# conditioning on a mixture random variable [closed]

Suppose we have the conditional distribution $P(X,Y|Z)$ and $Z$ follows a mixture distribution such that $P(Z)=q*P(Z_1) + (1-q)*P(Z_2)$ where $q \in (0,1)$. Then, how can I write the conditional distribution $P(X,Y|Z)$ in terms of $Z_1$ and $Z_2$?

My intuition tells me that $P(X,Y|Z) = q*P(X,Y|Z_1) + (1-q)*P(X,Y|Z_2)$, but I am not sure if this is correct, or why it is correct.

## closed as unclear what you're asking by Did, Shailesh, M. Vinay, JonMark Perry, Kushal BhuyanMay 28 '16 at 2:30

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• When one says that the distribution of $Z$ is a mixture, there is no random variables $Z_1$ and $Z_2$ defined, only some distributions $\mu$ and $\nu$ such that $P_Z=q\mu+(1-q)\nu$. Thus, the question of knowing whether the conditional distribution $P_T(\ \mid Z)$ can be written in terms of $P_T(\ \mid Z_1)$ and $P_T(\ \mid Z_2)$ (with $T=(X,Y)$) is absurd. – Did May 27 '16 at 21:35
• Is it possible if $Z_1$ and $Z_2$ map from the same measurable space $(\Omega, \sigma)$ to the Borel space? – stephani May 27 '16 at 21:40
• Sorry but when a major objection is raised against the very principle of a construction you have in mind, trying to push the problem under the rug as if nothing had happened is not a very fruitful reaction. As I said, there are tons of random variables $Z_1$ and $Z_2$ with prescribed distributions hence to expect that $P_T(\ \mid Z)$ coincide with $qP_T(\ \mid Z_1)+(1-q)P_T(\ \mid Z_2)$ for all of them is just ludicrous. By the way, I note that you obviously did not try to check the assertion in very simple cases (which would have saved time and helped you better grasp the problem). – Did May 27 '16 at 21:47