# Conditional distribution of a function of random variables

I have a question about conditional distribution. Suppose we have three independent random variables $X_1$, $X_2$, $X_3$.

Then we have mapping $Y_1=g(X_1, X_2)$. The mapping is not necessarily an invertible function, meaning it can be many to one eg. many combinations of $(X_1, X_2)$ have same $Y_1$ value. And we also have another mapping $Y_2=f(X_1,X_2,X_3)$.

My question is can we say the distribution functions follow $$P_{Y_2 \mid X_1,X_2}(y_2)=P_{X_3}(v)=P_{Y_2 \mid Y_1}(w)$$ where $v$ and $w$ are function of $x_1, x_2,y_2$. Please explain the answer. If $Y_1, Y_2$ are invertible function meaning one to one does the answer change?

Edit: Example $$Y_1=X_1+X_2$$ $$Y_2=X_1+X_2+X_3$$ where $X_1,X_2,X_3$ are independent. And $P_{X_3}$ is the distribution function of $X_3$. I think the above relation hold in this case. Does it hold generally? is my question.

• Two highly unusual notions in your question: "The mapping is not necessarily a function and can be many to many" Can you explain? "If Y1,Y2 are proper invertible function" Can you explain? – Did Aug 17 '13 at 16:42
• @Did I explained it in the question. Thanks. – triomphe Aug 17 '13 at 17:02
• Still one mystery: "can be many to many". Sure about that? – Did Aug 17 '13 at 17:04
• @Did not in the case I have, but I just thought to include that case too. it's ok with out that. thanks – triomphe Aug 17 '13 at 17:13

Some general facts which yield the answers to your questions:

• If the random variables $X$ and $Y$ are independent, then $E[u(X,Y)\mid Y]=v(Y)$, where $v$ is defined by $v(y)=E[u(X,y)]$ for every $y$.
• For every random variables $X$ and $Y$, $E[X\mid u(Y)]=E[E[X\mid Y]\mid u(Y)]$.
• For some unspecified random variable $Y$ and functions $u$ and $v$, there is no general formula for $E[u(Y)\mid v(Y)]$.

Hence none of the two idendities you suggest holds.

Edit: In the specific case $Y_1=X_1+X_2$ and $Y_2=X_1+X_2+X_3$:

• the conditional distribution of $Y_2$ conditionally on $(X_1,X_2)=(x_1,x_2)$ is the distribution of $x_1+x_2+X_3$,
• the distribution of $X_3$ is... well, the distribution of $X_3$,
• the conditional distribution of $Y_2$ conditionally on $Y_1=y_1$ is the distribution of $y_1+X_3$.
• I think what I mean is as a function of $P_{X_3}(\cdot)$... I changed the question. does it make any difference? – triomphe Aug 17 '13 at 17:53
• You seem to be thinking exclusively in terms of conditional densities--whose existence is not guaranteed in general. A general remark: to change the question after some answer is posted is not recommended (and I am sure you can guess the reasons). A specific remark: I am not sure to understand the revised version. Would you have a specific example in mind? – Did Aug 17 '13 at 18:11
• Thanks. sorry for changing the question, i only understood how to express it later. I included an example in the edit of the question. – triomphe Aug 17 '13 at 18:19