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Thread: Show integral equal to two things

  1. #1
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    Show integral equal to two things

    Hi

    I need to show that $\displaystyle \int_{\Omega}\int_{0}^{\infty} I_{\left[0,X(\omega)\right]} \, dx \, dP(\omega) $, where $\displaystyle I_{\left[a,b\right]} $ is the indicator function, i.e. $\displaystyle I_{\left[a,b\right]}(x)=1 $ if $\displaystyle x \in \left[a,b\right]$ and zero otherwise, is equal to both

    $\displaystyle E(X)$ and $\displaystyle \int_{0}^{\infty}(1-F(x)) \, dx $ , where $\displaystyle F(x)$ is the cdf of $\displaystyle X$.

    Hence deducing that

    $\displaystyle E(X)=\int_{0}^{\infty}(1-F(x)) \, dx $

    I tried:

    $\displaystyle \int_{\Omega}\int_{0}^{\infty} I_{\left[0,X(\omega)\right]} \, dx \, dP(\omega) = \int_{\Omega} \int_{0}^{X(\omega)} \, dx \, dP(\omega) = \int_{\Omega} X(\omega) \, dP(\omega) = E(X) $, since this is the books' definition of the expected value of $\displaystyle X$.

    Im not sure how to go about finding the other result.

    Thanks
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  2. #2
    Moo
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    For the other one, $\displaystyle \displaystyle 1-F(x)=P(X>x)=\mathbb{E}[I_{X\in]x,\infty[}]=\int_\Omega I_{]x,\infty[} dP_X$

    Then reverse the order of the integrals and you'll get the result.
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  3. #3
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    So I should just substitute this expression into the integral. In other words,

    $\displaystyle \int_{0}^{\infty} \int_{\Omega} I_{(x,\infty)} \, dP(\omega) \, dy $

    What do you mean by reversing the order? I should move the zero to infinity integral inside the $\displaystyle \Omega$ integral. This would case my ingegration to go from $\displaystyle x$ to $\displaystyle \infty $ right.
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  4. #4
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    $\displaystyle \int_{0}^{\infty}(1-F(x)) \, dx = \int_{0}^{\infty}\int_{\Omega} I_{X\in(x,\infty)} \, dP(\omega) \, dx = \int_{\Omega}\int_{0}^{\infty} I_{X\in(0,\infty)} \, dx \, dP(\omega) =$

    $\displaystyle = \int_{\Omega}\int_{0}^{X(\omega)} \, dx \, dP(\omega) = E(X) $


    Please correct me, thanks!
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  5. #5
    Moo
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    Quote Originally Posted by ecnanif View Post
    $\displaystyle \int_{0}^{\infty}(1-F(x)) \, dx = \int_{0}^{\infty}\int_{\Omega} I_{X\in(x,\infty)} \, dP(\omega) \, dx = \int_{\Omega}\int_{0}^{\infty} \bold{I_{X\in(0,\infty)}} \, dx \, dP(\omega) =$
    In the bold thing, it's rather $\displaystyle I_{\{x\in (0,X)\}}$, otherwise it's correct !
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  6. #6
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    Thanks a lot Moo!
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