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Specifically, say I have two events, A and B, and some distribution parameters $ \theta $, and I'd like to look at $P(A | B,\theta)$.

So, the simplest definition of conditional probability is, given some events A and B, then $P(A|B) = \frac{P(A \cap B)}{P(B)}$. So if there are multiple events to condition on, like I have above, could I say that $P(A | B,\theta) \stackrel{?}{=} \frac{P((A | \theta)\cap(B | \theta))}{P(B|\theta)}$ or am I looking at the in totally the wrong way? I tend to psych myself out when I deal with probability sometimes, I'm not really sure why.

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  • $\begingroup$ What's the union of $A$ and $B,\theta$? $\endgroup$
    – Ana S. H.
    Mar 21 '17 at 6:05
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You can do a little trick. Let $(B \cap \theta) = C$. Now you can write

$$P(A|B, \theta) = P(A|C).$$ The problem reduces to that of a conditional probability with only one condition: $$P(A|C) = \frac{P(A \cap C)}{P(C)}$$

Now fill in $(B \cap \theta)$ for $C$ again and you have it:

$$\frac{P(A \cap C)}{P(C)} = \frac{P(A \cap (B \cap \theta))}{P(B \cap \theta)}$$

And this is the result that you wanted to get to. Let's write this in exactly the form you had when you originally stated the question:

$$P(A|B , \theta) = \frac{ P(A \cap B \cap \theta) }{ P(B \cap \theta) }$$

As to your second question, why it is that probability freaks you out: it is one of the findings from psychological research that humans are not very good at probabilistic reasoning ;-). It was a bit hard for me to find a reference that I can point you to. But the work of Daniel Kahneman is certainly very important in this regard.

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I think you probably want this:

$$\rm{P}(A|B,\theta) = \frac{\rm{P}(A\cap B|\theta)}{\rm{P}(B|\theta)}$$

I often find it confusing thinking about how to manipulate probabilities. With multiple conditions, I find it easiest to think about it this way:

  • temporarily remove the condition(s) that you want to remain as conditions in your result. In this case write $\rm{P}(A|B)$, taking out $\theta$.
  • apply the normal rules. In this case $\rm{P}(A|B) = \rm{P}(A\cap B)/\rm{P}(B)$.
  • restore the condition(s) that were removed. In this case, restore $\theta$, to get the result $\rm{P}(A|B,\theta) = \rm{P}(A\cap B|\theta)/\rm{P}(B|\theta)$.
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  • $\begingroup$ Wouldn't P(A|B) = P(B and A)/P(B). So wouldn't something like this be correct? P(A|B,C) = P(C and B and A)/P(C and B) $\endgroup$ Mar 28 '15 at 19:05
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    $\begingroup$ @DashControl Yes, and if you expand TooTone's expression you will get the exact same result. They're the same thing :) $\endgroup$
    – Josh Chen
    Aug 18 '15 at 2:53
  • $\begingroup$ P(A|B,θ)= (P(A∩B|θ) * P(θ)) / (P(B|θ) * P(θ)) = P(A∩B∩θ) / P(B∩θ) $\endgroup$ Dec 27 '18 at 15:06

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