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Math Help - Quadratics with normally distributied coeffcients

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    Quadratics with normally distributied coeffcients

    Suppose that A, B, C, are independent random variables, each being uniformly distributed over (0, 1).

    (a) What is the joint cumulative distribution function of A, B, C?

    (b) What is the probability that all of the roots of the equation Ax2 + Bx + C = 0 are real?

    For (a) it's pretty clear that this defines a cube in \mathbb{R}_3. F_{a,b,c}(a,b,c)=a*b*c

    For part (b) the discriminate has to be greater then zero so I need to find P(B^2-4AC>0)=P(B^2>4AC)

    I'm not entirely sure how to go about finding this. I know I need to convert this into some expression of volume but I'm not sure how.
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    Re: Quadratics with normally distributied coeffcients

    Sorry there's an error in the title. The problem only says that the variables are independent not normally distributed. And of course I didn't check the spelling... (distributed coefficients...)
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    Re: Quadratics with normally distributied coeffcients

    Quote Originally Posted by bkbowser View Post
    Sorry there's an error in the title. The problem only says that the variables are independent not normally distributed. And of course I didn't check the spelling... (distributed coefficients...)
    you end up with the expression

    $\Large \displaystyle{\int_0^{\frac 1 4} \int_0^1 \int_{2\sqrt{AC}}^1} dB~dC~dA +\displaystyle{\int_{\frac 1 4}^1 \int_0^{\frac 1 {4A}} \int_{2\sqrt{AC}}^1} dB~dC~dA$

    I'm going to let you puzzle out how that was derived because it's a good exercise.
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    Re: Quadratics with normally distributied coeffcients

    Quote Originally Posted by romsek View Post
    you end up with the expression

    $\Large \displaystyle{\int_0^{\frac 1 4} \int_0^1 \int_{2\sqrt{AC}}^1} dB~dC~dA +\displaystyle{\int_{\frac 1 4}^1 \int_0^{\frac 1 {4A}} \int_{2\sqrt{AC}}^1} dB~dC~dA$

    I'm going to let you puzzle out how that was derived because it's a good exercise.
    If I combine the integrals A goes from 2\sqrt{AC} to 1 B goes from 0 to 1 but C is the funny one.

    I haven't been able to manipulate the inequality (B^2>4AC) to get something similar.

    You are just finding the volume of (A,B,C)\in\mathbb{R}_3 s.t. 0<A,B,C<1?

    Sorry I feel unprepared for this, I feel like we did set up boundaries for integration in Calculus as some function, but I don't recall any examples or a section in the text that I can flip to for more information.
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    Re: Quadratics with normally distributied coeffcients

    it's just the entire volume that meets the criteria $B^2 > 4AC$

    It will really benefit you to work this out yourself.
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    Re: Quadratics with normally distributied coeffcients

    Quote Originally Posted by romsek View Post
    it's just the entire volume that meets the criteria $B^2 > 4AC$

    It will really benefit you to work this out yourself.
    OK, I'll see what I can do.
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    Re: Quadratics with normally distributied coeffcients

    remember the max of A, B, C is 1
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