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Thread: A Simple Question About the Cumulative Density Function

  1. #11
    Actuary.com - Posting Master
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    Let's break this down more.

    1) You have a PDF, call it f(x), it is (2.5(200^2.5))/x^3.5 defined on range [200, infinity)

    2) To calculate your CDF, call it F(x), you would take the integral of your pdf between 200 and <some variable>, in this case you have it set up as F(x) = integral( (2.5(200^2.5))/t^3.5 dt from 200 to x]

    You should know that F'(x) = f(x), figuring out f'(x) is useful for other things (such as hazard/survival function but not for anything you're doing here so if that's what you're calculating... I have to ask why?).

    So let's do the integral that we have in 2), this is a simple power rule so you get 2.5*200^{2.5}/(-3.5+1) *t^{-2.5} from 200 to x which is basically the same as -200^{2.5}/t^{2.5} from 200 to x, so now plug in your limits to get:

    -200^{2.5}/x^{2.5} - (-200^{2.5}/200^{2.5}) = -200^{2.5}/x^{2.5} + 1 or if you want to re-write it 1 - 200^{2.5}/x^{2.5} which is your CDF.

    If you want, you can differentiate THAT to get your f(x) or 2.5*200^{2.5} / x^{3.5}

  2. #12
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    Quote Originally Posted by NoMoreExams View Post
    What function are you taking the derivative of? Is your function

    1) (2.5(200^2.5))/t^3.5

    2) Integral[ (2.5(200^2.5))/t^3.5 dt from 200 to x ]


    Those are 2 different functions.
    The first one.

  3. #13
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    Quote Originally Posted by aomara View Post
    The first one.
    See my other post. What are you hoping to accomplish by taking the derivative of a pdf?

  4. #14
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    Quote Originally Posted by NoMoreExams View Post
    Let's break this down more.

    1) You have a PDF, call it f(x), it is (2.5(200^2.5))/x^3.5 defined on range [200, infinity)

    2) To calculate your CDF, call it F(x), you would take the integral of your pdf between 200 and <some variable>, in this case you have it set up as F(x) = integral( (2.5(200^2.5))/t^3.5 dt from 200 to x]

    You should know that F'(x) = f(x), figuring out f'(x) is useful for other things (such as hazard/survival function but not for anything you're doing here so if that's what you're calculating... I have to ask why?).

    So let's do the integral that we have in 2), this is a simple power rule so you get 2.5*200^{2.5}/(-3.5+1) *t^{-2.5} from 200 to x which is basically the same as -200^{2.5}/t^{2.5} from 200 to x, so now plug in your limits to get:

    -200^{2.5}/x^{2.5} - (-200^{2.5}/200^{2.5}) = -200^{2.5}/x^{2.5} + 1 or if you want to re-write it 1 - 200^{2.5}/x^{2.5} which is your CDF.

    If you want, you can differentiate THAT to get your f(x) or 2.5*200^{2.5} / x^{3.5}
    Ok I see what I was doing wrong. Thank you very much.

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