regression coefficient of gross return $$R^i$$ on stochastic discount factor $$m$$, an object that is often called asset $$i$$’s beta

• $$\lambda_m$$ is minus the variance of $$m$$ divided by the mean of $$m$$, an object that is often called the price of risk.

To interpret this representation it helps to provide the following widely used example.

Example

Let $$c_t$$ be the logarithm of the consumption of a representative consumer or just a single consumer for whom we have data.

A popular model of $$m$$ is

$m_{t+1} = \exp(-\rho) \exp(- \gamma(c_{t+1} - c_t))$

where $$\rho > 0$$, $$\gamma > 0$$, and the log of consumption growth is governed by

$c_{t+1} - c_t = \mu + \sigma_c \epsilon_{t+1}$

where $$\epsilon_{t+1} \sim {\mathcal N}(0,1)$$.

Here

• $$\gamma >0$$ is a coefficient of relative risk aversion

• $$\rho >0$$ is a fixed intertemporal discount rate

$m_{t+1} = \exp(-\rho) \exp( - \gamma \mu - \gamma \sigma_c \epsilon_{t+1})$

In this case

$E m_{t+1} = \exp(-\rho) \exp \left( - \gamma \mu + \frac{\sigma_c^2 \gamma^2}{2} \right)$

and

$\operatorname{var}(m_{t+1}) = E(m) [ \exp(\sigma_c^2 \gamma^2) - 1) ]$

When $$\gamma >0$$, it is true that

• when consumption growth is high, $$m$$ is low

• when consumption growth is low, $$m$$ is high

According to representation (35.4), an asset with a gross return $$R^i$$ that is expected to be high when consumption growth is low has $$\beta_i$$ positive and a low expected return.

• because it has a high gross return when consumption growth is low, it is a good hedge against consumption risk. That justifies its low average return.

An asset with an $$R^i$$ that is low when consumption growth is low has $$\beta_i$$ negative and a high expected return.

• because it has a low gross return when consumption growth is low, it is a poor hedge against consumption risk. That justifies its high average return.

## 35.5. Mean-Variance Frontier¶

Now we’ll derive the celebrated mean-variance frontier.

We do this using a method deployed by Lars Peter Hansen and Scott Richard [HR87].

Note

Methods of Hansen and Richard are described and used extensively by [Coc05].

Their idea was rearrange the key equation (35.1), namely, $$E m R^i = 1$$, and then to apply a Cauchy-Schwarz inequality.

A convenient way to remember the Cauchy-Schwartz inequality in our context is that it says that an $$R^2$$ in any regression has to be less than or equal to $$1$$.

(Please note that here $$R^2$$ denotes the coefficient of determination in a regression, not a return on an asset!)

Let’s apply that idea to deduce

(35.5)$1= E\left(m R^{i}\right)=E(m) E\left(R^{i}\right)+\rho_{m, R^{i}}\frac{\sigma(m)}{E(m)} \sigma\left(R^{i}\right)$

where the correlation coefficient $$\rho_{m, R^i}$$ is defined as

$\rho_{m, R^i} \equiv \frac{\operatorname{cov}\left(m, R^{i}\right)}{\sigma(m) \sigma\left(R^{i}\right)}$

and where $$\sigma$$ denotes the standard deviation of the variable in parentheses

Equation (35.5) implies

$E R^{i}=R^{f}-\rho_{m, R^i} \frac{\sigma(m)}{E(m)} \sigma\left(R^{i}\right)$

Because $$\rho_{m, R^i} \in [-1,1]$$, it follows that $$|\rho_{m, R^i}| \leq 1$$ and that

(35.6)$\left|E R^i-R^{f}\right| \leqslant \frac{\sigma(m)}{E(m)} \sigma\left(R^{i}\right)$

Inequality (35.6) delineates a mean-variance frontier

(Actually, it looks more like a mean-standard-deviation frontier)

Evidently, points on the frontier correspond to gross returns that are perfectly correlated (either positively or negatively) with the stochastic discount factor $$m$$.

We summarize this observation as

$\begin{split} \rho_{m, R^{i}}=\left\{\begin{array}{ll} +1 & \implies R^i \text { is on lower frontier } \\ -1 & \implies R^i \text { is on an upper frontier } \end{array}\right. \end{split}$

The image below illustrates a mean-variance frontier.

The mathematical structure of the mean-variance frontier described by inequality (35.6) implies that

• all returns on frontier are perfectly correlated.

