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Thread: A Simple Question About the Cumulative Density Function

1. 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.  Reply With Quote

2. Originally Posted by bunkey 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.  Reply With Quote

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

4. Originally Posted by RossMoney34 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?  Reply With Quote

5. Originally Posted by bunkey 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  Reply With Quote

6. Originally Posted by NoMoreExams You should know that for a cont. function P(X = x) = 0
Now it's evident to me. Thank you!  Reply With Quote

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.  Reply With Quote

8. Originally Posted by aomara 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).  Reply With Quote

9. Originally Posted by NoMoreExams 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)?  Reply With Quote

10. Originally Posted by aomara 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.  Reply With Quote