Testing one asset pricing model against another a la Cochrane: a counterexample

Question

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”?)

2. 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!

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4. 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.

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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

• Cochrane, J. H. (2014). Week 5 Empirical methods notes. Business 35150 Advanced Investments, 225-247.

Answer 1

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).