Thus,

• Let $$R^m, R^{mv}$$ be two returns on the frontier.

• Then for some scalar $$a$$, a return $$R^{m v}$$ on the mean-variance frontier satisfies the affine equation $$R^{m v}=R^{f}+a\left(R^{m}-R^{f}\right)$$ . This is an exact equation with no residual.

• each return $$R^{mv}$$ that is on the mean-variance frontier is perfectly correlated with $$m$$

• $$\left(\rho_{m, R^{i}}=-1\right) \Rightarrow \begin{cases} m=a+b R^{m v} \\ R^{m v}=e+d m \end{cases}$$ for some scalars $$a, b, e, d$$,

Therefore, any return on the mean-variance frontier is a legitimate stochastic discount factor

• for any mean-variance-efficient return $$R^{m v}$$ that is on the frontier but that is not $$R^{f}$$, there exists a single-beta representation for any return $$R^i$$ that takes the form:

(35.7)$E R^{i}=R^{f}+\beta_{i, R^{m v}}\left[E\left(R^{m v}\right)-R^{f}\right]$
• The special case of a single-beta representation (35.7) with $$R^{i}=R^{m v}$$ is

$$E R^{m v}=R^{f}+1 \cdot\left[E\left(R^{m v}\right)-R^{f}\right]$$

## 35.6. Empirical Implementations¶

We briefly describe empirical implementations of multi-factor generalizations of the single-factor model described above.

The single-beta representation (35.7) is a special case with there being just a single factor.

Two representations are often used in empirical work.

One is a time-series regression of gross return $$R_t^i$$ on multiple risk factors $$f_t^j, j = a, b, \ldots$$.

Such regressions are designed to uncover exposures of return $$R^i$$ to each of a set of risk-factors $$f_t^j, j = a, b, \ldots,$$:

$\begin{split} R_{t}^{i}=a_{i}+\beta_{i, a} f_{t}^{a}+\beta_{i, b} f_{t}^{b}+\ldots+\epsilon_{t}^{i}, \quad t=1,2, \ldots, T\\ \epsilon_{t}^{i} \perp f_{t}^{j}, i=1,2, \ldots, I; j = a, b, \ldots \end{split}$

For example:

• a popular single-factor model specifies the single factor $$f_t$$ to be the return on the market portfolio

• another popular single-factor model called the consumption-based model specifies the factor to be $$m_{t+1} = \beta \frac{u^{\prime}\left(c_{t+1}\right)}{u^{\prime}\left(c_{t}\right)}$$, where $$c_t$$ is a representative consumer’s time $$t$$ consumption.

Model objects are interpreted as follows:

• $$\beta_{i,a}$$ is the exposure of return $$R^i$$ to factor $$f_a$$ risk

• $$\lambda_{a}$$ is the price of exposure to factor $$f_a$$ risk

The other representation entails a cross-section regression of average returns $$E R^i$$ for assets $$i =1, 2, \ldots, I$$ on prices of risk $$\lambda_j$$ for $$j =a, b, c, \ldots$$

Here is the regression specification:

$E R^{i} =\gamma+\beta_{i, a} \lambda_{a}+\beta_{i, b} \lambda_{b}+\cdots$

Testing strategies:

Time-series and cross-section regressions play roles in both estimating and testing beta representation models.

The basic idea is to implement the following two steps.

Step 1:

• Estimate $$a_{i}, \beta_{i, a}, \beta_{i, b}, \cdots$$ by running a time series regression: $$R_{t}^{i}$$ on a constant and $$f_{t}^{a}, f_{t}^{b}, \ldots$$

Step 2:

• take the $$\beta_{i, j}$$’s estimated in step one as regressors together with data on average returns $$E R^i$$ over some period and then estimate the cross-section regression

$\underbrace{E\left(R^{i}\right)}_{\text{average return over time series}}=\gamma+\underbrace{\beta_{i, a}}_{\text{regressor}\quad} \underbrace{\lambda_{a}}_{\text{regression}\text{coefficient}}+\underbrace{\beta_{i, b}}_{\text{regressor}\quad} \underbrace{\lambda_{b}}_{\text{regression}\text{coefficient}}+\cdots+\underbrace{\alpha_{i}}_{\text{pricing errors}}, i=1, \ldots, I; \quad \underbrace{\alpha_i \perp \beta_{i,j},j = a, b, \ldots}_{\text{least squares orthogonality condition}}$
• Here $$\perp$$ means __orthogonal to**

• estimate $$\gamma, \lambda_{a}, \lambda_{b}, \ldots$$ by an appropriate regression technique, recognizing that the regressors have been generated by a step 1 regression.

