Uncategorized – Page 5 – Research Notebook

How Many Assets Are Needed To Test a K-Factor Model?

November 23, 2017 by Alex

1. Motivation

Imagine you’re a financial economist who thinks that some risk factor, ${\color{white}i}f_t$ , explains the cross-section of expected returns. And, you decide to test your hunch. First, you regress the realized returns of $N$ different assets on ${\color{white}i}f_t$ to estimate each asset’s exposure to the risk factor, $\tilde{b}_n$ :

$\begin{equation*} r_{n,t} = \tilde{a}_n + \tilde{b}_n \cdot f_t + \tilde{e}_{n,t} \qquad t = 1, \, \ldots, \, T \, \, \text{for each } n \end{equation*}$

Then, you regress these same assets’ average returns on their exposures the risk factor, $\tilde{b}_n$ :

$\begin{equation*} \mathrm{E} [ r_n ] = \hat{\alpha} + \hat{\lambda} \cdot \tilde{b}_n + \hat{\epsilon}_n \qquad n = 1, \, \ldots, \, N \end{equation*}$

If ${\color{white}i}f_t$ is a priced factor, then the slope coefficient, $\hat{\lambda}$ , should be large and the intercept, $\hat{\alpha}$ , should be zero.

This two-stage methodology dates back to Fama and MacBeth (1973). And, in keeping with the original paper, many financial economists still use a small set of portfolios as the $N$ assets in their empirical analysis. For example, Fama and French (1993) use $N = 25$ portfolios created by sorting stocks based on size and book-to-market. In theory, this is fine. If you’ve found a priced risk factor, then exposure to ${\color{white}i}f_t$ should affect the average returns of all assets. There is no theoretical guidance about which assets to use.

But, here’s the thing: econometrically, the number of assets has to matter. We live in a world with lots of factors to choose from. This is the “anomaly zoo” coined in Cochrane (2011). If you consider a model with more risk factors than you have assets $K \geq N$ , then of course you can perfectly fit the data:

$\begin{equation*} \mathrm{E} [ r_n ] = 0 + {\textstyle \sum_{f \in \mathcal{K}}} \, \hat{\lambda}_f \cdot \tilde{b}_{n,f} + 0 \qquad n = 1, \, \ldots, \, N \leq K \end{equation*}$

After all, a system of $N$ linear equations with $K$ unknowns is guaranteed to have a solution if $K \geq N$ .

Clearly, you need at least $K$ assets to test a $K$ -factor model. This is obvious. But, you can elaborate on this idea and say something useful in the related setting where $K \leq N \leq F$ . This post answers the question: how many assets do you need when testing a $4$ -factor model chosen in a world with $F = 97$ candidate factors?

2. Problem Formulation

Suppose we live in a world with $N$ assets and $F$ candidate factors. Think about $F = 97$ (McLean and Pontiff; 2016), $F = 333$ (Harvey, Liu, and Zhu; 2016), or $F = 447$ (Hou, Xue, and Zhang; 2017). We are looking for the asset-pricing model with the fewest factors that perfectly explains the cross-section of expected returns:

(★) $\begin{equation*} \min_{\mathcal{K} \subseteq \mathcal{F}} \left\{ \, {\textstyle \sum_{f \in \mathcal{K}}} \, 1_{\{\hat{\lambda}_f \neq 0\}} \quad \text{s.t.} \quad \mathrm{E}[ r_n ] = {\textstyle \sum_{f \in \mathcal{K}}} \, \hat{\lambda}_f \cdot \tilde{b}_{n,f} \, \right\} \end{equation*}$

This is just a mathematical way of applying Occam’s razor to our model-selection problem.

Here’s what I want to know: if we find a solution to this problem, $\hat{\mathcal{K}}$ , how likely is it that $\hat{\mathcal{K}}$ is the only solution? In other words, if we find a $\hat{K}$ -factor model that solves Problem (★) and perfectly explains the cross-section of expected returns, should we celebrate or slow clap? Have we found the simplest model of the world? Or, is this just one of many such $\hat{K}$ -factor models that we were bound to uncover?

3. Similar Exposures

It turns out that the answer to this question is going to critically depend on the similarity of the $N$ assets’ exposures to the $F$ risk factors. To see why, just think about a world where $\mathcal{F}$ contains two copies of some risk factor in $\hat{\mathcal{K}}$ . Clearly, $\hat{\mathcal{K}}$ can’t be a unique solution to Problem (★) because you could just switch out the risk factor for its twin and have another $\hat{K}$ -factor model explaining the cross-section of expected returns.

Donoho and Huo (2001) gives a nice way of generalizing this notion of similarity to situations where $\mathcal{F}$ doesn’t contain multiple copies of the same factor. Specifically, they define the idea of mutual coherence:

$\begin{equation*} \rho_{\max} \overset{\scriptscriptstyle \text{def}}{=} \max_{1 \leq f < f' \leq F} \left\{ \, \frac{ \left| \sum_{n=1}^N \tilde{b}_{n,f} \cdot \tilde{b}_{n,f'} \right| }{ \sqrt{\sum_{n=1}^N \tilde{b}_{n,f}^2} \cdot \sqrt{\sum_{n=1}^N \tilde{b}_{n,f'}^2} } \, \right\} \end{equation*}$

Roughly speaking, you should think about mutual coherence as measuring the maximum correlation between the $N$ assets’ exposures to any pair of risk factors in $\mathcal{F}$ . A large value of $\rho_{\max}$ , means that the $N$ assets have really similar exposures to some pair of risk factors in $\mathcal{F}$ . And, in the extreme case where $\mathcal{F}$ contains two exact copies of the same risk factor, we have $\rho_{\max} = 1$ . By contrast, $\rho_{\max} = 0$ if there are more assets than risk factors and the risk factors were independent.

But, since we are thinking about a world where there are more candidate risk factors than assets $F > N$ , we know that $\rho_{\max} > 0$ . Even if the $F$ factors really are independent, some of them are going to have to look correlated in such a small sample. And, Welch (1974) characterizes exactly how correlated they will look:

(W) $\begin{equation*} \rho_{\max} \geq \sqrt{\frac{F - N}{N \cdot (F - 1)}} \qquad \text{given } F > N \end{equation*}$

4. Theoretical Minimum

Now, we can answer our original question: when can we be sure that a solution to Problem (★) is unique? i.e., if we find a $K$ -factor model that perfectly explains the cross-section of returns for $N$ assets, should we be surprised? Donoho and Elad (2003) show that, if $\hat{\mathcal{K}}$ is a solution to Problem (★) and

(DE) $\begin{equation*} \hat{K} < \frac{1}{2} \cdot \left( 1 + \frac{1}{\rho_{\max}} \right), \end{equation*}$

then $\hat{\mathcal{K}}$ is a unique solution. There are no other factor models that explain the cross-section of expected returns using $\hat{K}$ or fewer factors. Inserting the bound from Equation (W) into the bound from Equation (DE) and then solving for $N$ yields:

$\begin{equation*} N_{\min} \overset{\scriptscriptstyle \text{def}}{=} \left\lceil \frac{F \cdot (2 \cdot K - 1)^2}{(F - 1) + (2 \cdot K - 1)^2} \right\rceil \end{equation*}$

What’s this equation telling us? Suppose we find a $\hat{K}$ -factor model that perfectly explains the cross-section of expected returns—i.e., a factor model that solves Problem (★). If our empirical analysis used at least $N_{\min}$ assets, then we can be sure that we’ve found the simplest possible model. Whereas, if we only used $(N_{\min} - 1)$ assets, then there might be another asset-pricing model with the same number of factors that perfectly explains the cross-section of expected returns.

5. Plugging in Numbers

Let’s plug in some numbers to see if our formula for $N_{\min}$ makes any sense. Here’s the first exercise. Suppose that there really are only $F=97$ candidate risk factors like in McLean and Pontiff (2016). The figure to the right plots the minimum number of assets we’d need to include in our empirical analysis (y-axis) to identify a model with $K$ factors (x-axis). We need to study at least $N_{\min} = 9$ assets to be sure that we’ve found a unique $2$ -factor model; we need to study at least $N_{\min} = 21$ assets to be sure that we’ve found a unique $3$ -factor model; and, we need to study at least $N_{\min} = 33$ assets to be sure that we’ve found a unique $4$ -factor model. This is an interesting exercise because it says that we shouldn’t be surprised if someone found a $4$ -factor model that perfectly explained the cross-section of expected returns for the $25$ size- and value-sorted portfolios used in Fama and French (1993). In other words, since Gene and Ken already claimed the first $3$ factors, we can’t do cross-sectional asset pricing tests with just the Fama and French (1993) portfolios any more. Even if we found a $4$ -factor model that perfectly explained the cross-section of expected returns, there’d be no guarantee that it’d be unique.

