Equilibrium asset pricing and the cross section of expected returns

Abstract

In a mean-variance framework with a representative agent, any linear model for the cross section of expected returns can be supported as an equilibrium as long as the market portfolio is spanned by the factor mimicking portfolios. Any set of factors is admissible as long as the spanning condition is satisfied. Factors based on size, book-to-market, momentum, investment, profitability, behavioral biases, principal components, or any combination of these can be used as equilibrium factors. An equilibrium model with M risk factors can be reduced to a collection of M models where each model has a single risk factor, which is covariance with the market portfolio.

Appendix

Proof of Proposition 3

Suppose ${\varvec{\mu }}-{\mathbf {1}} r={\mathbf {D}}{\varvec{\tau }}$ and ${\varvec{\beta }}={\mathbf {Du}}$, where the columns of ${\mathbf {D}}=[{\mathbf {d}}_{1}\;{\mathbf {d}}_{2}\;\cdots \;{\mathbf {d}}_{M}]$ are linearly independent vectors, the elements of ${\varvec{\tau }}$ are non-zero, the vector ${\mathbf {u}}$ has at least one non-zero element such that ${\varvec{\beta }}^{\top }({\varvec{\mu }}-{\mathbf {1}}r)\ne 0$, and ${\mathbf {u}}$ is not proportional to ${\varvec{\tau }}$. Using the columns of ${\mathbf {D}}$, the Gram–Schmidt algorithm (Seber 2008) is used to create a set of mutually orthogonal vectors ${\mathbf {x}}_{1},{\mathbf {x}}_{2},\ldots ,{\mathbf {x}}_{M}$. Let ${\mathbf {x}}_{1}={\mathbf {d}}_{1}$ and let

$$\begin{aligned} {\mathbf {x}}_{2}&={\mathbf {d}}_{2}-{\mathbf {x}}_{1}({\mathbf {x}}_{1}^{\top }{\mathbf {x}}_{1})^{-1}{\mathbf {x}}_{1}^{\top }{\mathbf {d}}_{2},\\ {\mathbf {x}}_{3}&={\mathbf {d}}_{3}-{\mathbf {x}}_{1}({\mathbf {x}}_{1}^{\top }{\mathbf {x}}_{1})^{-1}{\mathbf {x}}_{1}^{\top }{\mathbf {d}}_{3}-{\mathbf {x}} _{2}({\mathbf {x}}_{2}^{\top }{\mathbf {x}}_{2})^{-1}{\mathbf {x}}_{2}^{\top } {\mathbf {d}}_{3},\\&\vdots \\ {\mathbf {x}}_{M}&={\mathbf {d}}_{M}-{\mathbf {x}}_{1}({\mathbf {x}}_{1}^{\top }{\mathbf {x}}_{1})^{-1}{\mathbf {x}}_{1}^{\top }{\mathbf {d}}_{M}-\cdots -{\mathbf {x}} _{M-1}({\mathbf {x}}_{M-1}^{\top }{\mathbf {x}}_{M-1})^{-1}{\mathbf {x}}_{M-1}^{\top }{\mathbf {d}}_{M}. \end{aligned}$$

In matrix form ${\mathbf {D}}=[{\mathbf {x}}_{1}\;{\mathbf {x}}_{2} \;\cdots \;{\mathbf {x}}_{M}]{\mathbf {C}}$, where ${\mathbf {C}}$ is the upper triangular matrix

