• Support PF! Buy your school textbooks, materials and every day products Here!

Bivariate probabilities

  • Thread starter boneill3
  • Start date
  • #1
127
0

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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
127
0
Sorry I didn't put the code in right

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

[/itex]
= 0.125
 
  • #3
127
0
it should be determine Pr (X+Y) < 1/2
 
  • #4
392
0
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
127
0
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

[itex]
\int_{0}^{\frac{1}{4}} \int_{y}^{\frac{1}{4}} (x+y) dxdy
[/itex]
 
  • #6
33,173
4,858
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

[itex]
\int_{0}^{\frac{1}{4}} \int_{y}^{\frac{1}{4}} (x+y) dxdy
[/itex]
That's not a unit square.
 
  • #7
127
0
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

[itex]
\int_{0}^{\frac{1}{4}}\int_{0}^{\frac{1}{4}} x+y\text{ } dx dy
[/itex]
 
  • #8
392
0
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
127
0
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

[itex]
\int_{0}^{1/2}\int_{0}^{\frac{1}{2}-x} x+y\text{ }dydx
[/itex]
 
  • #10
127
0
Sorry is the limit of y 1/2-x < y < 1/2

making the integral
[itex]
\int_{0}^{\frac{1}{2}}\int_{\frac{1}{2}-x}^{\frac{1}{2}}x+y\text{ }dydx
[/itex]
 
  • #11
392
0
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

[itex]
\int_{0}^{1/2}\int_{0}^{\frac{1}{2}-x} x+y\text{ }dydx
[/itex]
This one is right. Your later post (#10) is not.
 
  • #12
127
0
Thanks alot with your help guys
 

Related Threads for: Bivariate probabilities

  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
741
  • Last Post
Replies
0
Views
2K
Replies
6
Views
4K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
1
Views
549
  • Last Post
Replies
1
Views
709
Top