Now, let’s consider a second exercise. In a recent NBER working paper, Kozak, Nagel, and Santosh (2017) found used a machine-learning rule to identify a cross-sectional asset-pricing model with $\hat{K} = 33$ factors. For the sake of argument, let’s imagine that this $33$ -factor model perfectly explained the cross-section of expected returns. The figure to the left plots the minimum number of assets they’d need to include in their empirical analysis (y-axis) to be sure that they’d found the only such $33$ -factor model in a world with $F$ candidate factors (x-axis). There are roughly $2500$ stocks at the moment. So, if their $33$ -factor model had perfectly explained the cross-section of expected returns, then you should be really excited by this result if you think there are less than $F \approx 6000$ candidate factors.

Let’s do one last exercise before we call it quits. Notice that, as the number of candidate factors gets large, the minimum number of assets we’d need to identify a $K$ -factor model only depends on $K$ :

(1) $\begin{equation*} \lim_{F \to \infty} N_{\min} = (2 \cdot K - 1)^2 \end{equation*}$

This observation is cool because, by setting $N_{\min} = 2500$ and solving for $K$ , we can compute the size of the most complicated factor model that we can estimate with $2500$ stocks is $25 = \lfloor \sfrac{1}{2} \cdot( \sqrt{2500} + 1) \rfloor$ . No matter how many candidate factors there are, if we find a $25$ -factor model that perfectly explains the cross-section of expected returns, then we know it is unique if we use the universe of all stocks in our empirical analysis.

Neglecting The Madness Of Crowds

September 3, 2017 by Alex

Motivation

This post is motivated by two stylized facts about bubbles and crashes. The first is that these events are often attributed to the madness of crowds. In popular accounts, they occur when a large number of inexperienced traders floods into the market and mob psychology takes over. For some examples, just think about day traders during the DotCom bubble, out-of-town buyers during the housing bubble, or first-time investors during the Chinese warrant bubble.

The second stylized fact is that, even though bubbles and crashes have a large impact on the market, traders seems to ignore the risk posed by the madness of crowds during normal times. Gripped by “new-era thinking”, they often insist on justifying market events with fundamentals until some sudden price movement forces them to reckon with the madness of crowds. This phenomenon is referred to as “neglected risk” in the asset-pricing literature.

With these two stylized facts in mind, this post investigates how hard it is for traders to learn about aggregate noise-trader demand when the number of noise traders can vary over several orders of magnitude—i.e., when there’s a possibility that the crowd’s gone mad. I find something surprising: it makes sense for existing traders to neglect the madness of crowds during normal times. Here’s the logic. Noise traders push prices away from fundamentals. So, if you don’t see a large unexpected price movement away from fundamentals, then there must not be very many noise traders in the market. And, if there aren’t very many noise traders, then they can’t affect the equilibrium price very much. But, this means that there’s no way for you to learn about aggregate noise-trader demand from the equilibrium price, which means that there’s no reason for you to revise your beliefs about aggregate noise-trader demand away from zero.

To illustrate this point, I’m going to make use of a happy mathematical coincidence. It turns out that, if you assume changes in the number of noise traders are governed by a stochastic version of the logistic growth model (see here, here, or here for examples), then the stationary distribution for the number of noise traders will be Exponential. And, the right way to learn about the mean of a Gaussian random variable whose variance is drawn from an Exponential distribution is to use the LASSO, a penalized regression which delivers point estimates that are precisely zero whenever the unpenalized estimate is sufficiently small.

Inference Problem

Here’s is the inference problem I’m going to study. Suppose there’s a stock with price $P$ , and there are $N > 0$ noise traders present in this market. And, assume that this price is a linear function of three variables:

(1) $\begin{equation*} P = \alpha + \beta \cdot F + \gamma \cdot \{C - S\} \end{equation*}$

Above, $F$ denotes the stock’s fundamental value, $C \sim \mathrm{Normal}(0, \, N)$ denotes noise due to the madness of crowds, and $S \sim \mathrm{Normal}(0, \, \sigma^2)$ denotes noise due to random supply shocks. The negative sign on supply noise comes from the fact that more supply means lower prices. You can think about the supply noise as the result of hedging demand or rebalancing cascades. The source doesn’t matter. The key point is that this noise source has constant variance.

Crowd noise is different, though. Its variance is equal to the number of noise traders in the market, $N$ , and this population can change. Suppose there are $n = 1, \, \ldots, \, N$ noise traders, and each individual trader in this crowd has demand that’s iid normal:

(2) $\begin{equation*} c_n \overset{\scriptscriptstyle \text{iid}}{\sim} \mathrm{Normal}(0, \, 1) \end{equation*}$

Then, the aggregate demand of the entire crowd of noise traders has distribution:

(3) $\begin{equation*} C \overset{\scriptscriptstyle \text{def}}{=} {\textstyle \sum_{n=1}^N} \, c_n \sim \mathrm{Normal}(0, \, N) \end{equation*}$

In this setting, the rescaled pricing error, $\tilde{P} \overset{\scriptscriptstyle \text{def}}{=} \sfrac{1}{\gamma} \cdot \{P - \alpha - \beta \cdot F\}$ , is a normally distributed signal about the aggregate demand coming from the crowd of noise traders:

(4) $\begin{equation*} \tilde{P} = C - S \end{equation*}$

When the price is above its fundamental value, $\{P - \alpha - \beta \cdot F\} > 0$ , it must be because either the crowd of noise traders has high demand, $C > 0$ , or there is unexpectedly low supply, $S < 0$ . The question I want to answer below is: how hard is it for traders to figure out which noise source is responsible?

Bayes rule tells us how to learn about the crowd’s aggregate demand from the equilibrium pricing error:

(5) $\begin{equation*} \mathrm{Pr}(C|\tilde{P}) \propto \mathrm{Pr}(\tilde{P}|C) \times {\textstyle \int_0^{\infty}} \, \mathrm{Pr}(C|N) \cdot \mathrm{Pr}(N) \cdot \mathrm{d}N \end{equation*}$

Supply noise is normally distributed. This means we know how to calculate $\mathrm{Pr}(\tilde{P}|C)$ . So, if we knew the distribution of the number of noise traders in the market, then we could evaluate the remaining integral and solve for the most likely value for the crowd’s aggregate demand given the observed pricing error:

(6) $\begin{equation*} \hat{C}(\tilde{P}) \overset{\scriptscriptstyle \text{def}}{=} \underset{C \in \mathrm{R}}{\arg\min} \left\{ \, \log \mathrm{Pr}(\tilde{P}|C) + \log {\textstyle \int_0^{\infty}} \, \mathrm{Pr}(C|N) \cdot \mathrm{Pr}(N) \cdot \mathrm{d}N \, \right\} \end{equation*}$

Population Size

There are many ways that you could model the size of the noise-trader crowd. One way to go would be to use a population-dynamics model from the ecology literature, such as the stochastic version of the logistic growth model. This model is specifically designed to explain the unexpected booms and busts that we seen in wildlife populations. If we take this approach, then the number of noise traders, $N(t)$ , is governed by the following stochastic differential equation:

(7) $\begin{equation*} \mathrm{d}N = \theta \cdot \{ \mu - N \} \cdot N \cdot \mathrm{d}t - \delta \cdot N \cdot \mathrm{d}t + \varsigma \cdot N \cdot \mathrm{d}W \end{equation*}$

In the equation above, $\theta \cdot \{\mu - N\}$ denotes the rate at which the crowd of noise traders grows, $\delta > 0$ denotes the rate at which noise traders lose interest, $\mathrm{d}W$ is a Wiener process capturing random fluctuations in the number of noise traders in the crowd, and $\varsigma > 0$ denotes the volatility of these random fluctuations.

The key property of the logistic growth model is that it’s nonlinear. Population growth, $\theta \cdot \{ \mu - N \} \cdot N$ , is a quadratic function of the number of noise traders as shown in the figure below. This nonlinearity is what allows the model to generate population booms and busts. This nonlinearity will occur if existing noise traders try to recruit their friends to enter the market as well (see here, here, here, or here). $\mu > 0$ denotes the typical number of noise traders that could potentially be persuaded to join the crowd. And, $\theta > 0$ captures the intensity with which existing noise traders persuade their remaining $\{\mu - N\}$ friends to join.