Note that presumably the risk-free return $$E R^{f}=\gamma$$.

For excess returns $$R^{ei} = R^i - R^f$$ we have

$E R^{e i}=\beta_{i, a} \lambda_{a}+\beta_{i, b} \lambda_{b}+\cdots+\alpha_{i}, i=1, \ldots, I$

In the following exercises, we apply components of the theory.

Our basic tools are random number generator that we shall use to create artificial samples that conform to the theory and least squares regressions that let us watch aspects of the theory at work.

These exercises will further convince us that asset pricing theory is mostly about covariances and least squares regressions.

## 35.7. Exercises¶

import numpy as np
from scipy.stats import stats
import statsmodels.api as sm
from statsmodels.sandbox.regression.gmm import GMM
import matplotlib.pyplot as plt
%matplotlib inline

Lots of our calculations will involve computing population and sample OLS regressions.

So we define a function for simple univariate OLS regression that calls the OLS routine from statsmodels.

def simple_ols(X, Y, constant=False):

if constant:

model = sm.OLS(Y, X)
res = model.fit()

β_hat = res.params[-1]
σ_hat = np.sqrt(res.resid @ res.resid / res.df_resid)

return β_hat, σ_hat

### 35.7.1. Exercise 1¶

Look at the equation,

$R^i_t - R^f = \beta_{i, R^m} (R^m_t - R^f) + \sigma_i \varepsilon_{i, t}.$

Verify that this equation is a regression equation.

### 35.7.2. Exercise 2¶

Give a formula for the regression coefficient $$\beta_{i, R^m}$$.

### 35.7.3. Exercise 3¶

As in many sciences, it is useful to distinguish a direct problem from an inverse problem.

• A direct problem involves simulating a particular model with known parameter values.

• An inverse problem involves using data to estimate or choose a particular parameter vector from a manifold of models indexed by a set of parameter vectors.

Please assume the parameter values provided below and then simulate 2000 observations from the theory specified above for 5 assets, $$i = 1, \ldots, 5$$.

\begin{align*} E\left[R^f\right] &= 0.02 \\ \sigma_f &= 0.00 \\ \xi &= 0.06 \\ \lambda &= 0.04 \\ \beta_{1, R^m} &= 0.2 \\ \sigma_1 &= 0.04 \\ \beta_{2, R^m} &= .4 \\ \sigma_2 &= 0.04 \\ \beta_{3, R^m} &= .6 \\ \sigma_3 &= 0.04 \\ \beta_{4, R^m} &= .8 \\ \sigma_4 &= 0.04 \\ \beta_{5, R^m} &= 1.0 \\ \sigma_5 &= 0.04 \end{align*}

More Exercises

Now come some even more fun parts!

Our theory implies that there exist values of two scalars, $$a$$ and $$b$$, such that a legitimate stochastic discount factor is:

$m_t = a + b R^m_t$

The parameters $$a, b$$ must satisfy the following equations:

\begin{align*} E[(a + b R_t^m) R^m_t)] &= 1 \\ E[(a + b R_t^m) R^f_t)] &= 1 \end{align*}

### 35.7.4. Exercise 4¶

Using the equations above, find a system of two linear equations that you can solve for $$a$$ and $$b$$ as functions of the parameters $$(\lambda, \xi, E[R_f])$$.

Write a function that can solve these equations.

Please check the condition number of a key matrix that must be inverted to determine a, b

### 35.7.5. Exercise 5¶

Using the estimates of the parameters that you generated above, compute the implied stochastic discount factor.

## 35.8. Solutions¶

### 35.8.1. Solution to Exercise 1¶

To verify that it is a regression equation we must show that the residual is orthogonal to the regressor.

