Asset-Pricing Implications of Dimensional Analysis

1. Motivation

I have been trying to use dimensional analysis to understand asset-pricing problems. In many hard physical problems, it is possible to gain some insight about the functional form of the solution by examining the dimensions of the relevant input variables. In the canonical example of this brand of analysis, G.I. Taylor was able to tell the yield of the Trinity Test nuclear explosion from a few photographs via dimensional analysis (see Barenblatt 2003 and earlier post). So, maybe it is possible to better understand, say, the price impact of informed trading by studying the dimensions of this problem?

However, none of the asset-pricing problems I have looked at via dimensional analysis have yielded pretty solutions. It could be that the fundamental asset-pricing equations aren’t dimensionally consistent. Such equations do exist and they can be very helpful. For instance, Dolbear’s Law says that you can tell the outdoor temperature on a summer evening by counting the frequency of cricket chirps,

(1) $\begin{align*} \text{temperature in degrees celsius} = 37^{\circ} + \{ \, \text{cricket chirps every 15 seconds} \, \}. \end{align*}$

But, I don’t think this is what’s going on. Instead, my sense is that, because asset-pricing models are built by researchers trying to convey economic intuition rather than dictated by the physical constraints of a particular real-world problem, there aren’t any interesting unexplored symmetries hiding in the asset-pricing models for dimensional analysis to uncover. A good economist never includes superfluous variables when constructing a model, but there is often unexpected redundancy in our initial formulations of hard physical problems that we find in nature. This post explains my (perhaps wrong) intuition in more detail.

2. Period of Pendulum

Let’s start by looking at a physical problem where dimensional analysis actually helps. Consider the problem of modeling the period of a pendulum with length $\ell$ and mass $m$ . Suppose that in order to get the pendulum swinging, I initially pull it a distance of $a$ centimeters off to the side. In this setup, we can write the period of the pendulum as some function of these variables,

(2) $\begin{align*} p &= \mathrm{f}(\ell,m,a,g), \end{align*}$

together with the acceleration due to gravity, $g$ .

The key insight in dimensional analysis is that the pendulum shouldn’t behave differently if we measure its length in inches rather than centimeters. The marks on our ruler don’t matter. The period of the pendulum has dimensions of time, $\mathrm{dim}[p] = T$ . The length of the pendulum and the amplitude of its swing have dimensions of length, $\mathrm{dim}[\ell] = \mathrm{dim}[a] = L$ . The mass of the pendulum has dimensions of (wait for it…) mass, $\mathrm{dim}[m] = M$ . And, the force of gravity has dimensions, $\mathrm{dim}[g] = L \cdot T^{-2}$ . Suppose that we define new units of mass, length, and time so that $1$ new unit of mass is equal to $\mu$ old units of mass, $1 \overset{M}{\longrightarrow} \mu$ , $1$ new unit of length is equal to $\lambda$ old units of length, $1 \overset{L}{\longrightarrow} \lambda$ , and $1$ new unit of time is equal to $\tau$ old units of time, $1 \overset{T}{\longrightarrow} \tau$ . If our choice of units doesn’t affect the pendulum’s behavior, then we should be able to rewrite our old formula in these new units,

(3) $\begin{align*} {\textstyle \frac{p}{\tau}} &= \mathrm{f}\left( \, {\textstyle \frac{\ell}{\lambda}}, \, {\textstyle \frac{m}{\mu}}, \, {\textstyle \frac{a}{\lambda}}, \, {\textstyle \frac{g}{\sfrac{\lambda}{\tau^2}}} \, \right). \end{align*}$

Now comes the trick. Notice that these new units can be anything we want. So, let’s get clever and pick $\mu = m$ , $\lambda = \ell$ , and $\tau = \sqrt{\sfrac{\ell}{g}}$ . With these values the formula for the period of the pendulum becomes

(4) $\begin{align*} {\textstyle \frac{p}{\sqrt{\sfrac{\ell}{g}}}} &= \mathrm{f}(1,1,\sfrac{a}{\ell},1) = \mathrm{f}^\star(\sfrac{a}{\ell}), \end{align*}$

where $\mathrm{f}^\star(\cdot)$ is a new function of a dimensionless ratio, $\sfrac{a}{\ell}$ . Thus, we know that the period of a pendulum is

(5) $\begin{align*} p &= \sqrt{\sfrac{\ell}{g}} \times \mathrm{f}^{\star}(\sfrac{a}{\ell}). \end{align*}$

Without knowing anything except for the units that each variable is being measured in, we can see that 1) the period is unrelated to the mass and 2) the period of the pendulum is inversely proportional to the square-root of the force of gravity, $\sqrt{g}$ . Functional forms without physics! We now know how to compute the period of the same pendulum on Mars.

3. Price Impact

What happens if we try to use these same trick to understand price impact in the stock market? That is, how much does the price of a stock move if traders demand an extra $100$ shares on a particular day? Let’s use the standard terminology from information-based asset-pricing models and define price impact as a function of $3$ variables,

(6) $\begin{align*} \lambda &= \mathrm{f}(\sigma_v, \sigma_z, \gamma), \end{align*}$

where $\mathrm{dim}[\lambda] = D \cdot S^{-1}$ with $D$ denoting dollars and $S$ denoting shares. Suppose that informed traders know the fundamental value of the stock, $v$ , but uninformed traders don’t. Let $\sigma_v$ denote the volatility of the stock’s value from the perspective of uninformed traders, $\mathrm{dim}[\sigma_v] = D \cdot S^{-1}$ , and let $\sigma_z$ denote the volatility of asset-supply noise, $\mathrm{dim}[\sigma_z] = S$ . This is the noise term that keeps the asset’s price from being perfectly revealing. Finally, let $\gamma$ denote the risk aversion of the informed traders, $\mathrm{dim}[\gamma] = D^{-1}$ .