$$\begin{aligned} {\mathbf {C}}=\left[ \begin{array}{ccccc} 1 &{} ({\mathbf {x}}_{1}^{\top }{\mathbf {x}}_{1})^{-1}{\mathbf {x}}_{1}^{\top } {\mathbf {d}}_{2} &{} ({\mathbf {x}}_{1}^{\top }{\mathbf {x}}_{1})^{-1}{\mathbf {x}} _{1}^{\top }{\mathbf {d}}_{3} &{} \cdots &{} ({\mathbf {x}}_{1}^{\top }{\mathbf {x}} _{1})^{-1}{\mathbf {x}}_{1}^{\top }{\mathbf {d}}_{M}\\ 0 &{} 1 &{} ({\mathbf {x}}_{2}^{\top }{\mathbf {x}}_{2})^{-1}{\mathbf {x}}_{2}^{\top }{\mathbf {d}}_{3} &{} \cdots &{} ({\mathbf {x}}_{2}^{\top }{\mathbf {x}}_{2})^{-1} {\mathbf {x}}_{2}^{\top }{\mathbf {d}}_{M}\\ 0 &{} 0 &{} 1 &{} \cdots &{} ({\mathbf {x}}_{3}^{\top }{\mathbf {x}}_{3})^{-1}{\mathbf {x}} _{3}^{\top }{\mathbf {d}}_{M}\\ \vdots &{} \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} 0 &{} \cdots &{} 1 \end{array} \right] . \end{aligned}$$

Using ${\mathbf {C}}$ we can write ${\varvec{\mu }}-{\mathbf {1}} r=\mathbf {DC}^{-1}{\mathbf {C}}{\varvec{\tau }}$ and ${\varvec{\beta }}=\mathbf {DC} ^{-1}\mathbf {Cu}$, where $\mathbf {DC}^{-1}=[{\mathbf {x}}_{1}\;{\mathbf {x}} _{2}\;\cdots \;{\mathbf {x}}_{M}]$. Now consider two cases. For the first case, suppose each element of ${\mathbf {C}}{\varvec{\tau }}$ and each element of $\mathbf {Cu}$ is non-zero. In addition, for each $m=1,2,\ldots ,M$ suppose we divide the mth element of ${\mathbf {C}}{\varvec{\tau }}$ by the mth element of $\mathbf {Cu}$. If we obtain M different numbers then (13) is satisfied. In this case we are done. We simply need to change the notation to be consistent with the proposition, let ${\mathbf {B}}={\mathbf {C}}^{-1}$, ${\mathbf {p}}={\mathbf {C}}\tau$, ${\mathbf {q}}=\mathbf {Cu}$, and let ${\mathbf {e}}_{m}={\mathbf {x}}_{m}$ for $m=1,2,\ldots ,M$.

For the second case, suppose at least one of the elements of ${\mathbf {C}}{\varvec{\tau }}$ or $\mathbf {Cu}$ is zero or suppose (13) is not satisfied. In this case it is convenient to work with normalized vectors so let ${\hat{{\mathbf {x}}}}_{m}={\mathbf {x}}_{m}/\sqrt{{\mathbf {x}}_{m}^{\top }{\mathbf {x}} _{m}}$ for $m=1,2,\ldots ,M$. We can now write ${\varvec{\mu }}-{\mathbf {1}} r=[{\hat{{\mathbf {x}}}}_{1}\;{\hat{{\mathbf {x}}}}_{2}\;\cdots \;{\hat{{\mathbf {x}}}} _{M}]{\varvec{\pi }}$ and ${\varvec{\beta }}=[{\hat{{\mathbf {x}}}}_{1}\;{\hat{{\mathbf {x}}}}_{2}\;\cdots \;{\hat{{\mathbf {x}}}}_{M}]{\varvec{\eta }}$, where ${\varvec{\pi }}={\varvec{\Lambda }}{\mathbf {C}}{\varvec{\tau }}$, ${\varvec{\eta }}={\varvec{\Lambda }}{\mathbf {Cu}}$, and ${\varvec{\Lambda }}$ is a $M\times M$ diagonal matrix with $\sqrt{{\mathbf {x}} _{m}^{\top }{\mathbf {x}}_{m}}$ in the mth position on the main diagonal. Let ${\mathbf {Z}}$ be a $M\times M$ orthogonal matrix and consider the expressions ${\varvec{\mu }}-{\mathbf {1}}r=[{\hat{{\mathbf {x}}}}_{1}\;{\hat{{\mathbf {x}}}}_{2} \;\cdots \;{\hat{{\mathbf {x}}}}_{M}]{\mathbf {Z}}^{\top }\mathbf {Z}\varvec{\pi }$ and ${\varvec{\beta }}=[{\hat{{\mathbf {x}}}}_{1}\;{\hat{{\mathbf {x}}}}_{2}\;\cdots \;{\hat{{\mathbf {x}}}}_{M}]{\mathbf {Z}}^{\top }{\mathbf {Z}}{\varvec{\eta }}$.

