# Thread: joint probability density function

1. ## joint probability density function

If some could help me out with this problem I' really appreciate it, what I'm having trouble with on the second one is the bounds, If x plus y have to be less than or equal to 3 then shouldn't the bounds just be 0<x<2 and 1<y<3 because x can't be 3 and y cant be 0?

2. Originally Posted by jason0
If some could help me out with this problem I' really appreciate it, what I'm having trouble with on the second one is the bounds, If x plus y have to be less than or equal to 3 then shouldn't the bounds just be 0<x<2 and 1<y<3 because x can't be 3 and y cant be 0?

The bounds you mentioned won't work because for example they include the point (x, y) = (2, 3) where x+y = 5 > 3. (I'm not being very careful to distinguish between $\displaystyle >$ and $\displaystyle \geq$, etc., because it's not important for these kinds of problems, and the meaning should be clear.)

Geometrically the inequality x+y < 3 represents the region under the line x+y = 3 in the xy plane. So in D, this is the region under the diagonal going from the upper left to the lower right corner of the square.

Here's what I get:

$\displaystyle \int_1^3\int_0^{3-y}f(x,y)\,dx\,dy$
$\displaystyle =\frac{1}{12}\int_1^3\int_0^{3-y}(x+y)\,dx\,dy$
$\displaystyle =\frac{1}{12}\int_1^3\big[\frac{x^2}{2}+xy\big]_{x=0}^{x=3-y}\,dy$
$\displaystyle =\frac{1}{12}\int_1^3\bigg(\frac{(3-y)^2}{2}+(3-y)y\bigg)\,dy$
$\displaystyle =\frac{1}{24}\int_1^3\big((3-y)^2+2(3-y)y\big)\,dy$
$\displaystyle =\frac{1}{24}\int_1^3\big(y^2-6y+9+6y-2y^2\big)\,dy$
$\displaystyle =\frac{1}{24}\int_1^3\big(-y^2+9\big)\,dy$
$\displaystyle =\frac{1}{24}\big[-\frac{y^3}{3}+9y\big]_1^3$
$\displaystyle =\frac{1}{24}\bigg(-\frac{27}{3}+9(3)-(-\frac{1}{3}+9)\bigg)$
$\displaystyle =\frac{1}{24}\big(-9+27+\frac{1}{3}-9\big)$
$\displaystyle =\frac{1}{24}\bigg(\frac{28}{3}\bigg)$
$\displaystyle =\frac{7}{18}$