Deriving an Affine SDF from the Tangency Portfolio

# SDF as an affine transformation of the tangency portfolio


# SDF as an affine transformation of the tangency portfolio












I'm studying the paper “Deep Learning in Asset Pricing” by Chen et al, first published in 2019. 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 by Vlad Pyzhov (score 3, accepted)

https://quant.stackexchange.com/a/65637

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 by ln_greenspan (score 5)

https://quant.stackexchange.com/a/63777

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 by financial_physician (score 1)

https://quant.stackexchange.com/a/60933

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)

## Answer by StatePriceDensity (score 1)

https://quant.stackexchange.com/a/83570

Even though this thread is now very old, it is still one of the first results that pops up on google on this topic. Therefore, I decided to write a lengthy post about how I think one can arrive at this particular SDF formulation starting from "first principles".

The main ideas concerning SDFs and their properties that one needs to understand are discussed in the books by Kerry Back (2017) and John Cochrane (2005). I personally prefer the first book, as I find it more coherent and better organised. The relevant chapters are 3 and 5 (and potentially 6).

As someone not intimately familiar with this literature and learning the relevant concepts from books, it did not help that the authors of that paper (and many other authors) are not particularly precise in the use of their language, making it difficult to connect their statements back to what one can find in these books. For this reason, I begin with three points that initially caused me a lot of confusion before giving my explanation.

Some Upfront Comments

I) The SDF is not unique unless we assume a complete market. In other words, there can be many random variables $\tilde m$ such that your usual pricing equations, such as $$ p = E[\tilde m \tilde x]\\ $$

$$ 1 = E[\tilde m \tilde R]\\ $$

$$ 0 = E[\tilde m \tilde R^e]\\ $$

$$ 1/R_f=E[\tilde m] $$

hold. The paper assumes there to be no arbitrage opportunities, which in turn only implies that there is at least one strictly positive SDF. So, to put it exactly, there is no the SDF.

Nevertheless, the SDF formulation that is being proposed is one of a unique SDF in that it has the minimum standard deviation $\sigma[\tilde m]$ among all SDFs, and which, in theory, can be obtained by orthogonally projecting any SDF $\tilde m$ onto the span of assets. While the ways in which we can construct this minimum standard deviation or projection SDF from the data may not be unique, the resulting SDF with said property is unique among all possible SDFs in an incomplete market setting.

As this minimum standard deviation or projection SDF is spanned by the assets it can be viewed as a portfolio with with payoff $\tilde m$, price $E[\tilde m \times \tilde m]=E[\tilde m^2]$ and thus return $\tilde R_p = \tilde m/E[\tilde m^2]$, leading me to my second point.

II) Even though this SDF can be viewed as a replicable portfolio, the goal will not be to find a portfolio allocation with return $\tilde R_p$. In fact, from an investment management perspective, it would not be wise to invest in this way as $\tilde R_p$ is volatile and yields a lower expected return than even $R_f$, i.e. it is inefficient in a Markowitz mean-variance sense and has a negative Sharpe ratio (you can find the relevant proof in e.g. KB2017, and it is illustrated in Figure 5.3 in JC2005). Instead, the goal is to determine the allocation into the tangency portfolio (an all equity portfolio yielding the maximum Sharpe ratio).

How can we reconcile the fact that the SDF formulation considered relates to both an inefficient portfolio with a negative Sharpe ratio and a portfolio with the maximum possible Sharpe ratio at the same time? The answer is that we can construct an SDF as an affine function from any return (i.e. not just the tangency portfolio return) on the Markowitz mean-variance frontier (except $R_f$, see the books for details)! I will use this fact as a starting point below for explaining the formula presented in the paper.

To summarize, the tangency portfolio can be used as a starting point to construct a SDF, however, after applying an appropriate affine transformation to its return, the resulting SDF will not inherit the starting portfolio’s return profile (instead it will always be $\tilde R_p$ regardless of the mean-variance frontier return we start from initially)! While one might not be interested in constructing a portfolio that pays $\tilde m$ at price $E[\tilde m ^2]$, the relationship between this SDF and the tangency portfolio provides a useful conceptual structure for estimating the tangency portfolio weights (i.e. think of it as one possible way to construct a loss function for a problem which ultimately will yield the desired portfolio weights). So, interestingly, the problems of constructing an SDF from return data and estimating the weights of Markowitz-efficient portfolios turn out to be much closer related than one might initially expect.

