Let X and Y be real random variables with joint density $f$ (with respect to Lebesgue measure $\lambda ^2$ on $\mathbb R ^2$).For $x \in \mathbb R$,define
$f_X (x)=\int _\mathbb R f(x,y)\lambda (dy)$.
Clearly,$f_X (x)＞0$ for P$_X$-a.a. $x \in \mathbb R$ and $f_X ^{-1}$ is the density of the absolutely continuous part of the Lebesgue measure $\lambda$ with respect to P$_X$.
My question:The above is cited from the book that I am currently reading,and what confuses me is that why $f_X ^{-1}$ is the density of the absolutely continuous part of $\lambda$ with respect to P$_X$?Is $f_X ^{-1}$ itself a function at all?
Consider the square $[0,1] \times [0,1] \subset \mathbb R \times \mathbb R$ on which (X,Y) is uniformly distributed,then obviously $f(x,y)=1$ holds,by construction.Hence,we obtain directly that $f_X (x)=1$ for all $x \in [0,1]$,and therefore,$f_X ^{-1}$ is not even a function,and so it cannot be a density (as far as I can remember,every density is itself a nonnegative function).
Can someone give me a little clue?Thanks!
You are just mistaking the reciprocal for the inverse. In this statement $f_X^{-1}$ stands for $\frac 1 {f_X}$.