Find probability of $P(X Y < 1)$

Question

Let $X$ and $Y$ be two independent $\mathrm{Uniform}(0,2)$ random variables. Find $P(XY < 1)$

I started off by finding the pdf

$f_X(x)=\frac{1}{2} $ when $0<x<2$ Same for $Y$.

I then found their joint PDF via independence:

$f_{XY}(xy)=\frac{1}{4} $ when $0\leq y \leq 2 $ and $0 \leq x \leq 2 $ Otherwise $0$

$\int_0^2 \int_0^{1/y} \frac{1}{4} dx dy$ But this cannot be solved, so where did i go wrong?

Answer 1

Here is the graph of $P(x,y)$ in the region of interest ($x y < 1$).

$$P[x y <1] = \int\limits_{x=0}^2 \int\limits_{y=0}^{\min[2, 1/x]} {1 \over 4} dy\ dx = \frac{1}{4} (1+2 \log (2))$$

Your error was forgetting that the upper limit on $y$ was bounded by $2$.

Answer 2

Alternatively, you can consider the complementing blue area:

$\hspace{2cm}$

$$\begin{align}\mathbb{P}(XY<1)&=1-\mathbb{P}(XY>1)=\\ &=1-\int_{1/2}^2\int_{1/x}^2 \frac14dydx=\\ &=1-\int_{1/2}^2\left(\frac12-\frac1{4x}\right)dx=\\ &=1-\left(\frac12x-\frac14\ln x\right)\big{|}_{1/2}^2=\\ &=\frac14+\frac12\ln 2.\end{align}$$