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Thread: A Simple Question About the Cumulative Density Function

  1. #1
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    Question 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. #2
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    Quote 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.

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

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    Quote 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?

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    Quote 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

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    Quote 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!

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    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.

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    Quote 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 20th 2015 at 12:20 AM.

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    Quote 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)?

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    Quote 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.

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