Thus, when there are only a handful of noise traders, the crowd grows slowly because there aren’t many traders to do the recruiting. As the crowd gets larger, population growth increases. But, this growth eventually slows down again because, when there are already lots of noise traders in the market, it is hard to increase the size of the crowd because there aren’t many traders left to be recruited, $\{\mu - N\} \approx 0$ .

Because the logistic growth model has been studied for so long, the population-size distribution that it generates is well known. There are two regimes: $\theta \cdot \mu < \delta$ and $\theta \cdot \mu > \delta$ . If $\theta \cdot \mu < \delta$ , then population of noise traders will eventually die out, $\lim_{t\to\infty}N = 0$ . To see why, think about how the system behaves as the crowd size gets small. When $N = \epsilon \approx 0$ , the crowd grows at an almost linear rate, $\theta \cdot \mu \cdot \epsilon +\mathcal{O}(\epsilon^2)$ . So, $\theta \cdot \mu < \delta$ means that, when the crowd gets small enough, existing noise traders lose interest faster then they can recruit their friends, which leads to the end of the crowd, $N=0$ . By contrast, if the growth rate is larger than the rate of decay when $N$ is small, $\theta \cdot \mu > \delta$ , then the population of noise traders will never get all the way to $N=0$ . And, if we rescale the units so that $\sfrac{\varsigma^2}{2} = \theta \cdot \mu - \delta$ , then we will find that number of noise traders in the crowd will be governed by an Exponential distribution with rate parameter $\sfrac{\lambda^2}{2}$ :

(8) $\begin{equation*} N \sim \mathrm{Exponential}(\sfrac{\lambda^2}{2}) \qquad \text{where} \qquad \lambda = \sqrt{{\textstyle \frac{2}{\mu} \cdot \left\{ \frac{\theta \cdot \mu}{\theta \cdot \mu - \delta} \right\}}} \end{equation*}$

The figure above shows the probability-density function for an Exponential distribution. It illustrates how, when the rate parameter is larger, the size of the noise-trader crowd tends to be smaller. And, the functional form for $\lambda$ reveals that the rate parameter will be largest as $\theta \cdot \mu \searrow \delta$ —i.e., when the growth rate is barely larger than the decay rate when the crowd size is small. Makes sense.

Neglected Risk

We now have our distribution for the number of noise traders in the market. So, we can return to our original inference problem and try to solve for the most likely value of the crowd’s aggregate demand given the observed pricing error, $\hat{C}(\tilde{P})$ . It turns out that, if the variance of the crowd’s aggregate demand is drawn from an Exponential distribution, then it’s easy to solve the integral in Equation (6). Andrews and Mallows shows that:

(9) $\begin{equation*} \frac{\lambda}{2} \cdot e^{- \, \lambda \cdot |C|} = \int_0^\infty \, \underset{C|N \sim \mathrm{Normal}(0, \, N)}{ \left\{ \frac{1}{\sqrt{2 \cdot \pi \cdot N}} \cdot e^{-\, \frac{\{C-0\}^2}{2 \cdot N}} \right\} } \cdot \underset{N \sim \mathrm{Exponential}(\sfrac{\lambda^2}{2})}{ \left\{ \frac{\lambda^2}{2} \cdot e^{-\,\frac{\lambda^2}{2} \cdot N} \right\} } \cdot \mathrm{d}N \end{equation*}$

So, if we set $\sfrac{\lambda}{2} \cdot e^{-\lambda \cdot |C|} = {\textstyle \int_0^{\infty}} \, \mathrm{Pr}(C|N) \cdot \mathrm{Pr}(N) \cdot \mathrm{d}N$ , then the optimization problem in Equation (6) simplifies:

(10) $\begin{equation*} \hat{C}(\tilde{P}) = \underset{C \in \mathrm{R}}{\arg\min} \left\{ \, \frac{1}{2 \cdot \sigma^2} \cdot \{\tilde{P} - C\}^2 + \lambda \cdot | C | \, \right\} \end{equation*}$

This simplification is really cool because the optimization problem above is just the optimization problem for the LASSO with a penalty parameter of $\sigma^2 \cdot \lambda$ (see Park and Casella).

There’s something weird about this result, though. Using the LASSO implies that:

(11) $\begin{equation*} \hat{C}(\tilde{P}) = \mathrm{Sign}(\tilde{P}) \cdot \left\{ \, |\tilde{P}| - \sigma^2 \cdot \lambda \, \right\}_+ \end{equation*}$

Above, $\{ x \}_+ = x$ if $x > 0$ and $0$ . So, if the observed pricing error is relatively small, $|\tilde{P}| < \sigma^2 \cdot \lambda$ , then a fully-rational trader will walk away from the market believing that $\hat{C}(\tilde{P}) = 0$ . He will completely neglect the risk coming from the crowd of noise traders.

Here’s the logic behind this neglect. If the pricing error is small, then there must not be very many noise traders in the market. If there aren’t very many noise traders, then they can’t affect the equilibrium price very much. And, this means that there’s no way for traders to learn about aggregate noise-trader demand from the equilibrium price, which means that there’s no reason for them to revise their beliefs about noise-trader demand away from zero after seeing the price.

What’s more, this line of reasoning is consistent with the functional form for the LASSO’s penalty parameter, $\sigma^2 \cdot \lambda$ . This expression says that traders will ignore the madness of crowds in the face of more extreme pricing errors (larger values of $|\tilde{P}|$ ) when either the crowd of noise traders tends to be smaller ( $\lambda$ is larger) or it’s easier to explain pricing errors with supply noise ( $\sigma^2$ is larger). And, this basic intuition should carry over to other situations where the size of the crowd of noise traders has some other fat-tailed distribution rather than an Exponential distribution.

A Tell-Tale Sign of Short-Run Trading

January 26, 2017 by Alex

Motivation

Trading has gotten a lot faster over the last two decades. The term “short-run trader” used to refer to people who traded multiple times a day. Now, it refers to algorithms that trade multiple times a second.

Some people are worried about this new breed of short-run trader making stock prices less informative about firm fundamentals by trading so often on information that’s unrelated to companies’ long-term prospects. But, this is a red herring. By the same logic, market-neutral strategies should be making market indexes like the Russell 3000 less informative about macro fundamentals. And, no one believes this.

Short-run traders aren’t ignoring fundamentals; they’re learning about fundamentals before everyone else by studying order flow. And, this post shows why they also make trading volume look asymmetric and lumpy as a result. The logic is simple. If short-run traders get additional information from order flow, then they’ll use this information to cluster their trading at times when everyone else is moving in the opposite direction.

Benchmark Model

Consider a market with a single company that’s going to pay a dividend, $d_t$ , in future periods $t = 1, \, 2, \, \ldots$ And, suppose that there’s a unit mass of small agents, $i \in (0, \, 1]$ , who have noisy priors about these dividends,

$\begin{equation*} d_t \overset{\scriptscriptstyle \text{iid}}{\sim} \mathrm{N}( \mu_t^{(i)}, \, \sfrac{\sigma^2\!}{2} ) \quad \text{for some} \quad \sigma^2 > 0, \end{equation*}$

which are correct on average, $d_t = \int_0^1 \mu_t^{(i)} \cdot \mathrm{d}i$ . This assumption means that, by aggregating agents’ demand, equilibrium prices can contain information about dividends that isn’t known by any individual agent.

This unit mass of agents is split into two different groups: long-term investors and short-run traders. At time $t=0$ , each group of agents trades shares in its own separate fund, $f \in \{L, \, H\}$ , that offers frequency-specific exposure to the company’s dividends at times $t=1,\,2$ . The long-term investors, $i \in (0, \, \sfrac{1}{2}]$ , trade the low-frequency fund which has a payout:

$\begin{equation*} d_L \overset{\scriptscriptstyle \text{def}}{=} d_1 + d_2 \end{equation*}$

And, the short-run traders, $i \in (\sfrac{1}{2}, \, 1]$ , trade the high-frequency fund which has a payout:

$\begin{equation*} d_H \overset{\scriptscriptstyle \text{def}}{=} d_1 - d_2 \end{equation*}$

At time $t=0$ , each agent observes the equilibrium price of a frequency-specific fund, $p_f$ , and then chooses the number of shares to buy, $x_f^{(i)}$ , in order to maximize his expected utility at the end of time $t=2$ :

(1) $\begin{equation*} \max_{x_f^{(i)}} \mathrm{E}^{(i)}\left[ \, - e^{ \, - \alpha \cdot \{d_f - p_f\} \cdot x_f^{(i)} \, } \, \middle| \, p_f \, \right] \quad \text{for some} \quad \alpha > 0 \end{equation*}$

Above, $\mathrm{E}^{(i)}[\cdot|p_f]$ denotes agent $i$ ‘s conditional expectation, and $\alpha$ denotes his risk-aversion parameter.

