1. ## Joint probability density

If the joint probability density of X and Y is given by

f(x,y)=2 for x>0, y>0, x+y<1
and zero elsewhere

find P(X>2Y)

When I draw a picture with that restriction above( y<(1/2)x and y<1-x )
I get 1/3, but I want to solve this one by using double integral

2. Originally Posted by Judi
If the joint probability density of X and Y is given by

f(x,y)=2 for x>0, y>0, x+y<1
and zero elsewhere

find P(X>2Y)

When I draw a picture with that restriction above( y<(1/2)x and y<1-x )
I get 1/3, but I want to solve this one by using double integral
The double integral is just (this is tough without latex)

double integral over D 2 dydx

where the set D satisfies your 4 inequalities. Turning this double integral into an iterated integral to evaluate has to go back to the picture. From that picture I get the iterated integral will have two parts depending on whether x/2 < 1 - x or not:

int (0,2/3) int (0,x/2) 2 dydx + int (2/3,1) int (0,1-x) 2 dydx =

int (0,2/3) 2(x/2) dx + int (2/3,1) 2(1-x) dx.

This will equal 1/3 as you say.

3. Thank you!

4. Is there another way to do this one?
I am trying to understand what my professor did:

P(X>2Y)=2((2/3)(1/3)(1/2)+(1/3)(1/3)(1/2))=1/3

5. Originally Posted by Judi
Is there another way to do this one?
I am trying to understand what my professor did:

P(X>2Y)=2((2/3)(1/3)(1/2)+(1/3)(1/3)(1/2))=1/3
It looks like he/she just used the picture: that's 2 times the sum of the areas of the two triangles in the picture.

6. why is it times 2?
I can see two triangels inside the big triangle.
Why not just sum of the two triangles?
I am curious...

7. Originally Posted by Judi
why is it times 2?
I can see two triangels inside the big triangle.
Why not just sum of the two triangles?
I am curious...
2 is the constant in the density f(x,y) that converts area to probability. The total probability must equal 1; the total area covered when f(x,y) > 0 is 1/2. So multiply by 2 to convert area to probability.