# Marginal distribution question.

1. Apr 23, 2013

### Drao92

Hello,
i have the following conditional distribution
fXY(x,y)=2, of 0<x<1 and x<y<1
When i calculate the marginal distribution of y i dont know what interval of integration to choose.
I made the graph of x<y<1 and the points(x,y) for x<y are above x=y on that graph but i dont know how to write that interval.
fX is http://www.wolframalpha.com/input/?i=integral+from+x+to+1+from+2dy
In class we had somthing similar but with 0<y<sqrt(x) and we took y from 1-sqrt(x) to 1 but i didnt understand why and x was 0<x<1.

2. Apr 24, 2013

### chiro

Hey Drao92.

If you have to integrate out the x variable, try changing the order of integration to dxdy instead of dydx and integrate out the x variable.

Can you show us what the results are when you change the order of integration [i.e. the limits are [a,b] for y and [c,d] for x?]

3. Apr 24, 2013

### Drao92

Hi,
I dont know why you make me do this. I know how to solve a double integral.
I have problems with finding the interval of integration when 0<x<y<1.

4. Apr 24, 2013

### chiro

If you want to find the marginal you just integrate out the other variable. All I was doing was giving hints on how to do that.

5. Apr 24, 2013

### Drao92

dxdy is 2(d-c)(b-a)
dydx is the same.
Does it have a connection with the resul of fX which is 2-2x???

Last edited: Apr 24, 2013
6. Apr 24, 2013

### chiro

What limits did you get for integration for the y limits as a function of x? (Was it 0 to 1 - x by any chance?)

7. Apr 24, 2013

### Drao92

No, i took from x to 1
integral from x to 1 from 2dy=fX(x)
But i think it gives the same result as if we take from 0 to 1-x.

8. Apr 24, 2013

### chiro

In your limit you have x < y < 1 which means y > x. This means that the limits can not be from 0 to x since the y component is greater (in a standard x,y plot you are looking at the upper triangle not the lower triangle of the unit square).

Try re-calculating the limits again and see what you get.

9. Apr 24, 2013

### Drao92

Sorry, i did edit later my post.
Also, i have a question on 0 to 1-x;
if we consider y=1-x and we choose x=0.8 it results y=0.2 and in this case y<x. This is what i cant understand in finding interval limits with these conditions.

10. Apr 24, 2013

### chiro

Don't over-think this: Basically the joint distribution says you have a limit on the domain and we have to follow that.

In the case that you had y = 1 - x, this would be a particularly "slice" where if you had a distribution for the random variable X (or Y), then you would get with certainty the value of Y (or X).

Another way to think of it is that if your domain is a triangle (which it is for your question), then this distribution represents a specific "line" that passes through the triangle which is a uni-variate distribution in the same way that a line which is part of a two-dimensional figure is one-dimensional (you only need one parameter to describe a line).

It will make your life easier if you think of constraints.

Every time you add a condition (y < x, y = 1 - x, etc) you are basically making the number of possibilities smaller. The more conditions you have the easier something is to deal with. Understanding what those conditions mean geometrically (for example I outlined the line analogy) will make your life a lot easier when doing mathematics.

11. Apr 24, 2013

### Drao92

Thanks a lot for explanation.
Then i think the integration interval is 0 to y or 1-y to 1. if x<y<1 and 1>x>0
And if y<x<1 and 1>x>0 the integration interval is y to 1 or 0 to 1-y. Am i correct?
Im integrated out dx.
My formula for fY(y) is integral from fXY(x,y)dx. Is from course.
This > http://www.wolframalpha.com/input/?i=integral+from+0+to+y+from+2dx = distribution of y. I think on seminars when we took the limits 1-sqrt(x) to 1 for calculating the distribution of y we did somthing wrong... my teacher from seminar doesnt explain anything :|.

Last edited: Apr 24, 2013
12. Apr 24, 2013

### chiro

If you are integrating out y, then your limits have to be in terms of x not y.