I am interested in the density of Z ($Z=X*Y$), conditional on X>c: $Pr(Z=z|y>c)=f_{Z}(z|Y>c)$.
$X$ follows a Uniform distribution on the interval [0,k], i.e. $f_X (x)=\frac{1}{k}$ if $x \in [0,k]$, and 0 otherwise.
$Y$ follows a Pareto distribution, i.e. $f_Y (y)=\frac{a w^{a}}{y^{a+1}}$ if $y \in [w,+ \infty[$, and 0 otherwise.
X and Y are assumed to be independent.
Here is how I try to find $f_{Z}(z|Y>c)$:
(1) I define the individual densities of X and Y conditional on Y>c:
$f_X (x|Y>c)=f_X (x)$ since X and Y are independent.
$f_Y (y|Y>c)=\frac{f_Y{y}}{Pr(Y>c)}=\frac{a c^{a}}{y^{a+1}}$ if $y \in [c,+ \infty[$, and 0 otherwise.
(2) I use the following theorem to get the density of the product of two independent random variables (Z=X*Y):
$f_Z (z)= \int f_Y (y) f_X (\frac{z}{y}) \frac{1}{|y|} dy$
(3) I introduce the conditionality:
$f_Z (z|Y>c)= \int f_Y (y|Y>c) f_X (\frac{z}{y}) \frac{1}{|y|} dy$
(4) Then, I replace by the functions $f_Y$ and $f_X$:
$f_Z (z|Y>c)= \int_{c}^{k} \frac{a c^{a}}{y^{a+1}} \frac{1}{k} \frac{1}{y} dy$ if z on the interval [c,k], and $f_Z (z|Y>c)=0$ otherwise.
Here I am not sure about the definition of the interval of the integral, i.e. from c to k.
Also, should I do it this way or should I rather first define the density of V and then define the density of V conditional on the value of Y?
Consider \begin{equation} \begin{split} P(Z \gt z/ X > c) &= \frac{P(Z > z, X > c)}{P(X > c)} \\ &= \frac{P(XY > z, X > c)}{P(X > c)} \\ &= \frac{P(Y > z/c, X > c)}{P(X > c)} \text{(Not completely certain about this step)}\\ &= \frac{P(Y > z/c).P(X>c)}{P(X > c)} \\ &= P(Y > z/c) \\ &= 1-F_Y(z/c) \end{split} \end{equation} Hence \begin{equation} P(Z \leq z/X > c) = 1- P(Z \gt z/ X > c) = F_Y(z/c) \end{equation} Differentiate w.r.to $z$ both sides, then \begin{equation} f_Z(z/X > c) = f_Y(z/c) \frac{1}{c} \end{equation}