1
$\begingroup$

I am reading section section 14.6 of John Cochrane's lectures notes for the course Business 35150 Advanced Investments. On p. 239-240, he discusses testing one asset pricing model against another. I have quite some trouble following his arguments. Here is the essence:

  1. Example. FF3F. $$ E(R^{ei}) = \alpha_i + b_i\lambda_{rmrf} + h_i\lambda_{hml} + s_i\lambda_{smb} \tag{i} $$ Do we really need the size factor? Or can we write $$ E(R^{ei}) = \alpha_i + b_i\lambda_{rmrf} + h_i\lambda_{hml} \tag{ii} $$ and do as well? ($\alpha$ will rise, but will they rise “much”?)
  1. A common misconception: Measure $\lambda_{smb} = E(smb)$. If $\lambda_{smb} = 0$ (and “small”) we can drop it. Why is this wrong? Because if you drop $smb$ from the regression, $b_i$ and $h_i$ also change!

<...>

  1. Solution: (a) “ First run a regression of $smb_t$ on $rmrf_t$ and $hml_t$ and take the residual, $$ smb_t = \alpha_{smb} + b_s rmrf_t + h_s hml_t + \varepsilon_t \tag{iii} $$ Now, we can drop $smb$ from the three factor model if and only $\alpha_{smb}$ is zero. Intuitively, if the other assets are enough to price $smb$, then they are enough to price anything that $smb$ prices.

Below I present a counterexample showing that the proposed approach fails. Where is my mistake?

Suppose the following 2-factor asset pricing model actually holds: $$ E(R^{ei}) = \beta_{1,i}E(X_1) + \beta_{2,i}E(X_2). \tag{iv} $$ Construct a variable $X_3$ such that it is independent of the triplet $(R^{ei},X_1,X_2)$ and let $E(X_3)=\mu_3\neq 0$. Suppose we do not know what the true model is and instead of $(\text{iv})$ we (mistakenly) use $$ E(R^{ei}) = \tilde\beta_{1,i}E(X_1) + \tilde\beta_{2,i}E(X_2) + \tilde\beta_{3,i}E(X_3). \tag{v} $$ Let us apply Cochrane's method to assess whether we could do without $X_3$. That is, regress $X_3$ on $X_1$ and $X_2$ $$ X_{3,t} = \delta_0 + \delta_1 X_{1,t} + \delta_2 X_{2,t} + v_t \tag{vi} $$ and test $H_0\colon \delta_0=0$. By construction of $X_3$, we know that $$ X_{3,t} = \mu_3 + 0\times X_{1,t} + 0\times X_{2,t} + v_t, \tag{vi'} $$ and thus $\delta_0=\mu_3\neq 0$. If our test has enough power, $H_0$ will be rejected. By Cochrane's argument, $X_3$ belongs in the model. But we know that this is incorrect.

References

$\endgroup$
5
  • $\begingroup$ Related questions: 1, 2, 3, 4, 5. $\endgroup$ Nov 29, 2023 at 15:16
  • $\begingroup$ I'm not quite getting your counterexample. You say (iv) holds. That means the expected return of every (tradable) asset needs to have this two factor structure. You then suggest to look at another factor, $X_3$ which isn't related to the factors in (iv). But if such a factor exists, then (iv) doesn't hold? Like can you claim to construct $X_3$ if (iv) holds? $\endgroup$
    – Kevin
    Dec 7, 2023 at 12:45
  • $\begingroup$ @Kevin, what if $X_3$ has zero betas for $X_1$ and $X_2$? The $(\text{iv})$ model still holds for $X_3$. A question remains whether we can find such an $X_3$ that is also independent of $R^{ei}$. Are you suggesting that is impossible? $\endgroup$ Dec 7, 2023 at 12:50
  • $\begingroup$ If $X_3$ has zero betas, then $E[X_3]=0$ by (iv) and adding it to a factor model wouldn't really work? If (iv) truly holds, then the return of every tradable asset has to obey that factor structure. So we can't come up with a different portfolio whose return is independent of those factors and has additional pricing power. $\endgroup$
    – Kevin
    Dec 7, 2023 at 12:53
  • $\begingroup$ @Kevin, I might be missing something, but on the face of it this looks like a perfect explanation for my puzzle. You have taught me something – that's great! If you write it up and no one points out a mistake in your argument, the bounty is yours. And beyond that also my gratitude and appreciation! $\endgroup$ Dec 7, 2023 at 13:00

1 Answer 1

1
+50
$\begingroup$

I'll expand on the comments under the question. Suppose the following two-factor model holds $$ \mathbb E(R^{(i)}) = \beta_1^{(i)}\mathbb E(X_1) + \beta_2^{(i)}\mathbb E(X_2). \tag{iv} $$ That means the expected return of every tradable asset is fully explained by $X_1$ and $X_2$. These two factors capture all variation in expected returns (systematic risk).

Suppose we look at another asset (portfolio) with return $X_3$. By $(\text{iv})$, we know that the expected return needs to have a two factor structure and be fully explained of $X_3$'s betas with respect to $X_1$ and $X_2$. In an equation, we have $$ \mathbb E(X_3) = \beta_1^{(3)}\mathbb E(X_1) + \beta_2^{(3)}\mathbb E(X_2). $$ Different to the question, we cannot assume that the expectation of $X_3$ isn't explained by $X_1$ and $X_2$ because that would violate $(\text{iv})$.

If we do not know $(\text{iv})$ is true and use $X_1$, $X_2$ and $X_3$ as factors, we get \begin{align} \mathbb E(R^{(i)}) &= \beta_1^{(i)}\mathbb E(X_1) + \beta_2^{(i)}\mathbb E(X_2) + \beta_3^{(i)}\mathbb E(X_3) \\ &= (\beta_1^{(i)}+\beta_3^{(i)}\beta_1^{(3)})\mathbb E(X_1) + (\beta_2^{(i)}+\beta_3^{(i)}\beta_2^{(3)})\mathbb E(X_2), \end{align} which is again a two-factor model. Essentially, $X_1$ and $X_2$ span the full space already and adding a linear combination of both doesn't help (all with respect to expectations).

$\endgroup$
2
  • $\begingroup$ Thank you! On a related note, any thoughts on this question and the answers to it? $\endgroup$ Dec 10, 2023 at 14:21
  • $\begingroup$ @RichardHardy I actually agree with much that you write in your answer (+1). I'd also add the difference that FM run cross-sectional regressions of returns on characteristics whereas Cochrane's example is about adding tradable factors to a model. It's a slightly different approach. $\endgroup$
    – Kevin
    Dec 10, 2023 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.