By the logic of dimensional analysis, it shouldn’t matter whether we measure a stock’s value in dollars or euros and it shouldn’t matter whether we measure changes in demand in shares or tens of shares. So, suppose that $1$ new unit of value is equal to $\delta$ old units of value, $1 \overset{D}{\longrightarrow} \delta$ and that $1$ new unit of quantity is equal to $\psi$ old units of quantity, $1 \overset{S}{\longrightarrow} \psi$ . If our choice of units doesn’t affect market behavior, then we should be able to rewrite our old formula for price impact in these new units,

(7) $\begin{align*} \frac{\lambda}{\sfrac{\delta}{\psi}} &= \mathrm{f}\left(\frac{\sigma_v}{\sfrac{\delta}{\psi}}, \frac{\sigma_z}{\psi}, \frac{\gamma}{\sfrac{1}{\delta}}\right), \end{align*}$

just like before.

Now comes the trouble. If we get clever and choose our units to create a function of a single dimensionless variables, $\psi = \sigma_z$ and $\delta = \sfrac{1}{\gamma}$ , we find that:

(8) $\begin{align*} \lambda &= {\textstyle \frac{1}{\gamma \cdot \sigma_z}} \cdot \mathrm{f}^\star (\gamma \cdot \sigma_z \cdot \sigma_v). \end{align*}$

In the pendulum problem above, the single dimensionless quantity only involved some of the relevant variables; however, in the price-impact problem the dimensionless quantity involves all $3$ of the relevant variables. There is no progress. Before applying dimensional analysis we had an unknown function of $3$ variables. After applying dimensional analysis we still have an unknown function of $3$ variables. Dimensional analysis doesn’t provide any new insight about the functional form of the link between the quantity of interest (i.e., the price impact, $\lambda$ ) and any of the input parameters (i.e., the values $\sigma_v$ , $\sigma_z$ , or $\gamma$ ). I always seem to find this sort of non-result when applying dimensional analysis to asset-pricing problems.

4. Main Intuition

I think this particular non-result in the price-impact problem is suggestive of why dimensional analysis doesn’t help that much when trying to understand asset-pricing models more generally. What makes the canonical information-based asset-pricing papers great is that they pack a lot of economic intuition into a relatively simple model. There isn’t a lot of superfluous structure hanging around. When you look at the original formulation of the pendulum problem, there was a bunch of redundancy involved. The mass of the pendulum turned out to be irrelevant, and two of the variables, length and amplitude, turned out to have the exact same units. There is no such redundancy in the price-impact problem. As defined, the parameters $\sigma_v$ , $\sigma_z$ , and $\gamma$ are all needed to define the dimensionless quantity. The elegance of models like Kyle (1985) makes them unsuited to dimensional analysis.

To illustrate, consider changing the original price-impact problem slightly. Suppose that we, as econometricians, could directly observe the inverse of the dollar-demand volatility from noise traders, $\sfrac{1}{\sigma_y}$ , which has units of dollars $\mathrm{dim}[\sfrac{1}{\sigma_y}] = D^{-1}$ , instead of the demand volatility, which has units of shares $\mathrm{dim}[\sigma_z] = S$ . This is a less elegant model because it is needlessly complex. Demand volatility in shares now depends on both the equilibrium price and the shares demanded by noise traders. But, let’s go with it. In this new setup, the price impact is still an unknown function of $3$ variables,

(9) $\begin{align*} \lambda &= \mathrm{f}(\sigma_v, \sfrac{1}{\sigma_y}, \gamma), \end{align*}$

but now, because there is redundancy, we can make progress via dimensional analysis.

Again, suppose that $1$ new unit of value is equal to $\delta$ old units of value, $1 \overset{D}{\longrightarrow} \delta$ and that $1$ new unit of quantity is equal to $\psi$ old units of quantity, $1 \overset{S}{\longrightarrow} \psi$ . If our choice of units doesn’t affect market behavior, then we should be able to rewrite our old formula for price impact in these new units:

(10) $\begin{align*} \frac{\lambda}{\sfrac{\delta}{\psi}} &= \mathrm{f}\left(\frac{\sigma_v}{\sfrac{\delta}{\psi}}, \frac{\sfrac{1}{\sigma_y}}{\sfrac{1}{\delta}}, \frac{\gamma}{\sfrac{1}{\delta}}\right). \end{align*}$

If we choose our units to create a function of a single dimensionless variables, $\delta = \sigma_y$ and $\psi = \sfrac{\sigma_y}{\sigma_v}$ , we find that:

(11) $\begin{align*} \lambda &= \sigma_v \cdot \mathrm{f}^\star(\sigma_y \cdot \gamma). \end{align*}$

Before we had an unknown function with $3$ variables and now we have an unknown function of only $2$ variables. Progress! If we found situations where the product of informed traders’ risk aversion and dollar-demand noise was constant, $\sigma_y \cdot \gamma = \text{Const.}$ , then we could actually test this relationship with a regression,

(12) $\begin{align*} \log(\lambda) &= \alpha + \beta \times \log(\sigma_v) + \epsilon, \end{align*}$

and check whether or not $\beta = 1$ . But, the only reason that we could make progress in this alternative setting was that the model was written down clumsily in the first place.