Let $z_t$ denote the number of shares of the company’s stock that are available for purchase at time $t$ . We say that “markets clear” at time $t=0$ if the dividend payout from each available share at times $t=1,\,2$ has been unambiguously assigned to exactly one trader via their fund holdings:

(2) $\begin{align*} {\textstyle \int_0^{\sfrac{1}{2}}} x_L^{(i)} \cdot \mathrm{d}i + {\textstyle \int_{\sfrac{1}{2}}^1} x_H^{(i)} \cdot \mathrm{d}i &= z_1 \\ \text{and} \quad {\textstyle \int_0^{\sfrac{1}{2}}} x_L^{(i)} \cdot \mathrm{d}i - {\textstyle \int_{\sfrac{1}{2}}^1} x_H^{(i)} \cdot \mathrm{d}i &= z_2 \end{align*}$

Let $z_L \overset{\scriptscriptstyle \text{def}}{=} z_1 + z_2$ and $z_H \overset{\scriptscriptstyle \text{def}}{=} z_1 - z_2$ denote the number of available shares at each frequency.

An equilibrium then consists of a demand rule, $\mathrm{X}(\mathrm{E}^{(i)}[d_f|p_f], \, p_f) = x_f^{(i)}$ , and a price function, $\mathrm{P}(d_f, \, z_f) = p_f$ , such that 1) demand maximizes the expected utility of each agent given the price and 2) markets clear.

Because the equilibrium price of each fund only depends on its promised payout and the number of available shares, if agents knew the number of available shares, then they could reverse engineer a fund’s future payout at times $t=1, \, 2$ by studying its equilibrium price at time $t=0$ . And, an equilibrium in such a market wouldn’t be well-defined. So, to make sure that equilibrium prices at time $t=0$ aren’t fully revealing, let’s assume that the number of available shares in each period is a random variable:

$\begin{equation*} z_t \overset{\scriptscriptstyle \text{iid}}{\sim} \mathrm{N}( 0, \, \sfrac{1}{2}) \end{equation*}$

This means thinking about “available” shares as shares that haven’t already been purchased by noise traders.

The key fact about this benchmark model is that agents don’t use order-flow information at time $t=1$ to update their time $t=0$ beliefs. As a result, the model really isn’t about short-run traders in spite of how the variables are named. Crouzet/Dew-Becker/Nathanson shows that with some clever relabeling you could just as easily think about the low- and high-frequency funds as index and market-neutral funds, respectively.

Trading Volume

Although agents are only active at time $t=0$ in the benchmark model, each fund has to trade at times $t=1,\,2$ in order to deliver frequency-specific payouts. Let’s use $x_L$ and $x_H$ to denote aggregate demand:

$\begin{align*} x_L &\overset{\scriptscriptstyle \text{def}}{=} {\textstyle \int_0^{\sfrac{1}{2}}} x_L^{(i)} \cdot \mathrm{d}i \\ \text{and} \quad x_H &\overset{\scriptscriptstyle \text{def}}{=} {\textstyle \int_{\sfrac{1}{2}}^1} \, x_H^{(i)} \cdot \mathrm{d}i \end{align*}$

To deliver $d_L$ to every one of its shareholders, the low-frequency fund has to buy $x_L$ shares of the company’s stock between times $t=0$ and $t=1$ and then liquidate this position between $t=2$ and $t=3$ . And, to deliver $d_H$ to every one of its shareholders, the high-frequency fund has to buy $x_H$ shares between times $t=0$ and $t=1$ , sell $2 \cdot x_H$ shares between $t=1$ and $t=2$ , and then buy back $x_H$ shares between times $t=2$ and $t=3$ .

So, trading volume in the benchmark model is:

$\begin{align*} \mathit{vlm}_{0|1} &\overset{\scriptscriptstyle \text{def}}{=} |x_L| + |x_H| \\ \mathit{vlm}_{1|2} &\overset{\scriptscriptstyle \text{def}}{=} 2 \cdot |x_H| \\ \text{and} \quad \mathit{vlm}_{2|3} &\overset{\scriptscriptstyle \text{def}}{=} |x_L| + |x_H| \end{align*}$

The key thing to notice is that trading volume is symmetric, $\mathit{vlm}_{0|1} = \mathit{vlm}_{2|3}$ , because short-run traders don’t get any new information after time $t=0$ .

Model Solution

So, how many shares of each frequency-specific fund are agents going to demand in the benchmark model? To solve the model and answer this question, let’s first guess that the price function is linear:

$\begin{equation*} \mathrm{P}(d_f, \, z_f) = d_f - \sqrt{\sfrac{\sigma^2\!}{\mathit{SNR}}} \cdot z_f \quad \text{for some} \quad \mathit{SNR} > 0 \end{equation*}$

This guess introduces a new parameter, $\mathit{SNR}$ , which represents the signal-to-noise ratio of fund prices at time $t=0$ . If this parameter is large, then the time $t=0$ prices of the low- and high-frequency funds will reveal a lot of the information about the company’s time $t=1, \, 2$ dividend payouts.

Here’s the upshot of this guess. It implies that each fund’s price is a normally-distributed signal about its future payout, $d_f \sim \mathrm{N}(p_f, \, \sfrac{\sigma^2\!}{\mathit{SNR}})$ . And, with normally-distributed signals, we know how to compute agents’ posterior beliefs about $d_f$ after seeing $p_f$ :

$\begin{equation*} \mathrm{E}^{(i)}[d_f|p_f] = {\textstyle \left\{ \frac{1}{1 + \mathit{SNR}}\right\}} \cdot \mu_f^{(i)} + {\textstyle \left\{ \frac{\mathit{SNR}}{1 + \mathit{SNR}}\right\}} \cdot p_f \end{equation*}$

Above, $\mu_f^{(i)}$ denotes agent $i$ ‘s priors about a particular fund, either $\mu_L^{(i)} \overset{\scriptscriptstyle \text{def}}{=} \mu_1^{(i)} + \mu_2^{(i)}$ or $\mu_H^{(i)} \overset{\scriptscriptstyle \text{def}}{=} \mu_1^{(i)} - \mu_2^{(i)}$ . And, we can then use these posterior beliefs to compute agents’ equilibrium demand rule by solving the first-order condition of Equation (1) with respect to $x_f^{(i)}$ :

(3) $\begin{equation*} \mathrm{X}(\mathrm{E}^{(i)}[d_f|p_f], \, p_f) = \{1 + \mathit{SNR}\} \cdot {\textstyle \big\{ \frac{\mathrm{E}^{(i)}[d_f|p_f] - p_f}{\sigma} \big\}} \cdot {\textstyle \big\{ \frac{1}{\alpha \cdot \sigma} \big\}} \end{equation*}$

Finally, to verify that our original guess about a linear price function was correct, we can plug this equilibrium demand rule into the market-clearing conditions in Equation (2) and solve for $p_f$ :

$\begin{equation*} \mathrm{P}(d_f, \, z_f) = d_f - \alpha \cdot \sigma^2 \cdot z_f \end{equation*}$

The resulting price function is indeed linear, so our solution is internally consistent (…though not unique). And, by matching coefficients, we can solve for the equilibrium signal-to-noise ratio, $\mathit{SNR} = \{ \alpha \cdot \sigma \}^{-2}$ . In other words, fund prices at time $t=0$ reveal more information about a company’s dividend at times $t=1, \, 2$ when agents are less risk averse ( $\alpha$ small) or when they have more precise priors ( $\sigma$ small).

Order-Flow Info

How would this solution have to change if short-run traders could learn from time $t=1$ order flow?

