1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Did Aug 17 '13 at 16:42
  • $\begingroup$ @Did I explained it in the question. Thanks. $\endgroup$ – triomphe Aug 17 '13 at 17:02
  • $\begingroup$ Still one mystery: "can be many to many". Sure about that? $\endgroup$ – Did Aug 17 '13 at 17:04
  • $\begingroup$ @Did not in the case I have, but I just thought to include that case too. it's ok with out that. thanks $\endgroup$ – triomphe Aug 17 '13 at 17:13
2
$\begingroup$

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 $y_1+X_3$.
$\endgroup$
  • $\begingroup$ I think what I mean is as a function of $P_{X_3}(\cdot)$... I changed the question. does it make any difference? $\endgroup$ – triomphe Aug 17 '13 at 17:53
  • $\begingroup$ 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? $\endgroup$ – Did Aug 17 '13 at 18:11
  • $\begingroup$ Thanks. sorry for changing the question, i only understood how to express it later. I included an example in the edit of the question. $\endgroup$ – triomphe Aug 17 '13 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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