# Testing one asset pricing model against another a la Cochrane via change in $\hat\alpha' \text{cov}(\hat\alpha,\hat\alpha')^{-1}\hat\alpha$

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. 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 $$E(smb_t^{*}) = 0$$.
1. *Other solutions (equivalent)
(a) Drop $$smb$$, redo, test if $$\alpha' \text{cov}(\alpha)^{-1}\alpha$$ rises “too much.”
(b) Express the model as $$m = a − b_1 rmrf − b_2 hml − b_3 smb, 0 = E(m R^e)$$. A test on $$b_x$$ is a test of “can you drop the extra factor.”

How exactly can we do 5.a? On p. 238, Cochrane indicates that $$\hat\alpha' \text{cov}(\hat\alpha,\hat\alpha')^{-1}\hat\alpha\sim\chi^2_{N-1}$$ in the Fama-MacBeth approach and on p. 236 it is $$\sim\chi^2_{N-K-1}$$ in the cross-sectional approach. ($$N$$ is the number of assets or test portfolios, $$K$$ is the number of pricing factors.) If $$\chi^2_{full}:=\hat\alpha' \text{cov}(\hat\alpha,\hat\alpha')^{-1}\hat\alpha$$ corresponds to the full model and $$\chi^2_{restricted}:=\tilde\alpha' \text{cov}(\tilde\alpha,\tilde\alpha')^{-1}\tilde\alpha$$ corresponds to the restricted model, will the test statistic be $$\chi^2_{restricted}-\chi^2_{full}\sim\chi^2_{1}$$ or something similar? Is there a reference for this that I could look up?

### References

• Related questions: 1, 2, 3, 4, 5. Commented Nov 29, 2023 at 15:14