1

if the joint probability density of X and Y is given by

$f(x,y)=\frac{1}{y}$ for $0<x<y,0<y<1$

0 else where

find the probability that the sum of values X and Y will exceed 1/2. So this would be

$x+y>1/2$

But I am not sure how to set up an integral to solve this problem.

Fernando Martinez
  • 6,698
  • 19
  • 74
  • 108

1 Answers1

2

Your support, call it $A$, is given by $\color{blue}{A}=\{(x,y)\mid f(x,y)>0 \}$ and $\color{red}{D}$ being the region of probability that $(x, y)$ lie in $$\begin{align}P\left(( X,Y)\in D\right)&=P\left(X+Y>\frac{1}{2}\right) \\&=\iint\limits_{(x,y)\in \color{red}{D} ~\cap \color{blue}{A}} f(x,y)\, dx \, dy \\&= \iint\limits_{\substack{ \color{blue}{ 0<x<y \\ 0<y<1 }\\ \color{red}{x+y>1/2} }} \frac{1}{y} \,dx \,dy \end{align}$$

To translate this into an iterated integral, you can draw the region of integration: enter image description here

Notice that we had to separate the integration region into two regions: $E_1$ and $E_2$ in that manner if you integrate with respect to $x$ first. So you obtain: $$P\left( X+Y > \frac{1}{2}\right)= \underbrace{\int_{1/4}^{1/2} \int_{\color{red}{-y+1/2}}^{\color{blue}{y}}\frac{1}{y} \, dx \, dy}_{\text{region for }E_1}+\underbrace{\int_{1/2}^{\color{blue}{1}} \int_{\color{blue}{0}}^{\color{blue}{y}}\frac{1}{y}\, dx \,dy}_{\text{region for }E_2}$$

user84226
  • 393