Probability density of product of independent variables

Question

I have two independent variables $X$ and $Y$ and they have the same density function, $f(x) = 5x^4$ for $0 \leq x \leq 1$. I am wondering how would I find the probability density function of the product of $X$ and $Y$. I suppose I would first need to find the cumulative distribution function for $XY$ and then differentiate that? Not sure how that would be done.

Answer 1

You could solve it this way: \begin{align}f_{XY}(z) & = \int_\Bbb R f_{XY\mid Y=y}(z)\,f_{Y}(y)\,dy\tag 1\\[1ex]& = \int_\Bbb R \tfrac{1}{y}f_{X\mid Y=y}(\tfrac{z}{y})\,f_{Y}(y)\,dy \tag 2\\[1ex]&= \int_\Bbb R \tfrac{1}{y}f_{X}(\tfrac{z}{y})\,f_{Y}(y)\,dy \tag 3\\[1ex]&= 25\int_\Bbb R \frac{1}{y}\frac{z^4}{y^4}y^4\,\mathbf 1_{0\leq z/y\leq 1,0\leq y\leq 1} dy \tag 4\\[1ex]& = 25z^4\mathbf 1_{0\leq z\leq 1}\int_z^1 \frac{1}{y}\,dy\tag 5\\[1ex]& = 25 z^4 \ln(\tfrac{1}{z})~\mathbf 1_{0\leq z\leq 1}\tag 6\end{align} In the first equality we use the "density version" of the law of total probability, the second equality is due to transformation formula for probability densities and the third equation is due to independence of $X$ and $Y$. In the forth equality we substitute the function, and including the indicators for the joint support, which is equivalent to $\mathbf 1_{0\leq z\leq y\leq 1}$, so explaining why the domain for the integral is $(z..1)$.

Answer 2

HINT Here is one approach. Let $Z = XY$ then $$ \begin{split} F_Z(z) &= \mathbb{P}[Z \le z] \\ &= \mathbb{P}[XY \le z] \\ &= \int_0^1 \mathbb{P}[XY \le z | X = x] f_X(x)\ dx \\ &= \int_0^1 \mathbb{P}[Y \le z/x] f_X(x)\ dx \\ &= \int_0^1 F_Y(z/x) f_X(x)\ dx \\ \end{split} $$

Can you finish?