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A Simple Question About the Cumulative Density Function

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  • A Simple Question About the Cumulative Density Function

    Hey all, I was working on 7-13 of Hassets' Probability of Risk Management. The problem asked to calculate the difference between the 30th and 70th percentiles of X. The pdf was
    f(x) = 2.5*(200^2.5)/x^3.5 for x>200
    0 otherwise
    I got the answer correct: 93.06
    According to my calculation, x0.3 was 323.7289 and x0.7 was 230.6698 But I am really confused: how can the 70th percentile be less than the 30th??? Shouldn't you read the pdf from left to right?

    Also, sometimes the interval is closed and sometimes it's open. I know for example when it's 0<=x<=1, you do the integral from 0 to 1. But what about 0<x<1??

    Thank you guys and wish everyone a satisfying grade.

  • #2
    Originally posted by bunkey View Post
    Hey all, I was working on 7-13 of Hassets' Probability of Risk Management. The problem asked to calculate the difference between the 30th and 70th percentiles of X. The pdf was
    f(x) = 2.5*(200^2.5)/x^3.5 for x>200
    0 otherwise
    I got the answer correct: 93.06
    According to my calculation, x0.3 was 323.7289 and x0.7 was 230.6698 But I am really confused: how can the 70th percentile be less than the 30th??? Shouldn't you read the pdf from left to right?

    Also, sometimes the interval is closed and sometimes it's open. I know for example when it's 0<=x<=1, you do the integral from 0 to 1. But what about 0<x<1??

    Thank you guys and wish everyone a satisfying grade.
    Likelihood function (which is what a pdf is) does not give you the probability of an even. Plot the exponential function or gamma if you want to get more general, your claim of monot. increasing will disappear.

    Comment


    • #3
      It's not. The 30th percentile of this distribution is 230.7. The 70th percentile is 323.7.

      Comment


      • #4
        Originally posted by RossMoney34 View Post
        It's not. The 30th percentile of this distribution is 230.7. The 70th percentile is 323.7.
        Thanks! What about the the difference between x>1 and x>=1 in terms of probability?

        Comment


        • #5
          Originally posted by bunkey View Post
          Thanks! What about the the difference between x>1 and x>=1 in terms of probability?
          You should know that for a cont. function P(X = x) = 0

          Comment


          • #6
            Originally posted by NoMoreExams View Post
            You should know that for a cont. function P(X = x) = 0
            Now it's evident to me. Thank you!

            Comment


            • #7
              Hi everyone. This is probably a dumb question, but for this question the distribution function is F(x)= (2.5(200^2.5))/t^3.5 with the integral between 200 and x. I know that by taking the derivative of the equation we get -200^2.5/t^2.5, but when we solve using the integral, the equation comes out to 1-(200^2.5/x^2.5). Shouldn't it be -1 instead of 1? Plugging 200 in the equation should give me -1, right, or am I just missing something? Thanks in advance.

              Comment


              • #8
                Originally posted by aomara View Post
                Hi everyone. This is probably a dumb question, but for this question the distribution function is F(x)= (2.5(200^2.5))/t^3.5 with the integral between 200 and x. I know that by taking the derivative of the equation we get -200^2.5/t^2.5, but when we solve using the integral, the equation comes out to 1-(200^2.5/x^2.5). Shouldn't it be -1 instead of 1? Plugging 200 in the equation should give me -1, right, or am I just missing something? Thanks in advance.
                I have no idea what you are actually asking or trying to do. But why do you think plugging in 200 should give you -1. Also when you say between 200 and x, are you saying the integral is FROM 200 TO x?

                Also it should make sense to you that it's 1 - (200^2.5 / x^2.5) since this is a CDF, if your x is defined on range [200, infinity) what should your CDF be constrained to?

                Taking the derivative of your CDF, should get you back to what you started as your integrand. IF you are not getting it's because you forgot that -1*-1 = 1 (but maybe that's not what's confusing you).
                Last edited by NoMoreExams; May 20 2015, 01:20 AM.

                Comment


                • #9
                  Originally posted by NoMoreExams View Post
                  I have no idea what you are actually asking or trying to do. But why do you think plugging in 200 should give you -1. Also when you say between 200 and x, are you saying the integral is FROM 200 TO x?

                  Also it should make sense to you that it's 1 - (200^2.5 / x^2.5) since this is a CDF, if your x is defined on range [200, infinity) what should your CDF be constrained to?

                  Taking the derivative of your CDF, should get you back to what you started as your integrand. IF you are not getting it's because you forgot that -1*-1 = 1 (but maybe that's not what's confusing you).
                  I am saying the integral from 200 to x. The derivative of the original function comes to (-200^2.5/t^2.5). When you take the integral should it not come to (-1-(200^2.5/x^2.5) instead of (1-(200^2.5/x^2.5)?

                  Comment


                  • #10
                    Originally posted by aomara View Post
                    I am saying the integral from 200 to x. The derivative of the original function comes to (-200^2.5/t^2.5). When you take the integral should it not come to (-1-(200^2.5/x^2.5) instead of (1-(200^2.5/x^2.5)?
                    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.

                    Comment


                    • #11
                      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}

                      Comment


                      • #12
                        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.

                        Comment


                        • #13
                          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?

                          Comment


                          • #14
                            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.

                            Comment

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