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Basically, the title of the question is all there is.

quoting from bishop's pattern recognition and machine learning:

In both the Bayesian and frequentist paradigms, the likelihood functions $p(D/w)$ plays a central role. However, the manner in which it is used is fundamentally different in the two approaches. In a frequentist setting, w is considered to be a fixed parameter, whose value is determined by some form of 'estimator', and error bars on this estimate are obtained by considering the distribution of possible data sets $D$. By, contrast, from the Bayesian viewpoint htere is only a single data set $D$ (namely the one that is actually observed), and the uncertainty in the parameters is expressed through a probability distriution over $w$

What does $p(D/w)$ mean if $D$ is fixed?

figs_and_nuts
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    Blain Waan's answers to this broad question might be helpful, as well as this related question. – Durden Jun 18 '23 at 15:53
  • Essentially, in the Bayesian paradigm the $p(D|w)$ can be regarded, for a fixed $D$, as the weight over all possible $w$ parameters arising from the prior distribution $p(w)$, if you notice especially that the posterior is proportional to $p(D|w)p(w)$. – Fiodor1234 Jun 18 '23 at 17:37
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    No paradox there, $D$ is observed, but assumed to be the realisation of a random variable with density $p(D|w)$. – Xi'an Jun 18 '23 at 18:56
  • @Xi'an - Ohh. I thought, from the excerpt that I have given above, that the data is considered fixed in Bayesian as the parameter is considered fixed in the frequentist inference. And, since in frequentist we cannot talk about the probability of the parameter given data as it is fixed, I thought we couldn't talk about the probability of the data given parameter here in Bayesian – figs_and_nuts Jun 19 '23 at 07:04
  • Without a distribution on the data, the conditioning term would not make sense. – Xi'an Jun 19 '23 at 12:24

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