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Math Help - Uniform variate problem

  1. #1
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    Uniform variate problem

    Let X be a uniform variate defined on (-k,k). Determine k so that
    P(|X|<1) = P(|X|>2)
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  2. #2
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by wolfyparadise View Post
    Let X be a uniform variate defined on (-k,k). Determine k so that
    P(|X|<1) = P(|X|>2)

    3
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  3. #3
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    Could u please tell me how u got the answer

    Quote Originally Posted by matheagle View Post
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    How did u arise to this conclusion???
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  4. #4
    Moo
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    P(I'm here)=1/3, P(I'm there)=t+1/3
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    Hello,

    Draw a line where you represent -2,-1,0,1,2 and represent the areas where |X|<1 and |X|>2 and you'll see that for the areas to be equal, the domain has to stop to -3 and 3.


    More formally... :

    P(|X|<1)=\frac{1}{2k} \int_{-1}^1 ~dt=\frac 1k

    P(|X|>2)=\frac{1}{2k}\left(\int_{-k}^{-2} ~dt+\int_2^k ~dt\right)=\frac{1}{2k}\cdot (-2+k+k-2)=\frac{k-2}{k}

    Thus we must have \frac 1k=\frac{k-2}{k} \Rightarrow 1=k-2\Rightarrow k=3


    boldeagle : what's the point giving the solution without further information ? It doesn't help
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  5. #5
    MHF Contributor matheagle's Avatar
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    Fine, 1+2=3, since....

     2\int_0^a{dx\over 2k}=2\int_b^k{dx\over 2k}

    cancelling the constants leaves  \int_0^adx=\int_b^kdx

    Thus a=k-b hence k=a+b.
    Last edited by matheagle; December 26th 2009 at 06:47 AM.
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