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Math Help - Double integration problem

  1. #1
    Member garymarkhov's Avatar
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    Double integration problem

    k \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} (6-x-y)dx dy = 8k

    Can someone describe why this is true? I have no idea what steps to take to get this answer.
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  2. #2
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    It's a joke, right? Maybe the 8 was bent over in the original statement, " \infty"?

    That region is very much unbounded. You will be hard-pressed to drag a finite solution from it.
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  3. #3
    Member garymarkhov's Avatar
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    Quote Originally Posted by TKHunny View Post
    It's a joke, right? Maybe the 8 was bent over in the original statement, " \infty"?

    That region is very much unbounded. You will be hard-pressed to drag a finite solution from it.
    Whew, so I'm not crazy.

    No, it's not a joke. I've got it in my probability notes.

    The question is: find k such that

    f_{X,Y}(x,y)dx dy  = \left\{ \begin{array}{lr}(6-x-y), &0<x<2, 2<y<4\\ 0, &otherwise \end{array} \right.

    is a joint density. Find P(X+Y<3) and the marginal density of X.

    The first step in the solution was

    1= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f_{X,Y}(x,y)dx dy = k \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} (6-x-y)dx dy = 8k

    which makes no sense to me. Is there something I'm missing?
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  4. #4
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    This shows us the value of presenting the actual problem statement. The limits make all the difference.

    (-\infty,\infty) means (0,2) in the x-direction. It is no longer unbounded.

    Please evaluate the iterated integral with these very useful finite limits.

    For P(X + Y < 3), notice that this is nicely related to P(X < 3 - Y) and you should be almost done.
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    To clarify, your notation is bad.

    Consider for a moment a simpler function:

    f(x)= \begin{cases} 1/2, & \mbox{if } 3<x<5 \\ 0, & \mbox{elsewhere} \end{cases}

    If you are asked to evaluate this integral:

    \int_{-\infty}^{\infty} f(x) \, dx

    then you would have to split it into separate regions because it's a piecewise function, i.e.,

    \int_{-\infty}^{\infty} f(x) \, dx ~=~ \int_{-\infty}^3 f(x) \, dx + \int_3^5 f(x) \, dx +  \int_5^{\infty} f(x) \, dx

    =~ \int_{-\infty}^3 0 \, dx + \int_3^5 (1/2) \, dx +  \int_5^{\infty} 0 \, dx

    Naturally, the integrals on the sides equal zero, so we just write it as

    \int_{-\infty}^{\infty} f(x) \, dx ~=~ \int_3^5 (1/2) \, dx

    Extending this logic into two variables with your problem shows that you should be evaluating this:

    \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dx \, dy  ~=~ \int_2^4 \int_0^2 k(6-x-y) \, dx \, dy
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  6. #6
    Member garymarkhov's Avatar
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    Quote Originally Posted by TKHunny View Post
    This shows us the value of presenting the actual problem statement. The limits make all the difference.

    (-\infty,\infty) means (0,2) in the x-direction. It is no longer unbounded.

    Please evaluate the iterated integral with these very useful finite limits.

    For P(X + Y < 3), notice that this is nicely related to P(X < 3 - Y) and you should be almost done.
    Sorry, I should have included the whole problem statement.

    Can you expand on the P(X < 3 - Y) bit?
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  7. #7
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    It might be easier to worry about P(Y < 3 - X). You have new limits. Your Y-limits are no longer all of (2,4). You might get (2,3-X)? Since x runs in (0,2), that seems quite reasonable.
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  8. #8
    Member garymarkhov's Avatar
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    Quote Originally Posted by TKHunny View Post
    It might be easier to worry about P(Y < 3 - X). You have new limits. Your Y-limits are no longer all of (2,4). You might get (2,3-X)? Since x runs in (0,2), that seems quite reasonable.
    Sorry, I'm still not sure where to go with the solution. Do you mean I should evaluate  \int^4_2 \int^{3-y}_0 k(6-x-y) \, dx \, dy?
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  9. #9
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    Quote Originally Posted by garymarkhov View Post
    Sorry, I'm still not sure where to go with the solution. Do you mean I should evaluate  \int^4_2 \int^{3-y}_0 k(6-x-y) \, dx \, dy?
    Close, but you made a mistake with the limits of integration.

    Graph the region that the density is defined on. Then plot the line x+y=3. You will then be able to identify the region corresponding with x+y<3. You should integrate the joint density over this region only.

    Can you identify your error?
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  10. #10
    Member garymarkhov's Avatar
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    Quote Originally Posted by drumist View Post
    Close, but you made a mistake with the limits of integration.

    Graph the region that the density is defined on. Then plot the line x+y=3. You will then be able to identify the region corresponding with x+y<3. You should integrate the joint density over this region only.

    Can you identify your error?
    Is it that I need to restrict both integrals (see attachment)? So maybe \int^{3-x}_2 \int^{3-y}_0 k(6-x-y) \, dx \, dy
    Attached Thumbnails Attached Thumbnails Double integration problem-integration_graph.jpg  
    Last edited by garymarkhov; March 16th 2010 at 01:36 PM.
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  11. #11
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    Sorry for taking so long to respond.

    The region you displayed is correct! However, your endpoints are still not correct. :\

    This is the correct setup:

    \int_2^3 \int_0^{3-y} k(6-x-y) \, dx \, dy

    The error you originally made was that you took y from 2 to 4, but the region goes from 2 to 3.
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