Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
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

2 Answers 2

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.

share|improve this answer

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...)

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.