III) Except in a special case $\Sigma^{-1}\mu$ are not the weights of the tangency portfolio! Any textbook (e.g. KB2017) that goes through the derivation and discussion of the mean variance frontier with a risk-free asset will present a formula similar to $$ \pi_{\text{tang}} = \frac{1}{\mathbf{1}' \Sigma^{-1} \mu} \times \Sigma^{-1} \mu $$ for the weights of the tangency portfolio, meaning they are proportional up to a constant to $\Sigma^{-1}\mu$ but not equal to it. Moreover, this relationship is not unique to the tangency portfolio. In fact, this holds true for any other portfolio on the efficient frontier whose weights generally are $$ \pi(\mu_{\text{targ}}) = \frac{\mu_{\text{targ}} }{\mu ' \Sigma^{-1} \mu } \times \Sigma^{-1} \mu, $$ and $\mu_\text{targ}$ is the targeted excess return of the portfolio. Note that, $\pi$ represents the fractions of wealth invested into each individual risky asset and thus only sums up to one for an all equity allocation (as is the case with the tangency portfolio). The investment into the risk-free rate is given by $(1-\mathbf{1}'\pi)$. The two terms have intuitive interpretations: $\Sigma^{-1}\mu$ determines how to allocated between risky assets and is independent of the term to the left of $\times$ which is a scalar that determines how much to allocate between the risk-free rate and tangency portfolio.

Quite intuitively, if we have an all equity allocation, it must be that $\mathbf{1}'\pi=1$ and thus we scale $\Sigma^{-1}\mu$ by $\mathbf{1}' \Sigma^{-1} \mu$ in the case of the tangency portfolio yielding (5).

Ultimately, once we have estimated the vector $\Sigma^{-1}\mu$ we are just one step away from the tangency portfolio weights. However, if $\Sigma^{-1}\mu$ do not coincidently sum up to one without scaling and we invest by making up for the difference with an investment in $R_f$, we might still get the maximum Sharp ratio but a volatility and expected return different from the true tangency portfolio...

SDF Formulation Walkthrough

Keeping the comments above in mind, we can arrive at the formula in the paper by starting from the following theorem:

There is an SDF that is an affine function of a return $\tilde R_{mv}$ if and only if this return is on the mean-variance frontier and not equal to the constant-mimicking portfolio (or $R_f$ if it exists). We write $$ \tilde m = a + b \times \tilde R_{mv}. $$ In the context of this theorem, both books discuss in detail how starting from some $\tilde R_{mv}$ the scalars $a$ and $b$ can be chosen such that e.g. (2) - (4) will hold and that for this $\tilde R_{mv}$ must be on the mean-variance frontier.

In contrast, the paper takes the opposite approach: rather than starting from a given $\tilde R_{mv}$, it selects specific values for $a$ and $b$ such that the resulting $\tilde R_{mv}$ must be the return of the tangency portfolio for $\tilde m$ to be an SDF. One such way to achieve this would be to set $$ a=\frac{1}{R_f}\left(1+c \ E[\tilde R_{mv}]\right) $$ and $$ b=-\frac{c}{R_f} $$ where $$ c=\mathbf{1}\Sigma^{-1}E[\tilde R^e_{mv}]. $$ Substituting the above into (7) we obtain $$ \tilde m = \frac{1}{R_f}\left(1+c \ E[\tilde R_{mv}]\right)-\frac{c}{R_f}\ \tilde R_{mv} \\ = \frac{1}{R_f}\left(1-c \ \left( \tilde R_{mv}-E[\tilde R_{mv}]\right) \right)\\ = \frac{1}{R_f}\left(1-c \ \left( \tilde R^e_{mv}-E[\tilde R^e_{mv}]\right) \right). $$ It is easy to see that $\tilde m$ satisfies (4). However, we still need to confirm that $\tilde R_{mv}$ is on the mean-variance frontier for it to be a valid SDF according to the theorem. For this we first write $\tilde R_{mv}$ as a portfolio of risky assets $$ \tilde R_{mv} = w' \tilde R+(1-\mathbf{1}'w)R_f, $$ which simplifies to $w'\tilde R^e$ in the case of excess returns and thus $$ \tilde m = \frac{1}{R_f}\left(1-c \ w' \ \left( \tilde R^e -E[\tilde R^e]\right) \right). $$ Finally, substituting into (3), as is done in the paper, we find that $$ w=\frac{1}{c} \ \Sigma^{-1} \ E[\tilde R^e] = \frac{1}{\mathbf{1} \Sigma^{-1} \ E[\tilde R^e]} \ \Sigma^{-1} \ E[\tilde R^e] $$ which confirms that $\tilde R_{mv}$ must indeed be the tangency portfolio for $\tilde m$ to be a valid SDF, see (5). As a side-note, from (11) we can also directly arrive at the CAPM SDF as discussed in KB2017, formula (6.10).

While this comes close to what is presented in the paper, two additional steps are needed to arrive at their exact formulation. The first, as others in the thread have already pointed out, is that when working with excess returns, the SDF can be normalized by multiplying by a constant, here $R_f$, without affecting the weights $w$. See e.g. JC2005: "Then $0= E (\tilde mR^e)$ does not identify the mean of $\tilde m$, and we can normalize $a$ arbitrarily. I find it convenient to normalize to $E (\tilde m)= 1$... ."

The second and final step, again for convenience, is to define the variable $\omega$ by setting $\omega = c \times w$. Which gives us the formula from the paper $$ \tilde m = 1-\omega' \ \left( \tilde R^e -E[\tilde R^e]\right) . $$ Hope this helps. Any feedback and comments are welcome.