For markets to clear at time $t=1$ , the aggregate demand for the low-frequency fund plus the aggregate demand for the high-frequency fund has to equal the total number of available shares, $x_L + x_H = z_1$ . And, from Equation (3), we know that the aggregate demand for the low-frequency fund is related to the company’s total dividend payout:

$\begin{equation*} x_L = \sqrt{\sfrac{\mathit{SNR}}{\sigma^2}} \cdot \{ d_L - p_L \} \end{equation*}$

So, by looking at the time $t=1$ order flow, short-run traders can get a signal about the company’s time $t=2$ dividend since $d_2 = d_L - d_1$ :

$\begin{equation*} d_2 \sim \mathrm{N}\left( \{p_L - d_1\} - \sqrt{\sfrac{\sigma^2\!}{\mathit{SNR}}} \cdot x_H, \, \sfrac{\sigma^2\!}{\mathit{SNR}} \right), \end{equation*}$

And, this information about the time $t=2$ dividend is helpful since $\mathrm{Cov}[d_2, \, d_H] = - \,\sigma^2$ .

With this additional signal, the short-run traders who previously invested in the high-frequency fund would now rather trade the company’s stock directly at times $t=1, \, 2$ . Let $\tilde{x}_H$ denote their demand at time $t=2$ . When they observe a high price for the low-frequency asset at time $t=0$ , $p_L > 0$ , and a large dividend payout at time $t=1$ , $d_1 > 0$ , they know that $d_2$ is likely small. And, as a result, they’ll short more shares at time $t=2$ , $|\tilde{x}_H| > |x_H|$ . By contrast, when they observe a high price for the low-frequency asset at time $t=0$ , $p_L > 0$ , and a small dividend payout at time $t=1$ , $d_1 < 0$ , they know that $d_2$ is likely large. So, they’ll short fewer shares at time $t=2$ , $|\tilde{x}_H| < |x_{H}|$ .

Either way, trading volume is going to look asymmetric and lumpy as a result, with relatively more of the trading volume clustered at one of the end points. If $p_L > 0$ and $d_1 > 0$ , then relatively more of the trading volume will occur at between time $t=2$ and $t=3$ because $\mathit{vlm}_{0|1}$ is unchanged and:

$\begin{equation*} \mathit{vlm}_{2|3} = |x_L| + |x_H| < |x_L| + |\tilde{x}_H| \overset{\scriptscriptstyle \text{def}}{=} \widetilde{\mathit{vlm}}_{2|3} \end{equation*}$

Whereas, if $p_L > 0$ and $d_1 < 0$ , then relatively more of the trading volume will occur between time $t=0$ and $t=1$ because $\mathit{vlm}_{0|1}$ is unchanged and now $\mathit{vlm}_{2|3} > \widetilde{\mathit{vlm}}_{2|3}$ .

What’s more, to long-term investors who can’t see short-run order flow, short-run traders are going to add execution risk. The price at which the low-frequency fund executes its time $t=2$ orders will now vary. And, this variation will be related to the magnitude (but not the sign) of their time $t=0$ demand.

Finally, note that this analysis shows why it’s easier to model indexers and stock pickers than long-term investors and short-run traders. An equilibrium in either model has to contain a demand rule and a price function (e.g., see the setup in the benchmark model). But, an equilibrium in a model with multi-frequency trade also has to contain a rule for how long-term investors think short-run traders will affect their order execution. And, this rule is the crux of any model with long-term investors and short-run traders.

The Tension Between Learning and Predicting

January 24, 2017 by Alex

1. Motivation

Imagine we’re traders in a market where the cross-section of returns is related to $V \geq 1$ variables:

$\begin{align*} r_s = \alpha^\star + {\textstyle \sum_v} \, \beta_v^{\star} \cdot x_{s,v} + \epsilon_s^{\star}. \end{align*}$

In the equation above, $\alpha^\star$ denotes the mean return, and each $\beta_v^\star$ captures the relationship between returns and the $v$ th right-hand-side variable. Some notation: I’m going to be using a “star” to denote true parameter values and a “hat” to denote in-sample estimates of these values. e.g., $\hat{\beta}_v$ denotes an in-sample estimate of $\beta_v^\star$ . To make things simple, let’s assume that $\sum_s x_{s,v} = 0$ , $\sum_s x_{s,v}^2 = S$ , $\sum_s x_{s,v} \cdot x_{s,v'} = 0$ , and $\epsilon_s^{\star} \overset{\scriptscriptstyle \text{iid}}{\sim} \mathrm{N}(0, \, \sigma^2)$ .

Notice that learning about the most likely parameter values,

$\begin{align*} \{ \hat{\alpha}^{\text{ML}}, \, \hat{\beta}_1^{\text{ML}}, \, \ldots, \, \hat{\beta}_V^{\text{ML}} \} &= \arg \max_{\alpha, \, \beta_1, \, \ldots, \, \beta_V} \left\{ \, \mathrm{Pr}(\mathbf{r} | \{\alpha, \, \beta_1, \, \ldots, \, \beta_V\}) \, \right\}, \end{align*}$

is really easy in this setting because $\epsilon_s^{\star} \overset{\scriptscriptstyle \text{iid}}{\sim} \mathrm{N}(0, \, \sigma^2)$ . These maximum-likelihood estimates are just the coefficients from a cross-sectional regression,

$\begin{align*} r_s = \hat{\alpha}^{\text{ML}} + {\textstyle \sum_v} \, \hat{\beta}_v^{\text{ML}} \cdot x_{s,v} + \hat{\epsilon}_s^{\text{ML}}. \end{align*}$

So, finding $\{ \hat{\alpha}^{\text{ML}}, \, \hat{\beta}_1^{\text{ML}}, \, \ldots, \, \hat{\beta}_V^{\text{ML}} \}$ is homework question from Econometrics 101. Nothing could be simpler.

But, what if we’re interested in predicting returns,

$\begin{align*} \min_{\alpha, \, \beta_1, \, \ldots, \, \beta_V} \left\{ \, {\textstyle \frac{1}{S}} \cdot {\textstyle \sum_s} \, \left( r_s - [\alpha + {\textstyle \sum_v} \, \beta_v \cdot x_{s,v}] \right)^2 \, \right\}, \end{align*}$

rather than learning about the most likely parameter values? It might seem like this is the same exact problem. And, if we’re only considering $1$ right-hand-side variable, then it is the same exact problem. When $V = 1$ the best predictions come from using $\{\hat{\alpha}^{\text{ML}}, \, \hat{\beta}_1^{\text{ML}}\}$ . But, it turns out that when there’s $2$ or more right-hand-side variables (and an unknown mean), this is no longer true. When $V \geq 2$ we can make better predictions with less likely parameters. When $V \geq 2$ there’s a tension between the learning and predicting.

Why? That’s the topic of today’s post.

2. Maximum Likelihood

Finding the most likely (ML) parameter values is equivalent to minimizing the negative log likelihood. So, because we’re assume that $\epsilon_s^\star \overset{\scriptscriptstyle \text{iid}}{\sim} \mathrm{N}(0, \, \sigma^2)$ , this is just

$\begin{align*} - \, \log \mathrm{Pr}(\mathbf{r} | \{\alpha, \, \beta_1, \, \ldots, \, \beta_V\}) &= {\textstyle \frac{1}{2 \cdot (S \cdot \sigma^2)}} \cdot {\textstyle \sum_s} \left(r_s - [\alpha + {\textstyle \sum_v} \, \beta_v \cdot x_{s,v}] \right)^2 + \cdots \end{align*}$

where the “ $+ \cdots$ ” at the end denotes a bunch of terms that don’t include any of the parameters that we’re optimization over. Optimizing each parameter value then gives the first-order conditions below:

$\begin{align*} 0 &= {\textstyle \frac{1}{S}} \cdot {\textstyle \sum_s} \left( r_s - [\hat{\alpha}^{\text{ML}} + {\textstyle \sum_v} \, \hat{\beta}_v^{\text{ML}} \cdot x_{s,v}] \right) \cdot 1 \\ \text{and} \quad 0 &= {\textstyle \frac{1}{S}} \cdot {\textstyle \sum_s} \left( r_s - [\hat{\alpha}^{\text{ML}} + {\textstyle \sum_v} \, \hat{\beta}_v^{\text{ML}} \cdot x_{s,v}] \right) \cdot x_{s,v} \quad \text{for each } v = 1, \, \ldots, \, V. \end{align*}$

And, solving this system of $(V+1)$ equations and $(V+1)$ unknowns gives the most likely parameter values:

$\begin{align*} \hat{\alpha}^{\text{ML}} &= {\textstyle \frac{1}{S}} \cdot {\textstyle \sum_s} \, r_s \cdot 1 \\ \text{and} \quad \hat{\beta}_v^{\text{ML}} &= {\textstyle \frac{1}{S}} \cdot {\textstyle \sum_s} \, (r_s - \hat{\alpha}^{\text{ML}}) \cdot x_{s,v} \quad \text{for each } v = 1, \, \ldots, \, V. \end{align*}$

Clearly, the most likely parameter values are just the coefficients from a cross-sectional regression.

