# Thread: Joint Density f(X,Y) integration when there is no Y in the eqn

1. ## Joint Density f(X,Y) integration when there is no Y in the eqn

Hi I'm having difficulties with integration by parts when the main function of a joint distribution doesn't contain the Y. Here it is:

Let X and Y have a joint density given by f(X,Y)=a(1−X), 0≤X≤1,0≤Y ≤1
and zero elsewhere. Where, k is a constant.

(a) Find the value of a that makes this a probability density function. Use this
value for a in the pdf to work out the remaining questions.
(b) Find the marginal distribution of X and Y, i.e. f(X) and f(Y ).
(c) Find E(X), E(Y ), E(Y |X) and V ar(X|Y ).
(d) Find E[6X −12Y] and Var(2X −3Y)
(e) Find Cov(XY ) and the correlation ρxy
(f) Are X and Y independent? Show this using two alternative ways.

For part (a) I'm getting a value of a=-2 but I don't think that it's correct. I haven't moved any further in the question as all the others are dependent on this answer. Any help would be great!

2. ## Re: Joint Density f(X,Y) integration when there is no Y in the eqn

It seems to me you must have switched a negative somewhere. Notice that if a = -2 then for any (x,y) in the support of the distribution you have your joint density has a negative value. This can't happen.

3. ## Re: Joint Density f(X,Y) integration when there is no Y in the eqn

How would I go about integrating it? I always seem to end up with a negative number, so something is wrong with my process.

4. ## Re: Joint Density f(X,Y) integration when there is no Y in the eqn

I don't quite get what's causing the problems I guess. Integrate with respect to x and then with respect to y?

5. ## Re: Joint Density f(X,Y) integration when there is no Y in the eqn

Thank you. The step that you short cut through is the one I'm making the mistake on. Is it possible that you show all steps?

6. ## Re: Joint Density f(X,Y) integration when there is no Y in the eqn

The step where I integrate with respect to x?

7. ## Re: Joint Density f(X,Y) integration when there is no Y in the eqn

Wow, apparently I thought 1-1/2 = -1/2 weird. Big thanks. Another thing: In the case where I'm doing part a. Do I interpret the question as prove:
1. f(xy)>=0
2. \int_0^1 \int_0^1 f(xy)dxdy = 1

so that in part 2 - the part you explained to me, i would conclude 1 = a/2 Therefore a = 2?

8. ## Re: Joint Density f(X,Y) integration when there is no Y in the eqn

Originally Posted by crusoe
Wow, apparently I thought 1-1/2 = -1/2 weird.
No worries. It happens to the best of us. I once spent an additional hour working on a measure theory question because I couldn't get something to work out because I made the mistake of saying 1 + 1 = 1.

...

Big thanks. Another thing: In the case where I'm doing part a. Do I interpret the question as prove:
1. f(xy)>=0
2. \int_0^1 \int_0^1 f(xy)dxdy = 1

so that in part 2 - the part you explained to me, i would conclude 1 = a/2 Therefore a = 2?
It's not perfectly clear to me that the parts you're referencing are the same as they were in the first post but oh well. Essentially I think what you're saying is what you want to do. You need to show those two facts (always non-negative and integrates to 1) and that the value of a that gives you that is 2. You should be good to move on to the other parts after that.

9. ## Re: Joint Density f(X,Y) integration when there is no Y in the eqn

For part B of the question. How do I approach finding the marginal distribution f(X) and f(Y) when there isn't a Y in the equation?

For the marg distrn f(x) = inty a(1-x)dy I get = a(1-2x) as the marg distrbn. And I get a(1/2) as the marg distrbn of f(y)

10. ## Re: Joint Density f(X,Y) integration when there is no Y in the eqn

Originally Posted by crusoe
For part B of the question. How do I approach finding the marginal distribution f(X) and f(Y) when there isn't a Y in the equation?

For the marg distrn f(x) = inty a(1-x)dy I get = a(1-2x) as the marg distrbn. And I get a(1/2) as the marg distrbn of f(y)
Now that you know that a = 2 you should probably just replace any mention of a with 2. I don't quite understand what you're doing here:
nty a(1-x)dy I get = a(1-2x) as the marg distrbn

Remember that you're integrating with respect to y and anything without a y is just a constant. So 2(1-x) is just a constant so if you want you could do this:

11. ## Re: Joint Density f(X,Y) integration when there is no Y in the eqn

Originally Posted by Dason
No worries. It happens to the best of us...
With the exception of me, of course.

12. ## Re: Joint Density f(X,Y) integration when there is no Y in the eqn

Sorry. I'll try and make this more clear. Currently I'm working on these questions:

(b) Find the marginal distribution of X and Y, i.e. f(X) and f(Y ).
(c) Find E(X), E(Y ), E(Y |X) and V ar(X|Y ).

For (b) I did the following:

This simplifies to = a(1-2x) as the marginal distribution of X.

This simplifies to = a[1-1/2-0-0] = a(1/2) as the marginal distribution of Y.

I'm completely unsure if I can do those operations though. Especially after reading your most recent suggestion.

13. ## Re: Joint Density f(X,Y) integration when there is no Y in the eqn

Originally Posted by crusoe
Here is your problem. Remember that you're integrating with respect to y. This means x is a constant so

14. ## Re: Joint Density f(X,Y) integration when there is no Y in the eqn

Ok that makes sense for the marginal distribution of X, but what about Y? I get the following for Y

So my question regarding the marginal distribution of Y is: Is it ok for it to just take the value 1? I thought it should be something with a variable in it.

15. ## Re: Joint Density f(X,Y) integration when there is no Y in the eqn

Well if you really want to be technical you do have a variable in there. It's really 1 times the indicator function that y is between 0 and 1. Y has a uniform distribution on [0,1] is all that that says.