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Probability that roots of quadratic are real

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Let U1, U2, and U3 be independent random variables uniform on [0,1]. Find the probability that the roots of the quadratic U1x2+U2x+U3 are real.


    2. Relevant equations



    3. The attempt at a solution
    So we need to find P(U22>4U1U3), which involves evaluating some integral. The think the integrand would be 1 since we are dealing with uniform random variables. But beyond that, I need assistance in figuring out whether it should be a double integral or a triple integral and the limits of integration.
     
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  3. Jan 17, 2010 #2

    Dick

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    It's a triple integral, sure. But it's pretty straightforward. Integrate U1 and U3 from 0 to 1. That makes your only integral with nontrivial limits the U2 integral. What are the limits there?
     
  4. Jan 17, 2010 #3
    Are we trying to find the volume above the surface defined by [tex]U_2=2\sqrt{U_1U_3}[/tex] and inside [0,1]x[0,1]x[0,1]?
     
  5. Jan 17, 2010 #4

    vela

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    Yes.
     
  6. Jan 17, 2010 #5
    Why isn't it

    [tex] \int_0^1\int_0^1\int_{2\sqrt{U_1U_3}}^1 dU_2\,dU_1\,dU_3 [/tex]

    I can't figure out why my limits of integration for U2 are wrong.
     
  7. Jan 17, 2010 #6

    vela

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    Here's a hint: What's the lower limit equal to when U1=U3=1?
     
  8. Jan 17, 2010 #7

    Dick

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    Oh, heck. Thanks, vela. Requiring U2<=1 does create restrictions on the range of both U1 and U3. Backtrack and fix my stupid suggestion of integrating both U1 and U3 from 0 to 1 by requiring each variable in turn be less than or equal to 1. Sorry.
     
  9. Jan 17, 2010 #8
    Thanks to both of you for your help; I finally got it. I found that drawing a picture by taking slices through the U2-axis was also essential.
     
  10. Mar 2, 2011 #9
    Would it have killed you to post the solution? I can't figure out what the limits should be. I figured U1<1/(4U3) and U3<1/(4U1) but using those as the upper limits I get nowhere.
     
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