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Math Help - Continuous joint probabilities

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    Continuous joint probabilities

    Suppose x,y ~ f(x,y) = k(x-y) given 0=< X =< Y =< 1, otherwise it = 0
    What value of k makes this a density?

    I'm not quite sure what I'm being asked to find...I know it asks for k, but what constraints/definition is needing to be met in order for this to be a density? I just set up a double integral, each from 0 to 1, of k(x-y) but came out in the end with k*0, so I think this is somewhat incorrect...can anyone help me? I just need to know which direction to move in. Thanks!
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    Quote Originally Posted by mistykz View Post
    Suppose x,y ~ f(x,y) = k(x-y) given 0=< X =< Y =< 1, otherwise it = 0
    What value of k makes this a density?

    I'm not quite sure what I'm being asked to find...I know it asks for k, but what constraints/definition is needing to be met in order for this to be a density? I just set up a double integral, each from 0 to 1, of k(x-y) but came out in the end with k*0, so I think this is somewhat incorrect...can anyone help me? I just need to know which direction to move in. Thanks!
    You have to find the value of k such that \int_{x = 0}^{1} \int_{y = x}^{y = 1} k (x - y) \, dy \, dx = 1.

    Do you see why?
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    Would that be that, because it's a density, it needs to integrate to 1 overall?
    and since y is always greater than or equal to x, but less than 1, that gives the bounds
    Last edited by mistykz; February 26th 2009 at 08:44 AM.
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    Quote Originally Posted by mistykz View Post
    Would that be that, because it's a density, it needs to integrate to 1 overall?
    and since y is always greater than or equal to x, but less than 1, that gives the bounds
    Yes.
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    Alright, I understand it then. Thank you!
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    Quote Originally Posted by mistykz View Post
    Suppose x,y ~ f(x,y) = k(x-y) given 0=< X =< Y =< 1, otherwise it = 0
    What value of k makes this a density?

    I'm not quite sure what I'm being asked to find...I know it asks for k, but what constraints/definition is needing to be met in order for this to be a density? I just set up a double integral, each from 0 to 1, of k(x-y) but came out in the end with k*0, so I think this is somewhat incorrect...can anyone help me? I just need to know which direction to move in. Thanks!

    you just need to perform the double integral and set it equal to one.
    k\int_0^1\int_0^y (x-y) dxdy.
    I prefer integrating x first since that makes both lower bounds zero,
    but you can integrate in either order.
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