Now, for the sake of argument, let’s imagine there’s an oracle who knows the true parameter values, $\{ \alpha^\star, \, \beta_1^\star, \, \ldots, \, \beta_V^\star \}$ . With access to this oracle, we can compute our mean squared error when using the maximum-likelihood estimates to predict returns given any choice of true parameter values:

$\begin{align*} \mathrm{E} \left[ \, \left( r_s - [\hat{\alpha}^{\text{ML}} + {\textstyle \sum_v} \, \hat{\beta}_v^{\text{ML}} \cdot x_{s,v}] \right)^2 \, \middle| \, \{ \alpha^\star, \, \beta_1^\star, \, \ldots, \, \beta_V^\star \} \, \right] &= \sigma^2 + 1 \cdot (\sfrac{\sigma^2}{S}) + V \cdot (\sfrac{\sigma^2}{S}). \end{align*}$

The first term, $\sigma^2$ , captures the unavoidable error. i.e., even if we knew the true parameter values, we still wouldn’t be able to predict $\epsilon_s \overset{\scriptscriptstyle \text{iid}}{\sim} \mathrm{N}(0, \, \sigma^2)$ . And, the second and third terms, $1 \cdot (\sfrac{\sigma^2}{S})$ and $V \cdot (\sfrac{\sigma^2}{S})$ , capture the error that comes from using the most likely parameter values rather than the true parameter values.

3. A Better Predictor

With this benchmark in mind, let’s take a look at a variant of the James-Stein (JS) estimator:

$\begin{align*} \hat{\beta}_v^{\text{JS}} &\overset{\scriptscriptstyle \text{def}}{=} (1 - \lambda) \cdot \hat{\beta}_v^{\text{ML}} \quad \text{for each } v = 1, \, \ldots, \, V. \end{align*}$

In the definition above, $\lambda \in [0, \, 1]$ denotes a bias factor that shrinks the maximum-likelihood estimates towards zero whenever $\lambda > 0$ . So, with access to an oracle, we can compute our mean squared error when using the James-Stein estimates to predict returns just like before:

$\begin{align*} \mathrm{E} \left[ \, \left( r_s - [\hat{\alpha}^{\text{ML}} + {\textstyle \sum_v} \, \hat{\beta}_v^{\text{JS}} \cdot x_{s,v}] \right)^2 \, \middle| \, \{ \alpha^\star, \, \beta_1^\star, \, \ldots, \, \beta_V^\star \} \, \right] &= \sigma^2 + 1 \cdot (\sfrac{\sigma^2}{S}) \\ &\quad + V \cdot \left\{ \, (1 - \lambda)^2 \cdot (\sfrac{\sigma^2}{S}) + \lambda^2 \cdot ({\textstyle \frac{1}{V}} \cdot {\textstyle \sum_v} \, |\beta_v^\star|^2) \, \right\}. \end{align*}$

Now the third term is more complicated. If we use a more biased estimator, $\lambda \to 1$ and $|\hat{\beta}_v^{\text{JS}}| \to 0$ , then using the most likely parameter values rather than the true parameter values to predict returns is going to cause less damage. But, bias is going to generate really bad predictions whenever the true parameter value is large $|\beta_v^\star| \gg 0$ . The $(1 - \lambda)^2 \cdot (\sfrac{\sigma^2}{S})$ and $\lambda^2 \cdot (\frac{1}{V} \cdot \sum_v \, |\beta_v^\star|^2)$ terms capture these two opposing forces.

Comparing the maximum-likelihood and James-Stein prediction errors reveals that we should prefer the James-Stein estimator to the maximum-likelihood estimator if there’s a $\lambda \in (0, \, 1]$ such that:

$\begin{align*} (1 - \lambda)^2 \cdot (\sfrac{\sigma^2}{S}) + \lambda^2 \cdot ({\textstyle \frac{1}{V}} \cdot {\textstyle \sum_v} \, |\beta_v^\star|^2) < (\sfrac{\sigma^2}{S}). \end{align*}$

But, here’s the thing: if we have access to an oracle, then there’s always going to be some $\lambda > 0$ that satisfies this inequality. This is easier to see if we rearrange things a bit:

$\begin{align*} {\textstyle \frac{ \frac{1}{V} \cdot \sum_v \, |\beta_v^\star|^2 }{ (\sfrac{\sigma^2}{S}) } } < {\textstyle \frac{ 2 - \lambda }{ \lambda } }. \end{align*}$

So, there’s always a sufficiently small $\lambda$ such that this inequality holds. Thus, for any $\{ \alpha^\star, \, \beta_1^\star, \, \ldots, \, \beta_V^\star \}$ , there’s some $\lambda$ such that the James-Stein estimates gives better predictions than the most likely parameter values.

4. Abandoning the Oracle

Perhaps this isn’t a useful comparison? In the real world, we can’t see the true parameter values when deciding which estimator to use. So, we can’t know ahead of time whether or not we’ve picked a small enough $\lambda$ . It turns out that having to estimate $\lambda$ changes things, but only when there’s just $V=1$ right-hand-side variable. When there are $V \geq 2$ variables, James-Stein with an estimated $\lambda$ still gives better predictions.

To see where this distinction comes from, let’s first solve for the optimal choice of $\lambda$ when we still have access to the oracle. This will tell us what we have to estimate when we abandon the oracle. The optimal choice of $\lambda$ solves:

$\begin{align*} \lambda^\star &= \arg \min_{\lambda \in [0, \, 1]} \left\{ \, (1 - \lambda)^2 \cdot (\sfrac{\sigma^2}{S}) + \lambda^2 \cdot \left( {\textstyle \frac{1}{V}} \cdot {\textstyle \sum_v} \, |\beta_v^\star|^2 \right) \, \right\} \end{align*}$

So, if we differentiate, then we can solve the first-order condition for $\lambda^\star$ :

$\begin{align*} \lambda^\star &= {\textstyle \frac{ (\sfrac{\sigma^2}{S}) }{ \frac{1}{V} \cdot \sum_v \, |\beta_v^\star|^2 - (\sfrac{\sigma^2}{S}) } }. \end{align*}$

If the maximum-likelihood estimates are really noisy relative to the size of the true parameter values (i.e., $\sfrac{\sigma^2}{S}$ is close to $\frac{1}{V} \cdot \sum_v \, |\beta_v^\star|^2$ ), then using the most likely parameter values rather than the true parameter values is going to increase our prediction error a lot. So, we should use more biased parameter estimates.

But, notice what this formula is telling us. To estimate the right amount of bias, all we have to do is estimate the variance of the true parameter values. We don’t have to estimate every single one. So, we can estimate the variance of the true parameter values as follows,

$\begin{align*} {\textstyle \frac{1}{V-1}} \cdot {\textstyle \sum_v} \, |\hat{\beta}_v^{\text{ML}}|^2 &= (\sfrac{\sigma^2}{S}) + {\textstyle \frac{1}{V}} \cdot {\textstyle \sum_v} \, |\beta_v^\star|^2, \end{align*}$

where the factor of $(V - 1)$ on the left-hand side is a degrees-of-freedom correction. To see why we need this correction, recall that

$\begin{align*} \mathrm{E} \left[ \, \left( r_s - [\hat{\alpha}^{\text{ML}} + {\textstyle \sum_v} \, \hat{\beta}_v^{\text{ML}} \cdot x_{s,v}] \right)^2 \, \right] &= \sigma^2 + 1 \cdot (\sfrac{\sigma^2}{S}) + V \cdot (\sfrac{\sigma^2}{S}), \end{align*}$

so not all $V$ maximum-likelihood estimates can move independently. There is an adding-up constraint.

