Does the $y$ variable act as an indicator function such that $f_X(y)$ only has values for elements $\omega \in X\cap Y$ ?
Is $\int_Y f_x(y)dy$ the probability distribution of $X$ over the sample space of $Y$?
Forgive me if some of the terminology is inaccurate as I study this course in another language. This has been bothering me for some time and I would appreciate it in the extreme if somebody could enlighten me.
Backstory:
In one of the exercises of my probabilit we have two independent continous r.v.s' $X \geq 0$ and $Y \geq 0$ and we are supposed to show that their product:
$f_{XY}(x)=\int_0^\infty \frac{1}{y} f_X(\frac{x}{y})f_y(y)dy=\int_0^\infty \frac{1}{y} f_X(y)f_y(\frac{x}{y})dy$
I can get the first part of it by setting $Z=XY$ and $U=Y$ but what I have a hard time understanding is the second part what does $f_X(y)$ mean? The only thing I could think of was that you could:
$P(XY\leq x)= \int_Y P(Y \leq x/y|X=y)P(X=y)$
Then by independence the condition falls away and you would get the second term, but that does not seem right since $P(X=y) =0$ as they are both continous right?
It might be more accurate to write $P(XY \le x) = \int_0^\infty P(Y \le x/y) f_X(y) \, dy$, where $f_X$ is the marginal density of $X$.
This comes from the law of total expectation. If $1_{XY \le x}$ denotes the indicator random variable for the event $\{XY \le x\}$ then $$P(XY \le x) = E[1_{XY \le x}] = E[E[1_{XY \le x} \mid X]] = \int_0^\infty E[1_{XY \le x} \mid X=y] f_X(y) \, dy = \int_0^\infty P(Y \le x/y) f_X(y) \, dy.$$