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Math Help - Weird Density function

  1. #1
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    Weird Density function

    Hello.

    Could someone help me with this, please? Thanks!

    Given the following probability density function

    p(x) = (3/4)*(1-x^2) for |x|<1

    p(x) = 0 otherwise


    I need to find the probability that the random variable X falls between 0 and 1/2 (i.e. 0<X<1/2)

    Integrating from -1 to 1 the density function, I find its area equals -(1/2).

    Am I doing sometihng wrong?
    Last edited by paolopiace; December 7th 2007 at 03:44 PM. Reason: Clarified x^2
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  2. #2
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    Quote Originally Posted by paolopiace View Post
    Integrating from -1 to 1 the density function, I find its area equals -(1/2).
    Am I doing sometihng wrong?
    Yes you are. Look at the integral again.
    \int\limits_{ - 1}^1 {\left( {1 - x^2 } \right)dx}  = \frac{4}{3}
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  3. #3
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    Thanks Plato...

    I really don't know how I went wrong on that trivial integral...
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  4. #4
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    Always take advantage of the even functions.

    f(x)=(1-x^2) is even, since f(x)=f(-x). (You may confirm this easily.)

    Now \int_{ - 1}^1 {\left( {1 - x^2 } \right)\,dx} = 2\int_0^1 {\left( {1 - x^2 } \right)\,dx} = 2\left( {\left. {x - \frac{1}<br />
{3}x^3 } \right|_0^1 } \right) = \frac{4}<br />
{3}.

    This leads easy computations and prevent to get mistaken.
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