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Math Help - Derivative problem.

  1. #1
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    Derivative problem.

    I have a standard normal distribution. cdf stands for cumulative distr. function and pdf for probability distribution function.

    since the cdf is an intergral of pdf, the pdf is the derivative of cdf.

    I have the function of the form:

    f(x) = cdf(x)-cdf(-x)

    I want a 1st derivative of this function, but if I just use
    f'(x) = pdf(x) - pdf(-x) it is always zero, since the distribution is symmetrical.

    This makes no sense to me, the derivative should be zero for x where pdf(x)~0, but not zero near zero.

    Where's my mistake?

    Thanks in advance.
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  2. #2
    MHF Contributor matheagle's Avatar
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    The cdf is F_X(x)=P(X\le x) is any situation.

    So cdf(x)-cdf(-x) equals P(X\le x)-P(X\le -x)=P(-x<X\le x) where I hope x>0.

    In the continuous case this is \int_{-x}^x f_X(t)dt where f(t) is the density.

    By calc one the derivative is f_X(x)+f_X(-x)
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by egladil View Post
    I have a standard normal distribution. cdf stands for cumulative distr. function and pdf for probability distribution function.

    since the cdf is an intergral of pdf, the pdf is the derivative of cdf.

    I have the function of the form:

    f(x) = cdf(x)-cdf(-x)

    I want a 1st derivative of this function, but if I just use
    f'(x) = pdf(x) - pdf(-x) it is always zero, since the distribution is symmetrical..
    f'(x)=\frac{d}{dx} cdf(x) - \frac{d}{dx} cdf(-x)= \frac{d}{dx} cdf(x) - \frac{d}{dx}(-x) \left. \frac{d}{du} cdf(u)\right|_{u=-x}=pdf(x)+pdf(-x)=2\; pdf(x)

    Which we could have got to much faster by writing:

    f(x)=\int_{-x}^x pdf(u)\ du=2 \int_0^x pdf(u)\ du

    Then the fundamental theorem of calculus gives:

    f'(x)=2\; pdf(x)

    CB
    Last edited by CaptainBlack; July 12th 2009 at 07:12 AM.
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