Since ${\mathbf {Z}}$ is an orthogonal matrix and the columns of $[{\hat{{\mathbf {x}}}} _{1}\;{\hat{{\mathbf {x}}}}_{2}\;\cdots \;{\hat{{\mathbf {x}}}}_{M}]$ are orthogonal vectors of unit length, the columns of $[{\hat{{\mathbf {x}}}}_{1}\;{\hat{{\mathbf {x}}}}_{2}\;\cdots \;{\hat{{\mathbf {x}}}}_{M}]{\mathbf {Z}}^{\top }$ are also orthogonal vectors of unit length. Thus for the second case we can let $[{\mathbf {e}} _{1}\;{\mathbf {e}}_{2}\;\cdots \;{\mathbf {e}}_{M}]=[{\hat{{\mathbf {x}}}}_{1} \;{\hat{{\mathbf {x}}}}_{2}\;\cdots \;{\hat{{\mathbf {x}}}}_{M}]{\mathbf {Z}}^{\top }$. The problem then reduces to constructing an orthogonal matrix ${\mathbf {Z}}$ such that the elements of $\mathbf {Z{\varvec{\pi }}}$ and ${\mathbf {Z}}\varvec{\eta }$ are all non-zero and (13) is satisfied. To see that such a ${\mathbf {Z}}$ exists, ${\varvec{\eta }}$ is decomposed as ${\varvec{\eta }}={\varvec{\eta }}_{1} +{\varvec{\eta }}_{2}$, where ${\varvec{\eta }}_{1}={\varvec{\pi }}({\varvec{\pi }}^{\top }{\varvec{\pi }})^{-1}{\varvec{\pi }}^{\top }{\varvec{\eta }}$ and ${\varvec{\eta }} _{2}=({\mathbf {I}}-{\varvec{\pi }}({\varvec{\pi }}^{\top }{\varvec{\pi }})^{-1} {\varvec{\pi }}^{\top }){\varvec{\eta }}$. Since ${\varvec{\beta }}^{\top }({\varvec{\mu }} -{\mathbf {1}}r)\ne 0$, we have ${\varvec{\pi }}^{\top }{\varvec{\eta }}\ne 0$ and thus ${\varvec{\eta }}_{1}$ is a non-zero vector. Since ${\mathbf {u}}$ and ${\varvec{\tau }}$ are not collinear vectors, it follows that ${\varvec{\eta }}$ and ${\varvec{\pi }}$ are not collinear vectors, thus ${\varvec{\eta }}_{2}$ is a non-zero vector. The normalized dot product between ${\varvec{\pi }}$ and ${\varvec{\eta }}$ is ${\varvec{\pi }}^{\top }{\varvec{\eta }}/(\sqrt{{\varvec{\pi }} ^{\top }{\varvec{\pi }}}\sqrt{{\varvec{\eta }}^{\top }{\varvec{\eta }}})$, which is equal to the cosine of the angle between ${\varvec{\pi }}$ and ${\varvec{\eta }}$. Now let $\hat{{\mathbf {p}}}$ and $\hat{{\mathbf {q}}}$ be any two vectors with non-zero elements that satisfy (13), where the angle between $\hat{{\mathbf {p}}}$ and ${\hat{{\mathbf {q}}}}$ is the same as the angle between ${\varvec{\pi }}$ and ${\varvec{\eta }}$, i.e.,

