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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

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

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