Our assumptions about mutual orthogonality imply that

$E\left[\epsilon_{i,t}\right]=0,\quad E\left[\epsilon_{i,t}u_{t}\right]=0$

It follows that

\begin{split} \begin{aligned} E\left[\sigma_{i}\epsilon_{i,t}\left(R_{t}^{m}-R^{f}\right)\right]&=E\left[\sigma_{i}\epsilon_{i,t}\left(\xi+\lambda u_{t}\right)\right] \\ &=\sigma_{i}\xi E\left[\epsilon_{i,t}\right]+\sigma_{i}\lambda E\left[\epsilon_{i,t}u_{t}\right] \\ &=0 \end{aligned} \end{split}

### 35.8.2. Solution to Exercise 2¶

The regression coefficient $$\beta_{i, R^m}$$ is

$\beta_{i,R^{m}}=\frac{Cov\left(R_{t}^{i}-R^{f},R_{t}^{m}-R^{f}\right)}{Var\left(R_{t}^{m}-R^{f}\right)}$

### 35.8.3. Solution to Exercise 3¶

Direct Problem:

# Code for the direct problem

# assign the parameter values
ERf = 0.02
σf = 0.00 # Zejin: Hi tom, here is where you manipulate σf
ξ = 0.06
λ = 0.08
βi = np.array([0.2, .4, .6, .8, 1.0])
σi = np.array([0.04, 0.04, 0.04, 0.04, 0.04])
# in this cell we set the number of assets and number of observations
# we first set T to a large number to verify our computation results
T = 2000
N = 5
# simulate i.i.d. random shocks
e = np.random.normal(size=T)
u = np.random.normal(size=T)
ϵ = np.random.normal(size=(N, T))
# simulate the return on a risk-free asset
Rf = ERf + σf * e

# simulate the return on the market portfolio
excess_Rm = ξ + λ * u
Rm = Rf + excess_Rm

# simulate the return on asset i
Ri = np.empty((N, T))
for i in range(N):
Ri[i, :] = Rf + βi[i] * excess_Rm + σi[i] * ϵ[i, :]

Now that we have a panel of data, we’d like to solve the inverse problem by assuming the theory specified above and estimating the coefficients given above.

# Code for the inverse problem

Inverse Problem:

We will solve the inverse problem by simple OLS regressions.

1. estimate $$E\left[R^f\right]$$ and $$\sigma_f$$

ERf_hat, σf_hat = simple_ols(np.ones(T), Rf)
ERf_hat, σf_hat
(2.000000000000003e-02, 3.123283175179055e-17)

Let’s compare these with the true population parameter values.

ERf, σf
(0.02, 0.0)
1. $$\xi$$ and $$\lambda$$

ξ_hat, λ_hat = simple_ols(np.ones(T), Rm - Rf)
ξ_hat, λ_hat
(0.06036509315069172, 0.0799007653902787)
ξ, λ
(0.06, 0.08)
1. $$\beta_{i, R^m}$$ and $$\sigma_i$$

βi_hat = np.empty(N)
σi_hat = np.empty(N)

for i in range(N):
βi_hat[i], σi_hat[i] = simple_ols(Rm - Rf, Ri[i, :] - Rf)
βi_hat, σi_hat
(array([0.19879597, 0.39155972, 0.59712597, 0.80544824, 1.00554054]),
array([0.04005325, 0.03896574, 0.04133993, 0.04019361, 0.03982646]))
βi, σi
(array([0.2, 0.4, 0.6, 0.8, 1. ]), array([0.04, 0.04, 0.04, 0.04, 0.04]))

Q: How close did your estimates come to the parameters we specified?

### 35.8.4. Solution to Exercise 4¶

(35.8)\begin{align} a ((E(R^f) + \xi) + b ((E(R^f) + \xi)^2 + \lambda^2 + \sigma_f^2) & =1 \cr a E(R^f) + b (E(R^f)^2 + \xi E(R^f) + \sigma_f ^ 2) & = 1 \end{align}
# Code here
def solve_ab(ERf, σf, λ, ξ):

M = np.empty((2, 2))
M[0, 0] = ERf + ξ
M[0, 1] = (ERf + ξ) ** 2 + λ ** 2 + σf ** 2
M[1, 0] = ERf
M[1, 1] = ERf ** 2 + ξ * ERf + σf ** 2

a, b = np.linalg.solve(M, np.ones(2))
condM = np.linalg.cond(M)

return a, b, condM

Let’s try to solve $$a$$ and $$b$$ using the actual model parameters.

a, b, condM = solve_ab(ERf, σf, λ, ξ)
a, b, condM
(87.49999999999999, -468.7499999999999, 54.406619883717504)

### 35.8.5. Solution to Exercise 5¶

Now let’s pass $$\hat{E}(R^f), \hat{\sigma}^f, \hat{\lambda}, \hat{\xi}$$ to the function solve_ab.

a_hat, b_hat, M_hat = solve_ab(ERf_hat, σf_hat, λ_hat, ξ_hat)
a_hat, b_hat, M_hat
(87.99456291454673, -472.774452501456, 55.009393694969035)