Testing one asset pricing model against another a la Cochrane: why this works

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!

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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.
(b) “Drop smb” means the 25 portfolio alphas are the same with or without $$smb$$
(c) *Equivalently, we are forming an “orthogonalized factor” $$smb_t^* = \alpha_{smb} + \varepsilon_t = smb_t − b_s rmrf_f − h_s hml_t$$ This is a version of $$smb$$ purged of its correlation with $$rmrf$$ and $$hml$$. Now it is OK to drop $$smb$$ if $$E(smb^{*})$$ is zero, because the $$b$$ and $$h$$ are not aﬀected if you drop $$smb^*$$
(d) *Why does this work? Think about rewriting the original model in terms of $$smb^{*}$$, \begin{align*} R_t^{ei} &= \alpha_i + b_i rmrf_t + h_i hml_t + s_i smb_t + \varepsilon_t^i \\ &= \alpha_i + (b_i+s_i b_s) rmrf_t + (h_i+s_i h_s) hml_t + s_i (smb_t - b_s rmrf_t - h_s hml_t) + \varepsilon_t^i \\ &= \alpha_i + (b_i+s_i b_s) rmrf_t + (h_i+s_i h_s) hml_t + s_i smb_t^* + \varepsilon_t^i \end{align*} The other factors would now get the betas that were assigned to $$smb$$ merely because $$smb$$ was correlated with the other factors. This part of the $$smb$$ premium can be captured by the other factors, we don’t need $$smb$$ to do it. The only part that we need $$smb$$ for is the last part. Thus average returns can be explained without $$smb$$ if and only if $$E(smb_t^{*}) = 0$$.

Part 4. is unclear to me. I do not get why $$\text{(iii)}$$ is the relevant regression to run and $$\alpha_{smb}$$ in it the relevant coefficient to test. (Attempting to show that this approach fails, I provide a counterexample here.) I think I need a formal proof in addition to the intuition. I guess once I see the proof, Cochrane's intuition will become more intuitive to me, too. Could you help me understand this?

Also, does this apply as is if the factor we consider kicking out of the model is actually a characteristic, not a factor (such as the size of the firm rather than the firm's sensitivity to the $$smb$$ factor)?

Update: Another source that discusses this is chapter 13 of Cochrane "Asset Pricing" (2005), especially sections 13.4 and 13.6.

References

• Related questions: 1, 2, 3, 4, 5. Nov 29, 2023 at 15:16
• Does my answer to your other question help? The way I like to think of factor models is like basis vectors. If the model is true, they span the entire space. To see if we need a particular factor, we have to see if the remaining factors span this potential factor. If they do (if they price the factor''), it's redundant by linearity. Otherwise, if there's a residual (alpha), the potential factor possesses incremental pricing power (increases the span of attainable payoffs) and needs to be included in the model. Dec 8, 2023 at 0:53
• @Kevin, thank you for the comment. I will think about it. This needs time to sink in... :) Dec 8, 2023 at 6:59