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Math Help - one last joint distribution

  1. #1
    Member billym's Avatar
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    one last joint distribution

    joint density function of X and Y:

    f(x,y)=\left\{<br />
\begin{array}{lr}<br />
k(x+y)&0\le x+y\le 1; x,y\ge 0\\<br />
0&otherwise<br />
\end{array}<br />
\right.

    If I was looking for k, would \int_0^{1-y} \int_0^{1-x} k(x+y) dy dx=1 be the right way to start?
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  2. #2
    MHF Contributor matheagle's Avatar
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    NO
    The outer integral cannot contain variables...
    see my change below


    \int_0^{1} \int_0^{1-x} k(x+y) dy dx=1 is the right way to start
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  3. #3
    Member billym's Avatar
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    When it comes to finding the marginal densities, would I then use 1-x and 1-y as the upper limits?
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  4. #4
    MHF Contributor matheagle's Avatar
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    no, review calc 3
    IF you had variables on that outer integral HOW can you end up with a constant?
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  5. #5
    Member billym's Avatar
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    I understand that about finding the constant. What I mean is, once i have found the constant (3 I think), and want to find the marginal densities, would I then use:

    f_Y(y)=\int_0^{1-y} f(x,y) dx

    and

    f_X(x)=\int_0^{1-x} f(x,y) dy

    or do I have to stick with the initial values?
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  6. #6
    MHF Contributor matheagle's Avatar
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    These look ok

    Quote Originally Posted by billym View Post
    I understand that about finding the constant. What I mean is, once i have found the constant (3 I think), and want to find the marginal densities, would I then use:

    f_Y(y)=\int_0^{1-y} f(x,y) dx

    and

    f_X(x)=\int_0^{1-x} f(x,y) dy

    or do I have to stick with the initial values?
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