SDF as an affine transformation of the tangency portfolio

Question

I'm studying this paper. In the formulation of the theoretical setup they state:

Our goal is to explain the differences in the cross-section of returns $R$ for individual stocks. Let $R_{t+1, i}$ denote the return of asset $i$ at time $t+1 .$ The fundamental no-arbitrage assumption is equivalent to the existence of a stochastic discount factor (SDF) $M_{t+1}$ such that for any return in excess of the risk-free rate $R_{t+1, i}^{e}=R_{t+1, i}-R_{t+1}^{f},$ it holds $$ \mathbb{E}_{t}\left[M_{t+1} R_{t+1, i}^{e}\right]=0 \quad \Leftrightarrow \quad \mathbb{E}_{t}\left[R_{t+1, i}^{e}\right]=\underbrace{\left(-\frac{\operatorname{Cov}_{t}\left(R_{t+1, i}^{e}, M_{t+1}\right)}{\operatorname{Var}_{t}\left(M_{t+1}\right)}\right)}_{\beta_{t, i}} \cdot \underbrace{\frac{\operatorname{Var}_{t}\left(M_{t+1}\right)}{\mathbb{E}_{t}\left[M_{t+1}\right]}}_{\lambda_{t}} $$ where $\beta_{t, i}$ is the exposure to systematic risk and $\lambda_{t}$ is the price of risk. $E_{t}[.]$ denotes the expectation conditional on the information at time $t .$ The SDF is an affine transformation of the tangency portfolio. Without loss of generality we consider the SDF formulation $$ M_{t+1}=1-\sum_{i=1}^{N} \omega_{t,i} R_{t+1, i}^{e}=1-\omega_{t}^{\top} R_{t+1}^{e} $$

As sources they mention Chochrane's book (Asset Pricing) and Back's book (Asset Pricing and Portfolio Choice Theory) but I can't find a derivation of $a=1, b=-1$.

Q: How can the considered SDF $M_{t+1} = a + b \omega_{t}^{\top} R_{t+1}^{e}$ with $\omega_{t}^{\top} R_{t+1}^{e}$ the tangency portfolio, $a=1$ and $b=-1$ be derived?

Answer 1

Good question. As I've found in Erwin Hansen paper "Portfolio performance of linear SDF models: an out-of-sample assessment":

\begin{equation} \mathbb{E}[\hat{m}r^{e}] = 0 \end{equation} Now, for any constant c, the SDF $\hat{m}$ = $c\bar{m}$ also satisfies i.e. $\mathbb{E}[\bar{m}r^{e}] = 0$ From this example, it is clear that an infinite number of SDFs exist to satisfy condition simultaneously. This problem is solved by normalizing the value of the constant a in $M = a-b\omega r$ . As pointed out by Cochrane (2009), the choice of this normalization only depends on convenience. The first and simplest normalization consists of imposing a = 1. In this case, we say that the SDF is uncentered. The second normalization is $a = 1+b'\mathbb{E}[r]$, which corresponds to the centered SDF case.‡ After imposing a normalization on a, the set of parameters b is estimated by GMM using the pricing errors as ingredients to selected moment conditions, which we describe in the next section.

Answer 2

Coming back to my question after I replicated the paper for my thesis, where I found that my resulting SDF is always strictly positive and hovering around the value 1, just as expected given the formulation. Then, I also looked at their data and code and realized that this formulation is maybe just one way to "enforce" No-Arbitrage (NA). Because at the thesis presentation a professor asked me:

"How do you actually guarantee NA in your code?"

I don't, and they do not as well. Specifically, the law of one price (LOOP) + incomplete markets (IM) implies the existence of at least one SDF that satisfies

$$ \mathbb{E}_{t}\left[M_{t+1} R_{t+1, i}^{e}\right]=0 $$

whereas the stronger assumption of NA + IM is equivalent to the existence of at least one strictly positive SDF. Hence, they aim to estimate one of possibly many strictly positive SDFs. Given that they do not guarantee explicitly NA in their code, I assume that the choice

$$ M_{t+1}=1-\sum_{i=1}^{N} \omega_{t,i} R_{t+1, i}^{e}$$

was made such that the resulting SDF is very likely positive all the time and hence a suitable candidate SDF in IM with NA. Of course, since $\omega_{t}^{\top} R_{t+1}^{e}$ is a return, it may exceed 100% at some point in time, which would lead to a negative SDF. It is unlikely, but not impossible. Just my best guess at this moment.

Answer 3

Would it be appropriate to say that the scaling ($b$) on the weights ($w_{t,i}$) is irrelevant because you can just pick a scaling of $w_{t,i}$ that satisfies this requirement?

Furthermore, if $a \neq 1$ later in the paper they define $R^e_{t+1,i} = \beta_{t,i}F_{t+1} + \epsilon_{t+1,i}$ where $F_{t+1} = w_t^{\text{T}} R^e_{t+1}$ so we no longer have the $a$ term anyway when we make our approximation of $\hat{M}_{t+1}$ with $(\beta_t \beta_t)^{-1} \beta_t^T R^e_{t+1}$ we no longer have the $a$ term.

Do you know why regressing the returns on the loadings would result in $M_{t+1}$

(I'm hoping this comment furthers the discussion)