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I know that by the Bayesian formula p(θ|y) = p(y|θ) x p(θ) / p(y)

and understand that p(θ|y) is the probability of the parameter conditional on the data.

Could I say that this probability p(θ|y) is also conditional on the prior?

If not is there a way to express the posterior conditional on the prior?

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  • $\begingroup$ $p (\theta|\theta)$? $\endgroup$ – Dave Harris Feb 20 '17 at 1:13
  • $\begingroup$ I know what you are trying to say, but you want to rid yourself of the word conditional. Its not conditional on the prior because the prior isn't random. You are thinking of something similar to the envelope theorem for statistics. $\endgroup$ – Dave Harris Feb 20 '17 at 1:20
  • $\begingroup$ Let me ask it a bit differently. Say you utilize some previous information to construct an informative prior and you use this prior instead of a flat one. Then isnt the posterior conditional on that prior information ? $\endgroup$ – Nay Feb 20 '17 at 1:32
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    $\begingroup$ You're argument is regards to Bayesian updating. This is a complex literature. If you had two sets of data $y,y'$ and you combined them as $p(\theta|y;y')$ then that is the solution without Bayesian updating and as $p(\theta|y)$ with updating. There is an issue here of not accepting $y'$ intact. Consider that you began with data from British bankruptcy data to estimate US bankruptcy and changed the scale parameter to weaken it to allow for differences, then you are including non-sample data. There is also the subjective question as to why you accepted the other study intact. $\endgroup$ – Dave Harris Feb 20 '17 at 2:30
  • $\begingroup$ Maybe another way to think about this is to think about the question of a misspecified model. Imagine that your likelihood was misspecified, is your result conditional on the misspecification? No. It is contingent on the misspecification, though. I think the word you are thinking of is "contingent" rather than conditional. $\endgroup$ – Dave Harris Feb 20 '17 at 2:44
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In the narrow statistical sense, a conditional probability refers to a probability that depends on the outcome of another random variate. The posterior probability, however, depends on the entire prior probability distribution p(θ), not on a particular outcome θ. So, I would say strictly speaking: no.

This is, unless you see the prior itself as a random variate, which would be possible to argue I guess, albeit somewhat unusual. This reminded me about a an older discussion between Larry Wassermann, Andrew Gelman and others, on whether the p-value is a conditional probability, and if it is appropriate to write p(d>D|H0).

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