joint probability density function

**jason0** · April 7th 2010, 01:10 PM

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?

**undefined** · April 7th 2010, 03:35 PM

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 $>$ and $\geq$ , 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:

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

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