# Thread: Probability of a joint p.d.f.

1. ## Probability of a joint p.d.f.

$f_{X,Y}(x,y) = \{x + y , 0 \leq x \leq 1, 0 \leq y \leq 1; 0 otherwise$

Find $Pr(X+Y \leq 1)$

I understand i need to do $\int \int (x+y) dy dx$
But i can't work out what the limits of the integrals should be, any help would be great, thanks.

2. have you thought about doing the inner integral from 0 to 1 (since that's the range of y), and the outer integral from 0 to 1 (since that's the range of x?

Also, if you graph the pdf, you'll notice that it's the area of a circle...so why not find the probability of landing in a circle of radius 1 inscribed in a 1x1 box? This should be the area of that circle divided by the area of the square.

3. Originally Posted by Rapid_W
$f_{X,Y}(x,y) = \{x + y , 0 \leq x \leq 1, 0 \leq y \leq 1; 0 otherwise$

Find $Pr(X+Y \leq 1)$

I understand i need to do $\int \int (x+y) dy dx$
But i can't work out what the limits of the integrals should be, any help would be great, thanks.
A diagram makes things crystal clear. The support of the joint pdf is a square. You need to integrate the joint pdf over the lower region of this square that is bounded by the line y = 1 - x.

Originally Posted by financial
have you thought about doing the inner integral from 0 to 1 (since that's the range of y), and the outer integral from 0 to 1 (since that's the range of x?

Also, if you graph the pdf, you'll notice that it's the area of a circle...so why not find the probability of landing in a circle of radius 1 inscribed in a 1x1 box? This should be the area of that circle divided by the area of the square.
Sorry, but this is not correct.