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Thread: probability

  1. #1
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    probability

    If $\displaystyle f(x,y) = e^{-x-2y} $ find $\displaystyle P(X<Y) $.

    So is this equaled to $\displaystyle 1 - P(X>Y) = 1 - \int\limits_{0}^{\infty} \int\limits_{0}^{x} e^{-x-2y} \ dy \ dx = \frac{1}{3} $?
    Last edited by heathrowjohnny; Feb 22nd 2008 at 10:50 AM.
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  2. #2
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    never mind about the limits; is the general process correct?
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  3. #3
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    I get $\displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^x e^{-x-2y}~dy~dx$
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  4. #4
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    I took the complement.
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  5. #5
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    What were the domains of $\displaystyle X$ and $\displaystyle Y$ given in the problem?
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  6. #6
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    $\displaystyle -\infty < x < \infty $, $\displaystyle -\infty < y < \infty $.
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  7. #7
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    While the integral over the whole positive domain is $\displaystyle \frac{1}{2}$, the function is NOT even. For this to be a density function, $\displaystyle x,y \in\left[{\frac{{ - \ln (2)}}{3},\infty } \right)
    $ since $\displaystyle \int\limits_{\frac{{ - \ln (2)}}{3}}^\infty {\int\limits_{\frac{{ - \ln (2)}}{3}}^\infty {e^{ - x - 2y} } } ~dy~dx = 1$.

    Then again, there are infinitely many possible limits for $\displaystyle X$ and $\displaystyle Y$ that will work. I am not 100% confident in my Prob/Stats knowledge though, so don't quote me on this.
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  8. #8
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    yeah I knew the limits were wrong, I was just asking whether my method of finding the probability was correct (e.g. $\displaystyle P(X <Y) = 1 - P(X>Y) $)?
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  9. #9
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    Oh. You are correct.
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