Let and be independent positive random variables. Let =/. In what follows, all occurrences of , , are assumed to be positive numbers. Suppose that and are continuous, with known PDFs, $_$ and $_$. Provide a formula for $_{|}(|)$ in terms of $_$. That is, find and in the formula below.
$$f_{Z|Y}(z|y)=Af_X(B)$$
There is a general formula for these type of conditional PDFs:
$$f_{Z\mid Y}(z\mid y) = \frac{f_{Z,Y}(z,y)}{f(y)}$$
Rearranging $=/$ in terms of $X$ and substituting that into the previous formula I derived the following:
$$f_{X=YZ\mid Y}(yz\mid y) = \frac{f_{X,Y}(yz,y)}{f(y)} = \frac{f_{X}(yz)f_{Y}(y)}{f(y)} = f_{X}(yz)$$
I know for sure that $B=yz$, however, given the following equalities:
$$f_Z (z) = \int_{0}^\infty yf_Y (y)f_X (yz) \ dy = \int_{0}^\infty f_Y (y)f_{Z|Y} (z|y) \ dy$$
Then, my approach is wrong and $A=y$ instead of $A=1$.
Could anyone please explain why this is the case?
Apparently, the conditional probability of $Z$ given $Y$ as conditional CDF of $X$ proves the equalities above correct:
$$F_{Z | Y}(z | y) = P(Z<=z | Y=y) = P(X/Y<=z | Y=y) = P(X<=zy) = F_{X}(yz)$$
Then, differentiating by $\dfrac{d}{dz}$ on both sides and using the chain rule yields
$$f_{Z | Y}(z | y) = yfx(yz)$$
Again, confirming that $A=y$.
Any advice is greatly appreciated. Thanks!