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Bayes' rule appears to bevery simple at first sight, but when studied deeply I find it is difficult and confusing, especially in MCMC applications when multiple parameters need to be estimated.

For example, assuming $x,y,z,t$ are four parameters, which of the following three expressions are true (or true under some specifications)?

  1. $P(x)P(z\mid x,y)=P(x,z\mid y)$
  2. $P(x\mid y)P(z\mid x,y)=P(x,z\mid y)$
  3. $P(x\mid y)P(z\mid x,t)=P(x,z\mid y,t)$

I usually see formulas similar with 3, but I wonder why it hold. Could someone explain it in detail?

If there are any excellent books that could help me, pls list them.

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1 Answer 1

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The generalization of the formula $$\Pr(A \cap B) = \Pr(A \mid B)\Pr(B)$$ for a conditional probability $\Pr(\cdot \mid C)$ is the following one: $$\Pr(A \cap B \mid C) = \Pr(A \mid B\cap C)\Pr(B \mid C).$$

Your formula number 2 is analogous to the above formula (with $P(\cdot \mid y)$ substituted for $\Pr(\cdot \mid C)$), and it is the correct one.

But note that this is not Bayes' rule.

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  • $\begingroup$ +1 A great answer that needs no editing @StephaneLaurent $\endgroup$ Sep 23, 2012 at 11:42
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    $\begingroup$ I think you mean the No.2 formula is correct.\\Could you continue commenting on No.3 formula as I have questioned? $\endgroup$
    – jerry
    Sep 23, 2012 at 12:33
  • $\begingroup$ @jerry Yes, sorry and thanks, I meant number 2 (so I've done the correction in my answer). Formula 3 is generally wrong, could you show us the similar formula you have seen somewhere ? Maybe there were other assumptions. $\endgroup$ Sep 23, 2012 at 12:52
  • $\begingroup$ Sorry for my late reply. I paste an example with the similar formula in this thread <math.stackexchange.com/questions/211495/…> $\endgroup$
    – jerry
    Oct 12, 2012 at 6:58

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