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Math Help - Joint Uniform Distrubtion of 3 variables problem

  1. #1
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    Joint Uniform Distrubtion of 3 variables problem

    Hi all,

    I am working through a problem but I cant seem to see the method... the answer is in the back of the text book but no method!

    Q: Let U_1, U_2, U_3 be uniformly distributed on (0,1) . Find the prob that U_1x^2 + U_2x +U_3 have real roots. The answer in the back of the book is 1/9.

    So using the quadratic eqn, I deduce taht the prob distribution we need to find is given by:

    P(U_2^2 -4U_2U_3 >0)...... I am not sure how to evaluate this at all.

    Looking through the usual methods in the text book usually points to evaluating the area via direct integration but when I integrate:

    \int_0^1 \int_0^1 \int_0^1 U_2^2 -4U_2U_3 du_1 du_2 du_3 I get a negative answer which is obviously wrong... I suspect the limits need to be changed, but not sure where to start.....

    Thanks for reading
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  2. #2
    Moo
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    Hello,

    (by the way, you wrote U_2^2 instead of U_1^2)

    And in order to do this problem, you have to assume that the three rv's are independent.

    Your integral is that of E[U_1^2-4U_2U_3] so this is not what you want.

    In order to find the probability you're looking for, you need to integrate the joint pdf of the three rv's over the region where u_1^2-4u_2u_3>0

    So this gives \displaystyle \int_0^1 \int_0^1 \int_0^1 \bold{1}_{\{u_1^2-4u_2u_3>0\}} du_3 du_2 du_1

    So this gives you a new boundary for the integral with respect to u_3... continue and you will find the answer
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  3. #3
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    thanks Moo! will try that tomorrow... bit late where i am now!
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  4. #4
    Moo
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    I'm 1 hour ahead of you
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  5. #5
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    I guess its easy for a prob queen like you to solve these things late at night, us mere moretals though ....


    Quote Originally Posted by Moo View Post
    Hello,

    (by the way, you wrote U_2^2 instead of U_1^2)

    And in order to do this problem, you have to assume that the three rv's are independent.

    Your integral is that of E[U_1^2-4U_2U_3] so this is not what you want.

    In order to find the probability you're looking for, you need to integrate the joint pdf of the three rv's over the region where u_1^2-4u_2u_3>0

    So this gives \displaystyle \int_0^1 \int_0^1 \int_0^1 \bold{1}_{\{u_1^2-4u_2u_3>0\}} du_3 du_2 du_1

    So this gives you a new boundary for the integral with respect to u_3... continue and you will find the answer
    I don't follow Moo.... If I integrate over this region, with du_1,du_2 first, dont I just end up with \int_0^1 1 du_3 and is it here that I make the upper limit of integration to \int_0^{u_1^2 -4u_2u_3} du_3 ?? I am not sure... maybe I need to look back at my old multivariable integration notes first!
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  6. #6
    Moo
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    I'm not a queen, merely a cow

    If you integrate wrt u_1 first, you have u_1^2-u_2u_3>0 \Leftrightarrow u_1>\sqrt{u_2u_3} ( U_1 is uniform(0,1) so u_1>0 almost surely).

    So since the integral is 0 if u_1 doesn't satisfy this inequality, we'll have \displaystyle \int_0^1 \int_0^1 \int_{\sqrt{u_2u_3}}^1 1~du_1du_2du_3, which gives you \displaystyle \int_0^1\int_0^1 1-\sqrt{u_2u_3} ~ du_2du_3, which, I guess, you can finish off !
    (I just hope it really makes 1/9 lol)

    You can do it in any order you wish to do (thanks to Fubini-Tonelli's theorem)
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  7. #7
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    cheers for stepping me through that moo. I don't get 1/9.... this is my working:

    \displaystyle \int_0^1\int_0^1 1-\sqrt{u_2u_3} ~ du_2du_3

    \displaystyle = \int_0^1\ [u_2 -  \frac{2}{3}u_2^{3/2}u_3^{1/2}]_0^1 du_3

    \displaystyle =  \int_0^1\ 1 -  \frac{2}{3}u_3^{1/2} du_3

    \displaystyle = [ u_3 -  \frac{2}{3}.\frac{2}{3}u_3^{3/2} ]_0^1

    = 1- \frac{4}{9}


    = \frac{5}{9}

    I follow your working and theory so I have either made a mistake in the working here or theres a misprint in the book!

    In any case, thank you very much for the help in this!
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