Is $\Pr(y \ , \mu |, H_0)$ equal to $\Pr(y \ | \mu)\Pr(\mu | H_0)$ or $\Pr(y \ | \mu , H_0)\Pr(\mu | H_0)$?
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3$\begingroup$ The second because you have to have the condition $|H_0$ in both probabilities, unless of course some simplification can be done. $\endgroup$– JohnKCommented Jan 20, 2016 at 17:43
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1$\begingroup$ @JohnK, why not put that as an official answer (perhaps w/ a little elaboration)? $\endgroup$– gung - Reinstate MonicaCommented Jan 20, 2016 at 18:03
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2 Answers
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It is the second expansion that is the correct one because you need to keep the condition on $H_0$ on both probabilities. Once you have done that, you are allowed to treat the conditional probabilities as regular probabilities and all rules you know apply.
Having said that, often a simplifcation of
$$\Pr \left(A, B | C \right) = \Pr \left(A |B, C \right) \Pr \left(B |C \right) $$
is possible. If, say, you know that the events $A$ and $C$ are conditionally indendent (given $B$), then the first term on the RHS becomes simply $\Pr \left(A |B \right)$.
You will find these rules particularly useful in Bayesian statistics.
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Definitely the second one. Maybe the first one.
It's definitely the second one because $Pr(y|\mu,H_0) Pr(\mu|H_0) = \frac{Pr(y,\mu,H_0)}{Pr(\mu,H_0)} \frac{Pr(\mu,H_0)}{Pr(H_0)} = Pr(y,\mu|H_0)$.
It's the first one if we assume that $y$ is independent of $H_0$ given $\mu$, which gives us $Pr(y|\mu,H_0) = Pr(y|\mu)$.
