I've seen a trick for finding the p.d.f of $r(X,Y)$ where $X$ and $Y$ are r.v's by first calculating the cdf i.e $P(r(X,Y) \leq l)$ and then differentiating to find the pdf. So if $\Omega = \{(x,y) | r(x,y) \leq l)$ then $P(r(X,Y) \leq l) = \int_{\Omega} f_{X,Y}(x,y).$

However what if i want the pdf of $r(X,Y)$ conditioned on $X=t$ ? I'm thinking i might do something like this,

First calculate $f_{Y|X}(y|x=t)$ to get the conditional pdf. Then let $\Omega' = \{y | r(t,y) \leq l\}$ then the cdf we are interested in is $\int_{\Omega'} f_{Y|X}(y|x=t)$, this gives me the conditional cdf of $r(t,y)$, i can then differentiate to find the pdf.

Is this correct?


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