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Math Help - 3rd continous random variable question

  1. #1
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    3rd continous random variable question

    Consider the continuous random variable X with probability density function as:


    Find
    a) the value of C
    b) the moment generating function of X
    c) the second moment about origin of X

    I could do parts b and c, but I don't know how to find C unless it is in the f(x) function.
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  2. #2
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    Think of question A this way: what is the value of C that solves the equation \int  _0 ^C 2(1-x) \, dx = 1?
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  3. #3
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    Nevermind, I screwed up a negative. I got C=1. Thank you for your explanation!
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  4. #4
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    We have:

    \int _0 ^C 2(1-x) \, dx = 1

    \int _0 ^C 2 - 2x \, dx = 1

    2x - x^2 | _0 ^C = 1

    (2C - C^2) - (2(0) - 0^2) = 1

    2C - C^2 = 1

    C^2 - 2C + 1 = 0

    (C - 1)(C - 1) = 0

    C = 1

    Good job on getting that before I did.
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  5. #5
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    Thanks. I had screwed up a negative in my original calculations. I got it. You made it very easy.
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  6. #6
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    Okay, so for the moment generating function, I found the integral to be


    Since the denominator is t squared, and we're evaluating from 0 to 1, does this mean that the mgf doesn't exist?
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  7. #7
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    Quote Originally Posted by ban26ana View Post
    Okay, so for the moment generating function, I found the integral to be


    Since the denominator is t squared, and we're evaluating from 0 to 1, does this mean that the mgf doesn't exist?
    m(t) = E\left( e^{tX} \right) = \int_{-\infty}^{+\infty} e^{xt} f(x) \, dx = 2 \int_{0}^{1} e^{xt} (1 - x) \, dx = \frac{2(e^t - t - 1)}{t^2}.

    Confirm that E(X) = \frac{dm}{dt} evaluated at t = 0.
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