In many research papers, I came across following form of probability expression as shown by equation (A) and (B).
$F_{\Upsilon}(\gamma)= \text{Pr}(Y\leq\frac{\gamma}{X})$---(A)
where both $Y,X$ are random variables and all others are constant.
Further, eq.(A) is expressed as
$F_{\Upsilon}(\gamma)= \int_0^{\infty}\text{Pr}(Y\leq\frac{\gamma}{x})f_X(x)$ ---(B)
My question is how (B) is obtained from (A).
Any help in this regard will be highly appreciated.
I assume that the problem is formulated as:
$X$ and $Y$ are independent random variables with CDF's $F_X(x) = \text{Pr}(X\leq x)$, $F_Y(y) = \text{Pr}(Y\leq y)$, and PDF's $f_X(x) = F'_X(x)$, $f_Y(y) = F'_Y(y)$.
Find a CDF of a random variable $Z = XY$.
Then,
$$ \begin{aligned} F_Z(z) &= \text{Pr}(Z\leq z) = \text{Pr}(XY\leq z) = \text{Pr}\left(Y\leq\frac{z}{X}\right) = \\ &= \left[ \begin{array}{l} \text{we have to sum up the probability values for all possible values of }X \Leftrightarrow\\ \text{integrate over domain of }X\text{ with respect to density }f_X(x). \text{ For example, if } X >0 : \end{array}\right] = \\ &=\int_0^{+\infty}\underbrace{\text{Pr}\left(Y\leq\frac{z}{x}\right)}_{=F_Y\left(\frac{z}{x}\right)}f_X(x)dx = \\ &= \int_0^{+\infty}F_Y\left(\frac{z}{x}\right)f_X(x)dx \end{aligned} $$