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.
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!
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.
How would I go about integrating it? I always seem to end up with a negative number, so something is wrong with my process.
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?
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?
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.
...
sad but true.
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.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?
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)
Last edited by crusoe; 02-12-2011 at 11:48 PM.
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:
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.
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.
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