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Math Help - pdf of random variable

  1. #1
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    pdf of random variable

    Let f(x)=\frac{3}{16}x^{2}, -c<x<c

    A. Find c so that f(x) is the p.d.f. of a random variable X.
    B. Find \mu, \sigma^{2} and \sigma

    Could someone show me how to do this problem, or a problem similar to it? Thank you very much!
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  2. #2
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    Hello,
    Quote Originally Posted by logitech View Post
    Let f(x)=\frac{3}{16}x^{2}, -c<x<c

    A. Find c so that f(x) is the p.d.f. of a random variable X.
    B. Find \mu, \sigma^{2} and \sigma

    Could someone show me how to do this problem, or a problem similar to it? Thank you very much!
    A.
    The pdf of a rv has an integral equal to 1.
    So \int_{-c}^c f(x) ~dx=1
    Calculate this integral and find c

    B.
    For any bounded and continuous function h, we have : E(h(X))=\int_{-c}^c h(x)f(x) ~dx

    In particular :
    \mu=E(X)=\int_{-c}^c x f(x) ~dx

    and \sigma^2=E([X-\mu]^2)=\int_{-c}^c (x-\mu)^2 f(x) ~dx, which is also E(X^2)-\mu^2=\int_{-c}^c x^2 f(x) ~dx-\mu^2

    \sigma=\sqrt{\sigma^2}, that's all
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  3. #3
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    Following your extremely clear example, I found that c=2. Hopefully I calculated correctly in getting \mu=0, \sigma^{2}=\frac{12}{5} and \sigma=\frac{2\sqrt{15}}{5}
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  4. #4
    Moo
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    Quote Originally Posted by logitech View Post
    Hopefully I calculated correctly in getting \mu=0, \sigma^{2}=\frac{12}{5} and \sigma=\frac{2\sqrt{15}}{5}
    Bravo

    Following your extremely clear example, I found that c=2.
    Actually, this wasn't really an example. It was your exercise

    In other problems, you may have different boundaries. The boundaries for the integrals form the region where the pdf is not 0. That's what you have to remember.


    Sometimes, you can be given a function f(x)=cg(x), in the interval [a,b] and be asked to find c such that f is a pdf.
    In which case, integrate from a to b and find c such that the integral is 1.
    Same principle as above, though different question


    I hope it will help you in the future.
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  5. #5
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    I noticed it was the same question and I thank you for that because I understand it now. Thank you again for all of your help and hopefully I will have no problems on the other questions of that type.
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  6. #6
    MHF Contributor matheagle's Avatar
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    the mean is zero by inspection, this is a symmetric region and the density is an even function, calc 1.
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