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Math Help - Standard Deviation

  1. #1
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    Standard Deviation

    Let X be a random variable with probability density function:

    f(x)= (3/4)(1-x^2), -1 < x < 1
    0, otherwise

    What is the standard deviaton of x?
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  2. #2
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    Quote Originally Posted by chaddy View Post
    Let X be a random variable with probability density function:

    f(x)= (3/4)(1-x^2), -1 < x < 1
    0, otherwise

    What is the standard deviaton of x?
    You need to calculate E(X) and E(X^2):

    E(X) = \int_{-\infty}^{\infty} x\, f(x) \, dx = \frac{3}{4} \, \int_{-1}^{1} x \, (1 - x^2) \, dx = .....

    E(X^2) = \int_{-\infty}^{\infty} x^2\, f(x) \, dx = \frac{3}{4} \, \int_{-1}^{1} x^2 \, (1 - x^2) \, dx = .....

    Then variance = E(X^2) - [E(X)]^2 = .....
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    Okay, so <br />
E(X) = \frac{3}{4} \, \int_{-1}^{1} x \, (1 - x^2) \, dx = \frac{3}{8} \, \int_{-1}^{1} u \, du<br />
where  u = (1-x^2)
    Thus <br />
E(X) =  \frac{12}{8}<br />
    Is that right?
    Then <br />
E(X^2) = \frac{3}{4} \, \int_{-1}^{1} x^2 \, (1 - x^2) \, dx = \frac{3}{4} \, \int_{-1}^{1} x^2 \, dx + \frac{3}{4} \, \int_{-1}^{1} (1 - x^2) \, dx = 3  +  \frac{3}{4} \, \int_{-1}^{1} (1 - x^2) \, dx <br />
but I am not sure how to find that last integral?
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  4. #4
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    Quote Originally Posted by chaddy View Post
    Okay, so <br />
E(X) = \frac{3}{4} \, \int_{-1}^{1} x \, (1 - x^2) \, dx = \frac{3}{8} \, \int_{-1}^{1} u \, du<br />
where  u = (1-x^2)
    Thus <br />
E(X) =  \frac{12}{8}<br />
    Is that right?
    Then <br />
E(X^2) = \frac{3}{4} \, \int_{-1}^{1} x^2 \, (1 - x^2) \, dx = \frac{3}{4} \, \int_{-1}^{1} x^2 \, dx + \frac{3}{4} \, \int_{-1}^{1} (1 - x^2) \, dx = 3  +  \frac{3}{4} \, \int_{-1}^{1} (1 - x^2) \, dx <br />
but I am not sure how to find that last integral?
    When you're working with pdf's it's assumed you can integrate. So you need to thoroughly revise basic integration techniques.


    E(X) = \frac{3}{4} \, \int_{-1}^{1} x \, (1 - x^2) \, dx = \frac{3}{4} \, \int_{-1}^{1} x - x^3 \, dx = \frac{3}{4} \left[ \frac{x^2}{2} - \frac{x^4}{4}\right]_{-1}^{1} = 0. This result could be anticipated since you're integrating an odd function .....


    E(X^2) = \frac{3}{4} \, \int_{-1}^{1} x^2 \, (1 - x^2) \, dx = \frac{3}{4} \, \int_{-1}^{1} x^2 - x^4 \, dx = \frac{3}{4} \left[ \frac{x^3}{3} - \frac{x^5}{5}\right]_{-1}^{1}

     = \frac{3}{4} \left( \, \left( \frac{1}{3} - \frac{1}{5}\right) - \left( -\frac{1}{3} + \frac{1}{5} \right) \, \right) = \frac{2}{3} - \frac{2}{5} = \frac{4}{15}.
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