$$\begin{aligned} \frac{\hat{{\mathbf {p}}}^{\top }{\hat{{\mathbf {q}}}}}{\sqrt{{\hat{{\mathbf {p}}}}^{\top }{\hat{{\mathbf {p}}}}}\sqrt{{\hat{{\mathbf {q}}}}^{\top }{\hat{{\mathbf {q}}}}}} =\frac{{\varvec{\pi }}^{\top }{\varvec{\eta }}}{\sqrt{{\varvec{\pi }}^{\top }{\varvec{\pi }} }\sqrt{{\varvec{\eta }}^{\top }{\varvec{\eta }}}}. \end{aligned}$$

(47)

The vectors ${\mathbf {p}}$ and ${\mathbf {q}}$ are defined as

$$\begin{aligned} {\mathbf {p}}= & {} {\hat{{\mathbf {p}}}}\frac{\sqrt{{\varvec{\pi }}^{\top }{\varvec{\pi }}} }{\sqrt{{\hat{{\mathbf {p}}}}^{\top }{\hat{{\mathbf {p}}}}}}, \end{aligned}$$

(48)

$$\begin{aligned} {\mathbf {q}}= & {} {\hat{{\mathbf {q}}}}\frac{\sqrt{{\varvec{\eta }}^{\top }{\varvec{\eta }}} }{\sqrt{{\hat{{\mathbf {q}}}}^{\top }{\hat{{\mathbf {q}}}}}}. \end{aligned}$$

(49)

Note that ${\mathbf {p}}$ and ${\mathbf {q}}$ have non-zero elements that satisfy (13) and the angle between ${\mathbf {p}}$ and ${\mathbf {q}}$ is the same as in (47). Now decompose ${\mathbf {q}}$ as ${\mathbf {q}} ={\mathbf {q}}_{1}+{\mathbf {q}}_{2}$, where ${\mathbf {q}}_{1}={\mathbf {p}} ({\mathbf {p}}^{\top }{\mathbf {p}})^{-1}{\mathbf {p}}^{\top }{\mathbf {q}}$ and ${\mathbf {q}}_{2}=({\mathbf {I}}-{\mathbf {p}}({\mathbf {p}}^{\top }{\mathbf {p}} )^{-1}{\mathbf {p}}^{\top }){\mathbf {q}}$, and construct ${\mathbf {Z}}$ as

$$\begin{aligned} {\mathbf {Z}}=\left[ \frac{{\mathbf {p}}}{\sqrt{{\mathbf {p}}^{\top }{\mathbf {p}}} }\ \frac{{\mathbf {q}}_{2}}{\sqrt{{\mathbf {q}}_{2}^{\top }{\mathbf {q}}_{2}} }\ {\mathbf {V}}\right] \left[ \frac{{\varvec{\pi }}}{\sqrt{{\varvec{\pi }}^{\top }{\varvec{\pi }}}}\ \frac{{\varvec{\eta }}_{2}}{\sqrt{{\varvec{\eta }}_{2}^{\top }{\varvec{\eta }}_{2}}}\ {\mathbf {U}}\right] ^{\top } \end{aligned}$$

(50)

