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

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    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!

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    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.

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    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.

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    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?

    \int_0^1 \int_0^1 f(x,y) dx dy = \int_0^1 \int_0^1 a(1-x) dx dy= a \int_0^1 \int_0^1 (1-x) dx dy = a \int_0^1 1/2 dy = a/2

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    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?

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

    The step where I integrate with respect to x?

    \int_0^1 (1-x) dx = \left. x - (1/2)x^2 \right|_0^1 = (1 - 1/2) - (0 - 0) = 1/2

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    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?

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

    Quote Originally Posted by crusoe View Post
    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.

    ...

    sad but true.
    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.

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    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)
    Last edited by crusoe; 02-12-2011 at 11:48 PM.

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

    Quote Originally Posted by crusoe View Post
    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:
    \int_0^1 2*(1 - x) dy = 2*(1-x) \int_0^1 1 dy = 2*(1-x) * (1-0) = 2*(1-x)

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

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

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    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:

    f(x) = \int_y f(x,y) dy = \int_y a(1-x) dy = a \int_y (1-x) dy = a \left. 1y-x \right|_0^1 = a(1-x-0-x)
    This simplifies to = a(1-2x) as the marginal distribution of X.

    f(y) = \int_x f(x,y) dx = \int_x a(1-x) dx = a \int_x (1-x) dx = a \left. 1x-x^2/2 \right|_0^1
    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.

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

    Quote Originally Posted by crusoe View Post
    a \int_y (1-x) dy = a \left. 1y-x \right|_0^1 = a(1-x-0-x)
    Here is your problem. Remember that you're integrating with respect to y. This means x is a constant so
    2 \int_0^1 (1-x) dy = \left. 2 * (1*y - x*y) \right|_0^1

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    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

    f(y) =  \int_x a(1-x) dx = 2 \int_0^1 (1-x) dx = 2(\left. 1x-x^2/2) \right|_0^1 = 2(1-1/2-0-0) =1

    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.

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    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.

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