Give the density of the product of independent Uniform random variables

Question

I had a lot of trouble doing this problem on my exam. Can anybody please tell me why my attempt fails?

Let $X$~$Uniform(0,1)$ and $Y$~$Uniform(0,1)$ be independent random variables. Find the density of $Z = XY.$

Solution attempt: For $a\in(0,1):$ $F_Z(a) = P(Z<a) = P(XY < a) = P(Y < \frac{a}X) = \int_0^1\int_0^{\frac{a}x}f_X(x)f_Y(y)dydx = \int_0^1\int_0^{\frac{a}x}1dydx = \int_0^1(\frac{a}x)dx = (aln|x|)\rvert_0^1 = aln(1) - aln(0) = -aln(0)$ which is undefined..

Answer 1

You forgot the condition $\frac a x <1$ when you wrote the integral from $0$ to $\frac a x$. It should be $$\int_0^1\int_0^{\min\{1,a/x\}}1\,\mathrm d y\,\mathrm d x ~=~ \int_a^1\int_0^{\frac{a}x}1\,\mathrm d y\,\mathrm d x+ \int_0^a\int_0^{1}1\,\mathrm d y\,\mathrm d x$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Hereafter, $\ds{\bracks{\cdots}}$ is an $\ds{Iverson\ Bracket}$. \begin{align} & \color{#44f}{\int_{0}^{1}{\bracks{0< z/x <1} \over x}\dd x} = \bracks{z > 0}\int_{0}^{1} {\bracks{x > z} \over x}\dd x \\[5mm] = & \ \bracks{z > 0}\bracks{z < 1}\int_{z}^{1} {\dd x \over x} = \bbx{\color{#44f}{-\bracks{0 < z < 1}\ln\pars{z}}} \end{align}