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I am not a statistician or mathematician but am trying to learn.

My question: In Bayes Theorem, $p(C|X)=p(X|C)p(C)/p(X)$, what are the English terms for $p(X|C)$ and $p(C)/p(X)$?

In other words, is $p(C)$ the prior probability? What is the comparable word for $p(C)/p(X)$?

Is the likelihood part of the right side of this equation? If so, which part?

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  • $\begingroup$ Thank you for the discussion below. It is helpful. If the liklihood is p(X|C). p(C), does 1/p(x) have a term of art? I read, in Wikipedia I think, that it may have something to do with normalizing. Is that so? If so, what is being normalized? $\endgroup$ – johnhidley Apr 23 '14 at 23:18
  • $\begingroup$ I misunderstood what you wrote. I think this is what you actually said: p(X|C) is the likelihood, and p(C) is the prior. If so, does 1/p(x) have a term of art? I read, in Wikipedia I think, that it may have something to do with normalizing. Is that so? What is being normalized? $\endgroup$ – johnhidley Apr 23 '14 at 23:25
  • $\begingroup$ $1/p(X)$ scales the quantity $p(X|C)p(C)$ so that it is a probability on the [0,1] interval. $\endgroup$ – Sycorax says Reinstate Monica Apr 24 '14 at 3:28
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Assuming $X$ is your data and $C$ is the parameter, the likelihood is $p(X|C)$. $p(C)$ is the prior. I am not aware of a particular word for $p(C)/p(X)$.

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  • $\begingroup$ The posterior is proportional to the likelihood times the prior. $\frac{p(C)}{p(X)}$ is the reason it is proportional. $\endgroup$ – Ellis Valentiner Apr 23 '14 at 18:18
  • $\begingroup$ That's true, but I think OP is looking for a term of art to describe $p(C)/p(X)$. I'm not aware of "proportional" being used in that way. $\endgroup$ – Sycorax says Reinstate Monica Apr 23 '14 at 18:21
  • $\begingroup$ You're right. I don't think there is one. But it is just a constant. $\endgroup$ – Ellis Valentiner Apr 23 '14 at 18:23
  • $\begingroup$ It's not constant if $p(C)$ is a probability distribution other than the uniform. Frequentist inference assumes that it is constant and unknowable, but Bayesian inference does not. $\endgroup$ – Sycorax says Reinstate Monica Apr 23 '14 at 18:25
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If you write Bayes theorem as

$$ p(C|X) = \frac{ p(X|C)p(C) }{ p(X) } $$

then $p(C|X)$ is posterior distribution, $p(X|C)$ is likelihood and $p(C)$ is prior, while $p(X)$ is normalizing constant. In fact, $p(X)$ is

$$ p(X) = \int p(X|C)p(C) dC $$

and is used so that we obtain results in probability scale. However in fact if you use MCMC simulation for estimation, it is not needed and the formula reduces to

$$ p(C|X) \propto p(X|C)p(C) $$

As already mentioned by others, there is no reason to be interested in $p(C)/p(X)$ because normalizing our prior probability, i.e. our guess that we made before seeing the data, leads nowhere. I guess this is why it has no name.

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There's no standard term for p(C)/p(X). The denominator p(X) is called the marginal likelihood for your model.

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