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Given $P(G|E_1| E_2)$:

Would it be wrong to say $P(G|[E_1| E_2])=P([G|E_1]| E_2)$ ?

By this I mean considering that $E_1| E_2$ is an event that gives the condition for event $G$ / considering that $E_2$ is an event that gives the condition for event $G|E_1$?

If i could treat it as an event would i be able to say (by the definition of conditional probability)

$$P(G|[E_1| E_2])= \frac{P(G\cap [E_1| E_2])}{P(E_1| E_2)}$$

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    $\begingroup$ I don't think the notation $G|(E_1|E_2)$ is convention. I think you mean $P(G | (E_1 \cap E_2))$ or $P(G | E_1, E_2)$ $\endgroup$
    – David P
    Aug 24, 2016 at 2:40
  • $\begingroup$ How do you consider $E_1|E_2$ (or $G|E_1$) as an event? $\endgroup$
    – Andreas Blass
    Aug 24, 2016 at 2:40
  • $\begingroup$ Ah not sure, thats why i am asking? $E_1|E_2$ can be seen as the event that $E_1$ given $E_2$ has occuried? $\endgroup$
    – Brofessor
    Aug 24, 2016 at 2:53
  • $\begingroup$ @Brofessor If $E_1=\{1,2,3,4,5\}$ and $E_2=\{2,4,6\}$ how would you describe the proposed event $E_1|E_2$? $\endgroup$
    – David P
    Aug 24, 2016 at 3:01
  • $\begingroup$ I see, i think this was just an abuse of notation and a misunderstanding of events that are conditioned, thanks! my bad :\ $\endgroup$
    – Brofessor
    Aug 24, 2016 at 3:06

1 Answer 1

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Given $P(G\mid E_1 \mid E_2 )$

It is doubtful you'd ever be given such.   There is no such beast.

They can only ever be one conditioning symbol in a function.   Never more.   It is not a set operator like $\cap,\cup,\setminus$, but rather belongs to the function itself.   In a probability function it separates the event being measured and the condition it is being measured over.

My best interpretation of your intent is that you are looking for:

$$\mathsf P(G\mid E_1, E_2) ~=~ \dfrac{\mathsf P(G, E_1\mid E_2)}{\mathsf P(E_1\mid E_2)}~=~\dfrac{\mathsf P(G, E_1, E_2)}{\mathsf P(E_1, E_2)}$$

Equivalently $$\mathsf P(G\mid E_1 \cap E_2) ~=~ \dfrac{\mathsf P(G\cap E_1\mid E_2)}{\mathsf P(E_1\mid E_2)}~=~\dfrac{\mathsf P(G\cap E_1\cap E_2)}{\mathsf P(E_1\cap E_2)}$$

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  • $\begingroup$ I see! So my interpretation is incorrect, because the conditional probability is a probablistic function. Great that really cleared it up for me! $\endgroup$
    – Brofessor
    Aug 24, 2016 at 8:31

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