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Math Help - Continuous random variables and their PDFs

  1. #1
    Senior Member chella182's Avatar
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    Continuous random variables and their PDFs

    The question goes:

    Suppose the continuous random variable X has a probability density function (PDF)

    f_{X}(x)=\frac{c}{x^6}, x>1

    i) Show that for X to be a valid PDF, c must be equal to 5.
    ii) Calculate E(X^6e^{-2X}).

    For part i) I keep getting -5, not 5 for c, and I just can't fathom out part ii)
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by chella182 View Post
    The question goes:

    Suppose the continuous random variable X has a probability density function (PDF)

    f_{X}(x)=\frac{c}{x^6}, x>1

    i) Show that for X to be a valid PDF, c must be equal to 5.
    ii) Calculate E(X^6e^{-2X}).

    For part i) I keep getting -5, not 5 for c, and I just can't fathom out part ii)
    i) Remember that a pdf has to be a positive function. Hence the answer you're given is more likely to be correct than yours
    1=\int_1^{\infty} \frac{c}{x^6} ~dx=\int_1^\infty cx^{-6} ~dx
    so this is \left. \frac{c}{-5} \cdot\frac{1}{x^6}\right|_1^{\infty}

    =\lim_{x\to\infty} \frac{c}{-5}\cdot\frac{1}{x^6} {\color{red}-}\frac{c}{-5}\cdot 1=-\frac{c}{-5}=\frac c5

    just out from curiosity, which minus sign did you forget ?


    ii) recall that for any continuous function, \mathbb{E}(h(X))=\int_{\mathbb{R}} h(x)f(x) ~dx, where f is the pdf of the random variable X.

    This formula applies here and you get : \int_1^\infty x^6e^{-2x} \cdot \frac{5}{x^6} ~dx

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  3. #3
    Senior Member chella182's Avatar
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    Fully aware my answer was wrong, hence why I asked
    I didn't forget a minus sign either. I wrote it down wrong & confused myself.

    I've never seen that formula before (obviously been taught it in a different way), but cheers.
    Last edited by mr fantastic; May 21st 2009 at 07:22 PM. Reason: Changed the tone slightly
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