where ${\mathbf {V}}$ and ${\mathbf {U}}$ are $M\times (M-2)$ matrices. The matrix ${\mathbf {V}}$ has mutually orthogonal columns of unit length, where each column of ${\mathbf {V}}$ is orthogonal to ${\mathbf {p}}$ and ${\mathbf {q}}_{2}$, ${\mathbf {p}}^{\top }{\mathbf {V}}={\mathbf {0}}$ and ${\mathbf {q}}_{2}^{\top } {\mathbf {V}}={\mathbf {0}}$. Likewise, the matrix ${\mathbf {U}}$ has mutually orthogonal columns of unit length, where each column of ${\mathbf {U}}$ is orthogonal to ${\varvec{\pi }}$ and ${\varvec{\eta }}_{2}$, ${\varvec{\pi }}^{\top }{\mathbf {U}}={\mathbf {0}}$ and ${\varvec{\eta }}_{2}^{\top }{\mathbf {U}}={\mathbf {0}}$. Thus ${\mathbf {Z}}$ in (50) is a product of orthogonal matrices. Note that ${\mathbf {Z}}$ is the orthogonal transformation that maps ${\varvec{\pi }} /\sqrt{{\varvec{\pi }}}^{\top }{\varvec{\pi }}$ into ${\mathbf {p}}/\sqrt{{\mathbf {p}}^{\top }{\mathbf {p}}}$ and maps ${\varvec{\eta }}{2}/\sqrt{{\varvec{\eta }}}_{2}^{\top }{\varvec{\eta }}_{2}$ into ${\mathbf {q}}_{2}/\sqrt{{\mathbf {q}}_{2}^{\top }{\mathbf {q}}_{2}}$,

$$\begin{aligned} {\mathbf {Z}}\left[ \frac{\varvec{\pi }}{\sqrt{{\varvec{\pi }}^{\top }{\varvec{\pi }}}\ \frac{{\varvec{\eta }}{2}}{\sqrt{{\varvec{\eta }}{2}^{\top }{\varvec{\eta }}{2}} }}\right] =\left[ \frac{{\mathbf {p}}}{\sqrt{{\mathbf {p}}^{\top }{\mathbf {p}}} }\ \frac{{\mathbf {q}}_{2}}{\sqrt{\mathbf {q}}_{2}^{\top }{\mathbf {q}}_{2}}\right] \end{aligned}$$

(51)

From (48)–(49) we have ${\mathbf {p}} ^{\top }{\mathbf {p}}={\varvec{\pi }}^{\top }{\varvec{\pi }}$ and ${\mathbf {q}}^{\top }{\mathbf {q}}={\varvec{\eta }}^{\top }{\varvec{\eta }}$. Using (47) and the definitions of ${\varvec{\eta }}_{1}$ and ${\mathbf {q}}_{1}$, we have ${\mathbf {q}}_{1}^{\top }{\mathbf {q}}_{1}={\varvec{\eta }}_{1}^{\top }{\varvec{\eta }} _{1}$. Thus ${\mathbf {q}}_{2}^{\top }{\mathbf {q}}_{2}={\varvec{\eta }}_{2}^{\top }{\varvec{\eta }}_{2}$ and (51) implies $\mathbf {Z{\varvec{\pi }}}={\mathbf {p}}$ and ${\mathbf {Z}}{\varvec{\eta }}_{2}={\mathbf {q}}_{2}$. Using these two equations we can evaluate ${\mathbf {Z}}\varvec{\eta }$ to get

$$\begin{aligned} {\mathbf {Z}}{\varvec{\eta }}={\mathbf {Z}}{\varvec{\eta }}_{1}+{\mathbf {Z}}{\varvec{\eta }}_{2}={\mathbf {p}}({\varvec{\pi }} ^{\top }{\varvec{\pi }})^{-1}{\varvec{\pi }}^{\top }{\varvec{\eta }}+{\mathbf {q}} _{2}={\mathbf {q}}, \end{aligned}$$

where the last equality follows since ${\mathbf {p}}^{\top } {\mathbf {p}}={\varvec{\pi }}^{\top }{\varvec{\pi }}$ and the angle between ${\mathbf {p}}$ and ${\mathbf {q}}$ is the same as the angle between ${\varvec{\pi }}$ and ${\varvec{\eta }}$. This completes the proof. $\square$

Description of the empirical methodology The empirical methodology used in Sects. 5.4 and 5.5 is described here. The data are obtained from Ken French’s data library at Dartmouth College. Value-weighted monthly returns are downloaded for size decile portfolios, BTM decile portfolios, and the 25 Fama and French (1993) portfolios. Monthly returns are also downloaded for the riskless asset, the value-weighted market portfolio, and the SMB and HML factors. The sample period is July 1926 through December 2014, for a total of 1,062 months.

