What is the probability of X+Y being less than 1/2?

[boneill3]

Homework Statement

Homework Statement [/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

Homework Equations

The Attempt at a Solution

[ tex ]
\int_{0}^{\frac{1}{2}}\int_{0}^{\frac{1}{2}}x+y dx dy

[ /tex ]
= 0.125

 

[boneill3]

Sorry I didn't put the code in right

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

 

[boneill3]

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

 

[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.

 

[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

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

 

[Mark44]

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

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

That's not a unit square.

 

[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

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

 

[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.

 

[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

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

 

[boneill3]

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

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

 

[Billy Bob]

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

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

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

 

[boneill3]

Thanks a lot with your help guys