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In literature I sometimes stumple upon the remark, that choosing priors that depend on the data itself (for example Zellners g-prior) can be criticized from a theoretical point of view. Where exactly is the problem if the prior is not chosen independent from the data?

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2 Answers 2

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Generally, informative priors are typically viewed as your information about parameters (or hypotheses) before seeing the data. So any data-based prior is violating the likelihood principle since evidence from the sample is coming through the likelihood function and the prior.

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The $p$-values are wrong. Take a simple example. Test whether a population mean $\mu$ is equal to a particular value $\mu_0$ or not. Suppose the sample mean $\bar x$ is greater than $\mu_0$. Then it would be simply wrong to let the data guide you into testing only a one-sided alternative. Your $p$-value will be half of what it should be.

And just to be clear: The restriction $\mu \ge \mu_0$ implied by the one-sided alternative is a kind of empirical prior. (It throws away half of the possible values for $\mu$ a priori.)

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    $\begingroup$ Am astounded, flabbergasted and delighted that you are using a frequentist framework to illustrate a Bayesean concept. $\endgroup$
    – Alexis
    Commented Jun 14, 2015 at 17:11
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    $\begingroup$ Same. The first sentence made me go "what the hell?" $\endgroup$
    – user541686
    Commented Jun 14, 2015 at 18:00
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    $\begingroup$ I agree, this is an interesting way of starting a conversation in a Bayesian context. But do I understand it correct that this intuitive explanation also just describes a violation of the likelihood principle as @jaradniemi pointed out in his answer? $\endgroup$ Commented Jun 14, 2015 at 18:51
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    $\begingroup$ @Alexis a frequentist with domain expertise is a Bayesian in denial $\endgroup$ Commented Jun 14, 2015 at 19:25
  • $\begingroup$ @ssdecontrol: Haha, not quite sure if I understand what you're trying to say. Is that a tongue-in-cheek way of saying a frequentist with domain expertise can't exist? :P $\endgroup$
    – user541686
    Commented Jun 14, 2015 at 21:39

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