Equation (37) shows that the expected excess returns on the N portfolios are related cross-sectionally to the symmetric and skew-symmetric betas. Since the estimated prices of risk depend on the estimated betas, there is an errors-in-the-variables issue, as discussed in Shanken (1992). To address this issue, the estimation procedure is mapped into the generalized method of moments (GMM). This allows us to handle the moments that generate the betas at the same time as the moments that generate the prices of risk. A discussion of this empirical approach is in Cochrane (2001, pp. 240–242).

As in Sect. 5.4, let ${\mathbf {v}}_{n}$ be a $N\times 1$ vector with a 1 in the nth position and zeros elsewhere. The symmetric and skew-symmetric parts of ${\mathbf {v}}_{n}$ are ${\mathbf {v}}_{n,s}$ and ${\mathbf {v}}_{n,ss}$, respectively. Let $R_{{\mathbf {v}}_{n},t}={\mathbf {v}} _{n}^{\top }{\mathbf {R}}_{t}$ denote the nth portfolio’s return in month t, let $R_{{\mathbf {v}}_{n,s},t}={\mathbf {v}}_{n,s}^{\top }{\mathbf {R}}_{t}$ denote the return on the symmetric part in month t, and let $R_{{\mathbf {v}}_{n,ss} ,t}={\mathbf {v}}_{n,ss}^{\top }{\mathbf {R}}_{t}$ denote the return on the skew-symmetric part in month t. The return on the market portfolio in month t is $R_{mkt,t}$ and the riskless rate is $r_{t}$. We have the following moment conditions

$$\begin{aligned} {\mathbb {E}}\left[ R_{{\mathbf {v}}_{n,s},t}-a_{{\mathbf {v}}_{n,s}}-\beta _{{\mathbf {v}}_{n,s}}R_{mkt,t}\right]&=0, \end{aligned}$$

(52)

$$\begin{aligned} {\mathbb {E}}\left[ R_{{\mathbf {v}}_{n,ss},t}-a_{{\mathbf {v}}_{n,ss}}-\beta _{{\mathbf {v}}_{n,ss}}R_{mkt,t}\right]&=0, \end{aligned}$$

(53)

$$\begin{aligned} {\mathbb {E}}\left[ \left( R_{{\mathbf {v}}_{n,s},t}-a_{{\mathbf {v}}_{n,s}} -\beta _{{\mathbf {v}}_{n,s}}R_{mkt,t}\right) R_{mkt,t}\right]&=0, \end{aligned}$$

(54)

$$\begin{aligned} {\mathbb {E}}\left[ \left( R_{{\mathbf {v}}_{n,ss},t}-a_{{\mathbf {v}}_{n,ss}} -\beta _{{\mathbf {v}}_{n,ss}}R_{mkt,t}\right) R_{mkt,t}\right]&=0, \end{aligned}$$

(55)

$$\begin{aligned} {\mathbb {E}}\left[ R_{{\mathbf {v}}_{n},t}-r_{t}-\gamma _{1}\beta _{{\mathbf {v}} _{n,s}}-\gamma _{2}\beta _{{\mathbf {v}}_{n,ss}}\right]&=0. \end{aligned}$$

(56)

