Homework Help: Bivariate probabilities

1. May 8, 2009

boneill3

1. The problem statement, all variables and given/known data
The problem statement, all variables and given/known data[/b]
Let X and Y have the joint pdf
fXY(x,y) = x + y, 0<x<1, 0<y<1

determine Pr X+Y , 1/2

2. Relevant equations

3. The attempt at a solution
[ tex ]
\int_{0}^{\frac{1}{2}}\int_{0}^{\frac{1}{2}}x+y dx dy

[ /tex ]
= 0.125
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 8, 2009

boneill3

Sorry I didn't put the code in right

$\int_{0}^{\frac{1}{2}}\int_{0}^{\frac{1}{2}}x+y dx dy$
= 0.125

3. May 8, 2009

boneill3

it should be determine Pr (X+Y) < 1/2

4. May 9, 2009

Billy Bob

Do you mean P(X+Y<1/2)? Your integral gives P(X<1/2 and Y<1/2). Draw the region in the unit square where x+y<1/2.

5. May 9, 2009

boneill3

Yes I mean P(X+Y<1/2)?
The unit square would be

(0,0) (0,1/4) (1/4,0) and (1/4,1/4)

would the limits than be 0< x <= y < 1/4

And the integarl

$\int_{0}^{\frac{1}{4}} \int_{y}^{\frac{1}{4}} (x+y) dxdy$

6. May 10, 2009

Staff: Mentor

That's not a unit square.

7. May 10, 2009

boneill3

Isn't the unit square just the square with side length 1?

ie

(0,0) (0,1) (1,0) (1,1)

So we are looking at the area where x+y < 1/2

Wouldn't that be in the region (0,0) (0,1/4) (1/4,0) (1/4,1/4) of the unit square ?

so the lmits are

0 < x < 1/4 and 0<y<1/4

$\int_{0}^{\frac{1}{4}}\int_{0}^{\frac{1}{4}} x+y\text{ } dx dy$

8. May 10, 2009

Billy Bob

No. Draw {(x,y) : 0<x<1, 0<y<1, and x+y<1/2}. Think calc 1 or calc 3, or even college algebra, instead of probability theory. Find the boundary line x+y = 1/2, then shade one side of it.

9. May 11, 2009

boneill3

Is this the area bounded by the triangle with vertices (0,0) (0,1/2) (1/2,0) ?

with limits

0 < x < 1/2 0 < y < 1/2-x

$\int_{0}^{1/2}\int_{0}^{\frac{1}{2}-x} x+y\text{ }dydx$

10. May 11, 2009

boneill3

Sorry is the limit of y 1/2-x < y < 1/2

making the integral
$\int_{0}^{\frac{1}{2}}\int_{\frac{1}{2}-x}^{\frac{1}{2}}x+y\text{ }dydx$

11. May 11, 2009

Billy Bob

This one is right. Your later post (#10) is not.

12. May 11, 2009

boneill3

Thanks alot with your help guys