G. Monette and J. Fox provide in these slides a framework for the Type II Analysis of Variance/Deviance tests in terms of conditional hypothesis. My questions are:

  • In this frequentist approach, the "conditional hypothesis" $L_1\beta=0 \mid L_2\beta=0$ is only symbolic (isn't it?). Is there a Bayesian analogue to this approach such as inference on $L_1 \beta$ under the posterior distribution of $\beta$ taken conditionally to $L_2 \beta =0$ ?

  • Monette & Fox rigorously define the conditional hypothesis as a classical hypothesis $L_{1\mid 2}\beta =0$ for a certain matrix $L_{1\mid 2}$ but this matrix depends on the estimated asymptotic covariance matrix of the parameters $\beta$. That sounds strange. Does it actually depends on the true asymptotic covariance matrix and then $L_{1\mid 2}$ is only an estimate of a theoretical matrix ? Even for the true asymptotic covariance that sounds strange because it still depends on the choice of the estimating method.

In fact I have never seen the notion of conditional hypothesis before, is it presented in some textbooks ?


Still sick today but here are some thoughts. Consider a classical linear model $y = X\beta+\sigma\epsilon$ and the Jeffreys prior. Then the posterior distribution of $(\beta \mid \sigma)$ is ${\cal N}(\hat\beta, V)$ where $V$ is the asymptotic covariance matrix of the least-squares estimator $\hat\beta$, or (I do not remember), $V$ is this matrix up to a factor close to $1$. Then it is easy to see that $(L_1 \beta \mid L_2\beta=0)$ has the distribution of $L_{1 \mid 2} \beta$ under the conditional posterior distribution $(\beta \mid \sigma)$, where $L_{1 \mid 2}$ is a $V$-orthogonal complement as defined in Monette & Fox's slides. And the Wald statistic $Z_{1|2}$ should be related to the norm of $L_{1 \mid 2} \beta$.

For more general models the approach should asymptotically coincide with the Bayesian approach when $\hat\beta$ is taken to be the maximum-likelihood estimate.

Too sick to continue...


I really wonder about whether this an old or a recent approach. As shown in my answer to myself here, this is not the way used by SAS. But the "old" anova() R function uses this approach. Indeed, for a generalized least-squares model such as

glsfit <- gls(value ~ group*variable, data=ldat,  
            correlation=corSymm(form= ~ 1 | id),
            weights=varIdent(form = ~1 | variable))

the type II hypothesis Wald F-test statistic of the variable factor is provided by:

> anova(glsfit)
Denom. DF: 45 
               numDF   F-value p-value
(Intercept)        1 1401.9971  <.0001
group              4    2.3793  0.0658
variable           2   79.5687  <.0001
group:variable     8    1.4759  0.1929

(and for the group factor one has to exchange the order of the factors:

glsfit.reverse <- update(glsfit, model = value ~ variable*group)


Or is it a new theoretical justification of an old approach ?..

  • $\begingroup$ @Zen No, see my update. This is really close to the Bayesian interpretation. $\endgroup$ Mar 23, 2013 at 10:15
  • $\begingroup$ Good to know, Stéphane. Take care. $\endgroup$
    – Zen
    Mar 23, 2013 at 12:33

1 Answer 1

  • Yes, this is a slight and common abuse of notation. Recall that in Bayesian statistics there are no point hypothesis under a continuous model since they have probability $0$.

  • No, this is incorrect. The matrix $L_{1\vert 2}$ depends only on $L_1$ and $L_2$, see slide 5. Only the asymptotic distribution of the estimators of $\beta$ depends on the covariance matrix. This is a classical result.

The basic idea in the slides is more related to nested models, although they use a conditioning notation. See the following discussion for a distinction between notation and conditioning


  • $\begingroup$ 1) Yes but we can do Bayesian inference about $L_1\beta$, not necessarily a hypothesis test. You do not answer my question: would it be analogue ? For instance with a noninformative prior, if we would look whether the credibility interval contains $0$, would it be similar to the frequentist hypothesis test ?. 2) Slide 5 precisely says that $L_{1\mid 2}$ is defined with the help of $V$. $\endgroup$ Mar 23, 2013 at 7:26

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