Find $P\left(X+Y>\frac{1}{2}\right)$

Question

If the joint probability density of $X$ and $Y$ is given by

$$f(x,y)=\begin{cases}\dfrac{1}{y} && \text{if}\ 0 <x<y<1\\ 0 && \text{elsewhere}\end{cases}$$

Find the probability that the sum of the values of $X$ and $Y$ will exceed $\dfrac{1}{2}$

I'm having trouble with the first inequality of $0<x<y$. I have just started this Stats course and I do not understand how to set up the boundaries for this.

Answer 1

To solve these problems in general, you want to come up with a proper integral which will give the required boundaries. Given a value of $y$, what does $x$ need to be to have $X+Y> \frac{1}{2}$? Obviously $y> x> \frac{1}{2} -y$

So let: $$P(X+Y > \frac{1}{2}) = \int^a_b \int^{y}_{\frac{1}{2}-y} \frac{1}{y} \, \mathrm{d}x \, \mathrm{d}y$$

I'll leave finding $a,b$ and solving the integral as an exercise for you.