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I Am trying to derive the expression for the GRS test of the CAPM. I am following the book: The Econometrics Of Financial Markets by Campbell, Lo, McKinley (1997).

Define $Z_t$ as an $N×1$ vector of excess returns for N assets. We assume that the excess returns can be described by the following excess-return market model:

$$Z_t = \alpha + \beta Z_{mt} + \epsilon_t$$ We assume that excess returns are jointly normal, with: $$E[\epsilon_t]=0 $$ N×1 vector $$E[\epsilon_t \epsilon_t']=\Sigma$$

Accordingly, because excess returns are normally distributed conditionally on the excess return of the market and assuming they are temporally IID, given T observations, we get the following log-likelihood function:

$$L(\alpha,\beta,\Sigma)=-NTlog(2\pi)-T/2log(det(\Sigma))-1/2 \sum_{t=1}^{T} (Z_t-\alpha-\beta Z_{mt})'\Sigma^{-1}(Z_t-\alpha-\beta Z_{mt})$$

The partial first derivative w.r.t. alpha is: (1) $$\partial L/\partial \alpha=\Sigma^{-1}\sum_{t=1}^{T}(Z_t-\alpha-\beta Z_{mt}) $$

From which, by setting it equal to 0, we get the MLE of alpha:

$$\hat{\alpha}=\hat{\mu}-\hat{\beta}\hat{\mu_{m}}$$

Where $\hat{\mu}=1/T\sum_{t=1}^{T} Z_t$ and $\hat{\mu_m}=1/T\sum_{t=1}^{T} Z_{mt}$

The authors claim that the variance of the MLE estimator of alpha is $$Var[\hat{\alpha}]=1/T[1+\hat{\mu_m}^2/\hat{\sigma_m}^2]\Sigma$$ Where $\hat{\sigma_m}^2=1/T\sum_{t=1}^{T} (Z_{mt}-\hat{\mu_m})^2 $

So that the GRS test is simply the Wald statistics:

$$J= \hat{\alpha}'[var[\hat{\mu}]]^{-1}\hat{\alpha}=T[1+\hat{\mu_m}^2/\hat{\sigma_m}^2]^{-1}\hat{\alpha}'\Sigma^{-1}\hat{\alpha}$$

Of the null hypothesis that the alphas are jointly zero.

I know that the variance of the estimates can be derived using the inverse of the Fisher information matrix. However, if I compute the derivative of (1), namely the second derivative of the LogLik w.r.t. alpha, change sign and then take its expectation, I can not obtain the expression of the variance claimed by the authors. Can you help me with this last step , please?

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  • $\begingroup$ Hi: See P4 in this. qed.econ.queensu.ca/pub/faculty/abbott/econ351/351note04.pdf $\endgroup$
    – mark leeds
    Aug 16, 2019 at 0:19
  • $\begingroup$ Thank you. However, I should be able to get the same result via MLE, namely by computing the second derivative of LogL w.r.t. alpha, but it looks like I don't get the same result. Could you help me? $\endgroup$
    – Alchemy
    Aug 16, 2019 at 11:15

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Hi: This is an incomplete answer but I needed room. The Wald statistic for testing a linear constraint , $Rb = r$ is ,

$(Rb - r )^{\prime}[R(X^{\prime}X)^{-1} R^{\prime}]^{-1}(Rb - r)/s^2$

$X^{\prime}X$ can be obtained from P4 and, in your case, $R = 1$ and $b = \alpha$. But I still don't see how the expression for $(X^\prime X)^{-1}$ results in what you wrote. Hopefully someone else can help here because I don't see it. Note that $\Sigma$ is just a scale factor so don't worry about that.

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  • $\begingroup$ @Alchemy: I just realized that you have a GLS model rather than an OLS model. So this may be why their result doesn't match with the $(X^\prime X)^{-1}$ approach. I don't have time right now but you should google for "Wald test and GLS" and maybe something will result from that search. $\endgroup$
    – mark leeds
    Aug 16, 2019 at 15:45
  • $\begingroup$ @Alchemy: Like I said, I don't have time right now but I did look around and the reason for their use of MLE and asymptotics is because it's a GLS model so no closed form solution. See sections 5.1 and 5.2 of this link for a compact explanation of what's going on. I only glanced and it's pretty terse but hopefully it can be useful to you. Also, these types of articles ( overviews of the whole topic ) tend to be terse but there may be more detailed introductory explanations in the references. hedibert.org/wp-content/uploads/2014/04/… $\endgroup$
    – mark leeds
    Aug 16, 2019 at 15:58
  • $\begingroup$ @Alchemy: Here's a nice video of how variance of the GLS coefficients is derived. youtube.com/watch?v=JEQPis80KNw. But I still don't see how the authors obtain the expression you provided. $\endgroup$
    – mark leeds
    Aug 17, 2019 at 4:41

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