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In a number of statistics books and references you can find formulas such as:

$P\,(result~|~H_0)$ (Bortz, p. 114)

$P\,(reject~H_0\;|\;H_0~valid)$ (Wikipedia - Statistical hypothesis testing)

and you will probably find many others if you just google for 'hypothesis testing'.

Isn't this abusing the notation of $P$ and / or mixing two very different things into $P$?

My point is, if we talk about some result $R$ (e.g., the sequence $(0, 1, 1, 1, 1, 0, 1)$ of coin flips) over some space $\Omega$, then $P(R)$ is very well defined and clear to me.

However, $P(H_0)$ (which as I understand it is to be interpreted as "the probability that $H_0$ is true") must be from some totally unrelated meta-space $\Omega_{realworld}$ and the only thing I could relate it to would be the probability space over all possible probability spaces for $P$.

How can / does writing something like $P\,(R~|~H_0)$ make sense? How can I 'reason' with it? How could I apply, for example, Bayes' theorem?

left4bread
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  • http://stats.stackexchange.com/questions/167051/who-are-the-bayesians – Mark L. Stone Aug 30 '15 at 16:52
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    Both your examples are conditional on the truth of $H_0$ so $\Pr(H_0)$ doesn't come into it. You could write $\Pr_{H_0}(\mathrm{result})$ instead. – Scortchi - Reinstate Monica Aug 30 '15 at 17:40
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    This notation is not a conditional probability. It makes no implicit assertions about whether the right hand side (such as "$H_0$") is an event in a probability space. – whuber Aug 30 '15 at 17:52
  • @whuber, isn't this exactly what qualifies it as abuse of notation? In the same book (Bortz) $P(a|b)$ is only ever defined and / or mentioned as a conditional probability, same goes for every other source I can recall at the moment using it. – left4bread Aug 31 '15 at 07:03
  • @Scortchi, what if I, for example, naively try to apply Bayes' rule? $P(R|H_0)=P(R)*P(H_0|R)*P(H_0)^{-1}$? In my opinion it pretends to be something which it isn't. – left4bread Aug 31 '15 at 07:05
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    I agree it's an abuse. However, it's not a serious one, for two reasons. One is that you can always view a constant as a (limiting) version of a random variable: put all the (prior) probability on $H_0$. The other is that the meaning is usually evident: if the context does not include a probability distribution for $H_0$, then this notation is not a conditional probability. – whuber Aug 31 '15 at 14:59

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