Joint pdf/marginal/expectation related

[markov4]

Homework Statement

Homework Equations

Well i know the double integral formula for the joint pdf

The Attempt at a Solution

So firstly we're told to find the joint pdf of X and Y. I'm not sure where to begin. So how do we find the function that we take the double integral over? I'm suspecting 1/2*b*h will be used and I know the joint pdf must equal 1, so do we equate this somehow to the formula for the area of a triangle? Could someone just point me in the right direction please? I suspect my bounds will be 0->1 and 0->x right? Or instead of the second one, 0->1-x ?

Thanks in advance!

 

[lanedance]

so as there is uniform probabilty the area of any subset of the triangle will be directly proportional to the

say the area is A = (1/2).b.h

then the integral over the whole triangle
\int \int dxdy f_{X,Y}(x,y) = 1

The integrand is joint pdf f_{X,Y}(x,y) = f_{X,Y}(X=x,y=y)

f_{X,Y}(x,y) dx dy is the probability of finding X between x & x+dx and Y between y &y+dy

dxdy represents an area element. Knowing that and using the comments bove regrading the relation of probabilty and area, what can you say about f_{X,Y}(x,y)?

 

[lanedance]

the bounds you have said look close, try setting up the whole double integral for the area of the triangle & i'll have a look at it, here's some latex code to help below (click on it)

A = \int_{a}^{b} dx \int_{f(x)}^{g(x)} dy

 

[markov4]

dxdy represents an area element. Knowing that and using the comments bove regrading the relation of probabilty and area, what can you say about f_{X,Y}(x,y)?

So knowing that the area of a triangle is (1/2)*b*h, and knowing that our joint pdf must integrate to one and it also incorporates the area under a triangle..then wouldn't f_{X,Y}(x,y) be equal to (1/2)*b*h <--as in that's the function we take the double integral over, to get it to equal one? Where b=x and h=y (?).

That or we just set the LHS of the double integral formula equal to our equation for the area of the triangle, and find the integrals of dx and dy on the right side, then substitute through for b and h? Although i think the first way i described it makes more sense.

 

[lanedance]

not quite... if the area of the trangle is A = (1/2)*b*h, and you let the joint pdf be constant equal to f_{X,Y}(x,y) = A within the triangle, zero outside

then if you integrate over the triangle
\int \int dx dy f_{X,Y}(x,y) = \int \int dx dy A = A \int \int dx dy = A(A) = A^2
which is not equal to one


so you actually need f_{X,Y}(x,y) = \frac{1}{A} within the triangle, and zero outside. This normalises the joint pdf correctly. The joint pdf is a constant function as the probabilty of picking any point in the triangle is uniform.

 

[markov4]

I think i see now..thanks!