Thus, if we have to estimate the right amount of bias to use, then we should choose:

$\begin{align*} \hat{\lambda} &= {\textstyle \frac{ (\sfrac{\sigma^2}{S}) }{ \frac{1}{V-1} \cdot \sum_v \, |\hat{\beta}_v^{\text{ML}}|^2 } }. \end{align*}$

Notice that, when we have to estimate the right amount of bias, $\hat{\lambda} > 0$ only if $V \geq 2$ . If $V = 1$ , then the denominator is infinite and $\hat{\lambda} = 0$ . After all, if there’s only $1$ right-hand-side variable, then the equation to estimate the variance of the true parameter values has the same first-order condition as the equation to estimate $\hat{\beta}_1^{\text{ML}}$ . With this estimated amount of bias, our prediction error becomes

$\begin{align*} \mathrm{E} \left[ \, \left( r_s - [\hat{\alpha}^{\text{ML}} + {\textstyle \sum_v} \, \hat{\beta}_v^{\text{JS}} \cdot x_{s,v}] \right)^2 \, \right] &= \sigma^2 + 1 \cdot (\sfrac{\sigma^2}{S}) + (1 - \lambda)^2 \cdot (\sfrac{\sigma^2}{S}), \end{align*}$

which is always less than the maximum-likelihood prediction error whenever $V \geq 2$ .

5. What This Means

My last post looked at one reason why it’s harder to predict the cross-section of returns when $V \geq 2$ : Bayesian variable selection doesn’t scale. If we’re not sure which subset of variables actually predict returns, then finding the subset of variables that’s the most likely to predict returns means solving a non-convex optimization problem. It turns out that solving this optimization problem means doing an exhaustive search over the powerset containing all $2^V$ possible subsets of variables. And, this just isn’t feasible when $V \gg 2$ .

But, this scaling problem isn’t the only reason why it’s harder to predict the cross-section of returns when $V \geq 2$ . And, this post points out another one of these reasons: even if you could solve this non-convex optimization problem and find the most likely parameter values, these parameter values wouldn’t give the best predictions. When $V \geq 2$ , there’s a fundamental tension between making good predictions and learning about the most likely parameter values in the data-generating process for returns. So, when $V \geq 2$ traders are going to solve the prediction problem and live with the resulting biased beliefs about the underlying parameter values. What’s more, the $\hat{\lambda}$ with the best out-of-sample fit in the data is going to quantify how much the desire to make good predictions distorts traders’ beliefs.

Why Bayesian Variable Selection Doesn’t Scale

January 19, 2017 by Alex

1. Motivation

Traders are constantly looking for variables that predict returns. If $x$ is the only candidate variable traders are considering, then it’s easy to use the Bayesian information criterion to check whether $x$ predicts returns. Previously, I showed that using the univariate version of the Bayesian information criterion means solving

(●) $\begin{align*} \hat{\beta} &= \arg \min_{\beta} \big\{ \, \underset{\text{Prediction Error}}{{\textstyle \frac{1}{S}} \cdot {\textstyle \sum_s} (r_s - \beta \cdot x_s)^2} + \underset{\text{Penalty}}{ \lambda \cdot 1_{\{ \beta \neq 0 \}} } \, \big\} \qquad \text{with} \qquad \lambda = {\textstyle \frac{1}{S}} \cdot \log(S) \end{align*}$

after standardizing things so that $\hat{\mu}_x, \, \hat{\mu}_r = 0$ and $\hat{\sigma}_x^2 = 1$ . If the solution is some $\hat{\beta} \neq 0$ , then $x$ predicts returns. Notation: Parameters with hats are in-sample estimates. e.g., if $x_s \overset{\scriptscriptstyle \text{iid}}{\sim} \mathrm{N}(0, \, 1)$ , then $\frac{1}{S} \cdot \sum_s x_s = \hat{\mu}_x \sim \mathrm{N}(0, \, \sfrac{1}{S})$ .

But, what if there’s more than $1$ variable? There’s an obvious multivariate extension of (●):

(⣿) $\begin{align*} \{\hat{\beta}_1, \, \ldots, \, \hat{\beta}_V \} &= \arg \min_{\beta_1, \, \ldots, \, \beta_V} \left\{ \, {\textstyle \frac{1}{S}} \cdot {\textstyle \sum_s} ( r_s - {\textstyle \sum_v} \, \beta_v \cdot x_{s,v} )^2 + \lambda \cdot {\textstyle \sum_v} 1_{\{ \beta_v \neq 0 \}} \, \right\}. \end{align*}$

So, you might guess that it’d be easy to check which subset of $V \geq 1$ variables predicts returns by evaluating (⣿). But, it’s not. To evaluate the multivariate version of the Bayesian information criterion, traders would have to check $2^V$ different parameter values. That’s a combinatorial nightmare when $V \gg 1$ . Thus, traders can’t take a strictly Bayesian approach to variable selection when there are lots of variables to choose from.

Why is evaluating (⣿) so hard? That’s the topic of today’s post.

2. Non-Convex Problem

Let’s start by looking at what makes (●) so easy. The key insight is that you face the same-sized penalty no matter what $\hat{\beta} \neq 0$ you choose when solving the univariate version of the Bayesian information criterion:

$\begin{align*} \lambda \cdot 1_{\{ 0.01 \neq 0 \}} = \lambda \cdot 1_{\{ 100 \neq 0 \}} = \lambda \qquad \text{or, put differently} \qquad {\textstyle \frac{\mathrm{d}\lambda}{\mathrm{d}\beta}} = 0. \end{align*}$

So, if you’re going set $\hat{\beta} \neq 0$ , then you might as well choose the value that minimizes your prediction error:

$\begin{align*} \arg \min_{\beta \neq 0} \left\{ \, {\textstyle \frac{1}{S}} \cdot {\textstyle \sum_s} (r_s - \beta \cdot x_s)^2 + \lambda \cdot 1_{\{ \beta \neq 0 \}} \, \right\} &= \arg \min_{\beta} \left\{ \, {\textstyle \frac{1}{S}} \cdot {\textstyle \sum_s} (r_s - \beta \cdot x_s)^2 + \lambda \, \right\} \\ &= \arg \min_{\beta} \left\{ \, {\textstyle \frac{1}{S}} \cdot {\textstyle \sum_s} (r_s - \beta \cdot x_s)^2 \, \right\} \\ &= \hat{\beta}^{\text{OLS}}. \end{align*}$

Thus, to evaluate (●), all you have to do is check $2$ parameter values, $\beta = 0$ and $\beta = \hat{\beta}^{\text{OLS}}$ , and see which one gives a smaller result. Practically speaking, this means running an OLS regression, $r_s = \hat{\beta}^{\text{OLS}} \cdot x_s + \hat{\epsilon}_s$ , and checking whether or not the penalized residual variance, $\hat{\sigma}_\epsilon^2 + \lambda$ , is smaller than the raw return variance, $\hat{\sigma}_r^2$ .

Most explanations for why (⣿) is hard to evaluate focus on the fact that (⣿) is a non-convex optimization problem (e.g., see here and here). But, the univariate version of the Bayesian information criterion is also a non-convex optimization problem. Just look at the region around $\beta = 0$ in the figure to the left, which shows the objective function from (●). So, non-convexity can only part of the explanation for why (⣿) is hard to evaluate. Increasing the number of variables must be add a missing ingredient.

3. The Missing Ingredient

DATA + CODE

If there are many variables to consider, then these variables can be correlated. Correlation. This is the missing ingredient that makes evaluating (⣿) hard. Let’s look at a numerical example to see why.

