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2

Not necessarily for Bayes factors. But Bayes factors require conversion to posterior probabilities for proper inference, and posterior probabilities most definitely can require a Bonferroni-style multiplicity adjustment of sorts. Here is the explanation: If the hypotheses are independent a priori, then probability that all nulls are simultaneously true ...

0

$X_1,\ldots,X_n$ is a sample from $U(\mu-\theta,\mu+\theta)$ distribution. So the likelihood function given $x=(x_1,\ldots,x_n)$ is \begin{align} L(\mu,\theta\mid x)&=\prod_{i=1}^n\frac{1}{2\theta}1_{\mu-\theta<x_i<\mu+\theta} \\&=\frac{1}{(2\theta)^n}1_{\mu-\theta<x_{(1)},x_{(n)}<\mu+\theta}\,, \end{align} where $x_{(1)}=\min\limits_{... 1 Let us consider the example, suppose we are testing $$H_0: p = 0.5\\H_1: p<0.5$$ and suppose that$\bar{x} = 0.4$. $$L(X, p) = p^{\sum X_i}(1-p)^{n - \sum X_i}$$ maximizing it wrt$p$subject to$p \le 0.5$yileds$\hat{p} = \bar{x} < 0.5$and hence $$\Lambda = \frac{0.4^{\sum X_i}0.6^{n - \sum X_i}}{0.5^n} \ge k \iff \sum X_i \ln 0.4 + (n - \sum X_i)\... 0 For the record, here is a somewhat more extensive proof. It also contains some background information. Maybe this is helpful for others studying the topic. The main idea of the proof is to show that Jaynes' conditions 1 and 2 imply that$$P(D_{m_k}|H_iX)=P(D_{m_k}|X),$$for all but one data set$m_k=1,\ldots,m\$. It then shows that for all these data sets, ...

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The autour of the survey package Thomas Lumley has written a book using the package, but I cannot find symbolical nesting mentioned there. The docu page you cites also says If the models are symbolically nested, so that the relevant parameters can be identified just by manipulating the model formulas, anova is equivalent to ...

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