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Math Help - Double integral help

  1. #1
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    Double integral help

    Can someone look at the work I have supplied on calculating a double integral.

    The answer I get is 0 and it is supposed to be 1/6. ThanksDouble integral help-scan0001.jpg
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  2. #2
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    Are you sure you have the problem right? Looking at the work you have written, there is no overlap between the conditioned part (Y1<2Y2) and the probability you are trying to find (Y1>Y2). Sketch it out (not the function, just the bounds they are interested in) and you will see what I mean.

    Also, this might be better served in the statistics section.
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  3. #3
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    I just need help with the integration part. Disregard everything except the calculus/integration stuff.
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  4. #4
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    Not sure what there is to check here; the setup isn't quite right. You need to divide by the conditioned part (since that is your new sample space), before you find the intersection (again, from reading the problem not seeing where your intersection is supposed to come from).

    Also, if you are integrating with respect to y_2 first, it would go from y_1/2 to infinity.
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  5. #5
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    I guess i'm not being clear here. Don't worry about the stats AT ALL, just the calculus.

    Calculate the double integral of e^-(y1+y2)dy2dy1 using the limits I have provided. The answer to that specifice integral is supposed to be 1/6 and thats not what I get. I think I get 0. Can someone help with this integral and show where I have made a mistake.
    AGAIN, DO NOT CONSIDER THE STATS PART OF THIS PROBLEM -- JUST THE INTEGRAL
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  6. #6
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    If it's just a calculus problem, not sure why you are getting 0. Your work is correct:

    \frac{e^{-2x}}{2}-\frac{2e^{-\frac{3}{2}}}{3}|_{0}^{\infty} = -\frac{1}{2}-(0-\frac{2}{3})=\frac{1}{6}
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  7. #7
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    Wow, I forgot to plug in the lower limit... Stupid mistake... Thanks for your time.
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