Suppose there are only $S = 9$ stocks and $V = 3$ variables. The diagram above summarizes the correlation structure between these variables and returns. The red bar is the total variation in returns. The blue bars represent the portion of this variation that’s related to each of the $3$ variables. If you can draw a vertical line through a pair of bars (i.e., the bars overlap), then the associated variables are correlated. So, because the first $2$ blue bars don’t overlap, $x_1$ and $x_2$ are perfectly uncorrelated in-sample:

$\begin{align*} \widehat{\mathrm{Cor}}[x_1, \, x_2] &= 0. \end{align*}$

Whereas, because the first $2$ blue bars both overlap the third, $x_1$ and $x_2$ are both correlated with $x_3$ :

$\begin{align*} \widehat{\mathrm{Cor}}[x_1, \, x_3] = \widehat{\mathrm{Cor}}[x_2, \, x_3] &= 0.41. \end{align*}$

Finally, longer overlaps denote larger correlations. So, $x_3$ is the single best predictor of returns since the third blue bar has the longest overlap with the top red bar:

$\begin{align*} \widehat{\mathrm{Cor}}[r, \, x_1] = \widehat{\mathrm{Cor}}[r, \, x_2] &= 0.62 \\ \widehat{\mathrm{Cor}}[r, \, x_3] &= 0.67. \end{align*}$

And, this creates a problem. If you had to pick only $1$ variable to predict returns, then you’d pick $x_3$ :

$\begin{align*} {\textstyle \frac{1}{9}} \cdot {\textstyle \sum_s} ( r_s - \hat{\beta}_3^{\text{OLS}} \cdot x_{s,3} )^2 + \lambda = 0.80 < 0.86 &= {\textstyle \frac{1}{9}} \cdot {\textstyle \sum_s} ( r_s - \hat{\beta}_1^{\text{OLS}} \cdot x_{s,1} )^2 + \lambda \\ &= {\textstyle \frac{1}{9}} \cdot {\textstyle \sum_s} ( r_s - \hat{\beta}_2^{\text{OLS}} \cdot x_{s,2} )^2 + \lambda. \end{align*}$

But, $\{x_1, \, x_2 \}$ is actually the subset of variables that minimizes (⣿) in this example:

$\begin{align*} {\textstyle \frac{1}{9}} \cdot {\textstyle \sum_s} ( r_s - {\textstyle \sum_{v \in \{1,2\}}} \hat{\beta}_v^{\text{OLS}} \cdot x_{s,v} )^2 + 2 \cdot \lambda = 0.72. \end{align*}$

In other words, the variable that best predicts returns on its own isn’t part of the subset of variables that collectively best predict returns. In other examples it might be, but there’s no quick way to figure out which kind of example we’re in because (⣿) is a non-convex optimization problem. Until you actually plug $\{x_1, \, x_2 \}$ into (⣿), there’s absolutely no reason to suspect that either variable belongs to subset that minimizes (⣿). Think about it. $x_1$ and $x_2$ are both worse choices than $x_3$ on their own. And, if you start with $x_3$ and add either variable, things get even worse:

$\begin{align*} {\textstyle \frac{1}{9}} \cdot {\textstyle \sum_s} ( r_s - \hat{\beta}_3^{\text{OLS}} \cdot x_{s,3} )^2 + \lambda = 0.80 < 0.90 &= {\textstyle \frac{1}{9}} \cdot {\textstyle \sum_s} ( r_s - {\textstyle \sum_{v \in \{3,1\}}} \, \hat{\beta}_v^{\text{OLS}} \cdot x_{s,v} )^2 + 2 \cdot \lambda \\ &= {\textstyle \frac{1}{9}} \cdot {\textstyle \sum_s} ( r_s - {\textstyle \sum_{v \in \{3,2\}}} \, \hat{\beta}_v^{\text{OLS}} \cdot x_{s,v} )^2 + 2 \cdot \lambda. \end{align*}$

With many correlated variables, you can never tell how close you are to the subset of variables that best predicts returns. To evaluate (⣿), you’ve got to check all $2^V$ possible combinations. There are no shortcuts.

4. Birthday-Paradox Math

If correlation makes it hard to evaluate (⣿), then shouldn’t we be able to fix the problem by only considering independent variables? Yes… but only in a fairytale world where there are an infinite number of stocks, $S \to \infty$ . The problem is unavoidable in the real world where there are almost as many candidate variables as there are stocks because independent variables are still going to be correlated in finite samples.

Suppose there are $V \geq 2$ independent variables that might predict returns. Although these variables are independent, they won’t be perfectly uncorrelated in finite samples. So, let’s characterize the maximum in-sample correlation between any pair of variables. After standardizing each variable so that $\hat{\mu}_{x,v} = 0$ and $\hat{\sigma}_{x,v}^2 = 1$ , the in-sample correlation between $x_v$ and $x_{v'}$ when $S \gg 1$ is roughly:

$\begin{align*} \hat{\rho}_{v,v'} \sim \mathrm{N}(0, \, \sfrac{1}{S}). \end{align*}$

Although $\lim_{S \to \infty} \sfrac{1}{S} \cdot (\hat{\rho}_{v,v'} - 0)^2 = 0$ , our estimates won’t be exactly zero in finite samples.

CODE

Since the normal distribution is symmetric, the probability that $x_v$ and $x_{v'}$ have an in-sample correlation more extreme than $c$ is:

$\begin{align*} \mathrm{Pr}[ |\hat{\rho}_{v,v'}| > c ] &= 2 \cdot \mathrm{Pr}[ \hat{\rho}_{v,v'} > c ] = 2 \cdot \mathrm{\Phi}(c \cdot \!\sqrt{S}). \end{align*}$

So, since there are ${V \choose 2} \leq \frac{1}{2} \cdot V^2$ pairs of variables, we know that the probability that no pair has a correlation more extreme than $c$ is:

$\begin{align*} \mathrm{Pr}[ \max |\hat{\rho}_{v,v'}| \leq c ] &\leq \big( \, 1 - 2 \cdot \mathrm{\Phi}(c \cdot \!{\textstyle \sqrt{S}}) \, \big)^{\frac{1}{2} \cdot V^2}. \end{align*}$

Here’s the punchline. Because $V^2$ shows up as an exponent in the equation above, the probability that all pairwise in-sample correlations happen to be really small is going to shrink exponentially fast as traders consider more and more variables. This means that finite-sample effects are always going to make evaluating (⣿) computationally intractable in real-world settings with many variables, even when the variables are truly uncorrelated as $S \to \infty$ . e.g., the in-sample correlation of $\hat{\rho}_{1,3} = \hat{\rho}_{2,3} = 0.41$ from the previous example might seem like an unreasonably high number for independent random variables, something that only happens when $S=9$ . But, the figure above shows that even when there are $S = 50$ stocks, there’s still a $50{\scriptstyle \%}$ chance of observing an in-sample correlation of at least $0.41$ when considering $V = 20$ candidate variables.

5. What This Means

Market efficiency has been an “organizing principle for 30 years of empirical work” in academic finance. The principle is based on on a negative feedback loop: predictable returns suggest profitable trading strategies, but implementing these strategies eliminates the initial predictability. So, if there are enough really smart traders, then they’re going to find the subset of variables that predicts returns and eliminate this predictability. That’s market efficiency.

Even for really smart traders, finding the subset of variables that predicts returns is hard. And, this problem gets harder when there are more candidate variables to choose from. But, while researchers have thought about this problem in the past, they’ve primarily focused on the dangers of p-hacking (e.g., see here and here). If traders regress returns on $V = 20$ different variables, then they should expect that $1$ of these regressions is going to produce a statistically significant coefficient with a $p\text{-value} \leq 0.05$ even if none of the variables actually predicts returns. So, researchers have focused on correcting p-values.

But, this misses the point. Searching for the subset of variables that predicts returns isn’t hard because you have to adjust your p-values. It’s hard because it requires a brute-force search through the powerset of all possible subsets of predictors. It’s hard because any optimization problem with a hard-thresholding rule like (⣿) can be re-written as an integer programming problem,

$\begin{align*} \min_{{\boldsymbol \delta} \in \{0, \, 1\}^V} \left\{ \, {\textstyle \frac{1}{S}} \cdot {\textstyle \sum_s} (r_s - {\textstyle \sum_v} [\hat{\beta}_v^{\text{OLS}} \cdot x_{s,v}] \cdot \delta_v)^2 \quad \text{subject to} \quad k \geq {\textstyle \sum_v} \delta_v \, \right\}, \end{align*}$

which means that it’s NP-hard (e.g., see here, here, or here). It’s hard because, in the diagram above finding the subset of $30$ variables that predicts returns is equivalent to finding the cheapest way to cover the red bar with blue bars at a cost of $\lambda$ per blue bar, which means solving the knapsack problem.

p style=”text-indent: 15px;”>So, the fact that Bayesian variable selection doesn’t scale is a really central problem. It means that, even if there are lots and lots of really smart traders, they may not be able to find the best subset of variables for predicting returns. You’re probably on a secure wifi network right now. This network is considered “secure” because cracking its $128$ -bit passcode would involve a brute-force search over $2^{128}$ parameter values, which would take 1 billion billion years. So, if there are over V = 313 predictors documented in top academic journals, why shouldn’t we consider the subset of variables that actually predicts returns “secure”, too? After all, finding it would involve a brute-force search over $2^{313}$ parameter values. We might be able to approximate it. But, the exact collection of variables may as well be in a vault somewhere.

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