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There is a similarly worded question here but it doesn't quite answer my question.

I have three variables, $x, y, \text{and } z$, each with their own unique probability density functions. I want to consider a new variable described by some complicated function of these three variables that is not 1-to-1. As an example, something like this:

$u = sin(x+y)-cos(z)$

or any other non-unique mapping. I want to figure out the probability density for this variable. I know the densities $X(x), Y(y), \text{and } Z(z)$ so I also know the joint distribution ( $f(x,y,z) = X(x)Y(y)Z(z)$ because they are independent ).

1) Can I analytically figure out the distribution U(u)?

2) If not, can I find the distribution numerically?

3) If I can't say anything useful about this distribution, why?

Any references are also appreciated--I haven't had much luck finding anything on this though I may just be searching the wrong terms.

NothingQuenchier
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  • Did you try to Google for jacobian random variabele transform? –  Aug 30 '17 at 16:18
  • Yes but as far as I can tell that's only useful for 1-to-1 transformations, right? The case I'm suggesting would not give a square Jacobian. – NothingQuenchier Aug 30 '17 at 16:19
  • see section 2.3 of https://faculty.math.illinois.edu/~r-ash/Stat/StatLec1-5.pdf, there is a trick (W=Y), so in your case you can try $u=sin(x+y)-cos(z),v=x,w=y$. –  Aug 30 '17 at 16:22
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    I don't see the problem with 2) - By "finding the distribution numerically", do you just refer to being able to sample from it? That should be straightforward, just sample from them individually and apply the function – Sam Aug 30 '17 at 17:24
  • See, *inter alia* (there are many threads about this!) https://stats.stackexchange.com/questions/180715, https://stats.stackexchange.com/questions/138763, and https://stats.stackexchange.com/search?q=density+jacobian. – whuber Aug 30 '17 at 17:48
  • Thanks everyone--I think the method in the UIUC notes and one of the bottom links is what I needed. @Sam, I wasn't sure if there was any problem with 2, but I'm still new to sampling and wasn't sure if my intuition was correct. – NothingQuenchier Aug 30 '17 at 18:12

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