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It's obvious that when $X,Y,Z$ are independent, we have $$P\{(X|Y)|Z\} = P\{X | (Y \cap Z)\},$$ but if we only have $Y$ and $Z$ are independent, does this equation still holds?


OK, a bit about how this comes. I saw an attempt to calculate $P\{X|Y\}$ goes like this $$ P\{X|Y\} = P\{X|Y \cap Z\}P\{Z\} + P\{X|Y \cap Z^c\}P\{Z^c\}. $$ My interpretation of this is that \begin{align*} & P\{X|Y\} = P\{(X|Y) \cap Z\} + P\{(X|Y) \cap Z^c\} \\ & \; = P\{(X|Y) | Z\}P\{Z\} + P\{(X|Y) | Z^c\}P\{Z^c\} \\ & \; = P\{X|Y \cap Z\}P\{Z\} + P\{X|Y \cap Z^c\}P\{Z^c\}. \end{align*}

So I was wondering what is the condition to have

$$ P\{ (X|Y) | Z\} = P\{X|Y \cap Z \} $$

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What are $X$, $Y$, and $Z$? Events? Random variables? – Dilip Sarwate May 31 '12 at 15:51
They are events – ablmf May 31 '12 at 15:56
Regarding your edit: see my answer. Your two expressions are actually the same thing. – leonbloy May 31 '12 at 17:26
up vote 3 down vote accepted

The only interpretation I can give to the (slightly strange for me) $P\{(X|Y)|Z\}$ notation is "probability of event X given ocurrence of event Y and event Z". Which is precisely $P\{X | Y \cap Z\}$; or, as it's more simply and commonly writen, $P(X | Y Z)$.

No need to ask about independence here.

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These events are not exactly the "same thing". In general, if you have a probability distribution $P$ and an event $A$, you can condition on $A$ to get a new probability distribution $P'$.

So in this case you could either condition on $Z$ and to get distribution $P'$, and then condition $P'$ on $Y$ to get distribution $P''$, or you could go directly from $P$ to $P''$ by conditioning on $Y \cap Z$.

In either case, you get the same probability distribution at the end.

(If $Y \cap Z$ has probability zero then this may be more complicated...)

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