There are a total of 5N moment conditions since (52)–(56) each hold for $n=1,2,\ldots ,N$. To estimate the model, the expectation operator ${\mathbb {E}}$ is replaced with the time series average $(1/T)\sum _{t=1}^{T}$, where T is the sample size. The left-hand sides of (52)–(56) are then stacked into a $5N\times 1$ vector ${\mathbf {g}}({\mathbf {b}};T)$, where ${\mathbf {b}}$ is a $\left( 4N+2\right) \times 1$ vector of parameters to be estimated. The vector ${\mathbf {b}}$ includes $a_{{\mathbf {v}}_{n,s}}$, $a_{{\mathbf {v}}_{n,ss}}$, $\beta _{{\mathbf {v}}_{n,s}}$, and $\beta _{{\mathbf {v}}_{n,ss}}$ for $n=1,2,\ldots ,N$ plus $\gamma _{1}$ and $\gamma _{2}$. To derive estimates for the parameters, the system $\mathbf {cg}({\mathbf {b}};T)={\mathbf {0}}$ is solved, where ${\mathbf {c}}$ is the $\left( 4N+2\right) \times 5N$ matrix

$$\begin{aligned} {\mathbf {c}}=\left[ \begin{array}{cc} {\mathbf {I}} &{} {\mathbf {0}}\\ {\mathbf {0}} &{} {\varvec{\beta }}_{s}^{\top }\\ {\mathbf {0}} &{} {\varvec{\beta }}_{ss}^{\top } \end{array} \right] . \end{aligned}$$

(57)

In (57), ${\mathbf {I}}$ is a $4N\times 4N$ identity matrix, ${\varvec{\beta }}_{s}$ is a $N\times 1$ vector of symmetric betas, $\varvec{\beta }_{ss}$ is a $N\times 1$ vector of skew-symmetric betas, and each ${\mathbf {0}}$ represents a conformable block of zeros. Note that the solution to $\mathbf {cg}({\mathbf {b}};T)={\mathbf {0}}$ gives ordinary least squares (OLS) estimates for the parameters. The first 4N moment conditions exactly identify $a_{{\mathbf {v}}_{n,s}}$, $a_{{\mathbf {v}}_{n,ss}}$, $\beta _{{\mathbf {v}}_{n,s}}$, and $\beta _{{\mathbf {v}}_{n,ss}}$ for $n=1,2,\ldots ,N$. The last N moment conditions are overidentified since there are two parameters and N moments. When the last N moments are weighted by ${\varvec{\beta }}_{s}$ and ${\varvec{\beta }}_{ss}$, as in $\mathbf {cg} ({\mathbf {b}};T)={\mathbf {0}}$, we obtain the OLS first-order conditions for estimating $\gamma _{1}$ and $\gamma _{2}$.

Robust standard errors for the estimated parameters follow directly from the GMM framework. Let ${\mathbf {d}}$ denote the matrix of partial derivatives with respect to the parameters, i.e., ${\mathbf {d}}=\partial {\mathbf {g}}({\mathbf {b}} ;T)/\partial {\mathbf {b}}^{\top }$. The covariance matrix of standard errors for the estimated parameters is

$$\begin{aligned} \textit{Var}({\hat{{\mathbf {b}}}})=\frac{1}{T}\left( \mathbf {cd}\right) ^{-1} \mathbf {cSc}^{\top }\left( \left( \mathbf {cd}\right) ^{-1}\right) ^{\top }, \end{aligned}$$

where ${\mathbf {S}}$ is the long-run covariance matrix for the moments. The technique in Newey and West (1987), with four lags, is used to construct ${\mathbf {S}}$. Using a different number of lags, such as three, five, or six, did not have a material impact on the empirical results.

To estimate the pricing model in (43), an otherwise identical empirical framework is used except that now there are more moment conditions and parameters. In (37), there are two types of betas and two prices of risk, which gives 5N moments and $4N+2$ parameters. In contrast, in (43), there are three types of betas and three prices of risk, which gives 7N moments and $6N+3$ parameters. Other than this difference, the estimation technique is the same once we use generalized symmetry and generalized skew-symmetry for the betas instead of the standard notions of symmetry and skew-symmetry.

Note that the SMB and HML factors, which were downloaded, are used to calculate the loadings in Table 3, but these factors are not used in Sects. 5.4 and 5.5. $\square$