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

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

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

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    If the condition was $0 < x <1$ and $0 < y < 1$, then the 'interesting' area would be the the interior of the unit square between $(0,0)$ and $(1,1)$. Now the real condition is just an additional $x < y$. How can you describe the points of the unit square that have $y$-coordinate bigger than the $x$ coordinate? Maybe look at the function "y=x" and how it divides the plane into 2 parts? – 2017-02-24

1 Answers 1

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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.