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Homework Help: Stats - Continious random distribution (PDF)

  1. Oct 4, 2008 #1
    1. The problem statement, all variables and given/known data
    A function f( x ) had the following form : f ( x) = kx^-(k+1) where 1 < x < infinity

    a) for what values of k is f a pdf ?

    2. Relevant equations
    See attachments (the theorem is included)

    3. The attempt at a solution
    started out by using the first equation in the attachment, for stuck with kx^-(k+1) = 0 and Got stuck there... well the only way k will satisfy the equation is that if it is equal to 0... but i'm not sure if that is the answer.

    Tried using the 2nd equation, integration was long, and i encountered logs, K turned out to have a single value of 0.00693

    ok i figured that my first attempt was right and that k is > = 0 . But i just want to confirm this, and i don't really feel like deleting my post after writing it out

    Attached Files:

    • heo.JPG
      File size:
      10.2 KB
  2. jcsd
  3. Oct 5, 2008 #2


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    Hi mohdhm! :smile:
    erm … are you sure about k = 0? :wink:
  4. Oct 5, 2008 #3
    What is your function? Is it (kx)^{-k-1} or k * x^{-k-1}
  5. Oct 5, 2008 #4
    to dirk_mec1: it is the 2nd function you listed, sorry for not being clear.

    tiny-tim: well k=0 works for all positive values of x and satisfies the first equation. So i guess i'm Sure. also, any positive number k will yield a positive answer which also satisfies the equation. but i have this... conflicting hunch, i want to use the integral, but i tried and it was just a big mess, am i on the right track ?
  6. Oct 5, 2008 #5

    D H

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    Just because a function is semi-positive definite does not mean it qualifies as a PDF. The area under the function must be one. In your example, if k=0, f(x;k) reduces to f(x)=0. The area under this function is definitely not one.

    This function integrates easily with respect to x. Treat k as a constant when doing the integral.
    Last edited: Oct 5, 2008
  7. Oct 5, 2008 #6
    lol... yeah i did treat K as a constant. Well i can give this integral yet another go, might post my work too since i haven't done integrals for a while
  8. Oct 5, 2008 #7
    DH you are absolutely wrong, there is no way that integral with bounds of negative infinity to positive infinity, can work out. The whole thing just blows up when i get to the logs part. i get a negative value when converting bases, and the whole thing turns into a mess. So what's up guys, can anyone drop me a hint that can lead me on the right path ?
  9. Oct 6, 2008 #8


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    Hi mohdhm! :smile:
    erm … the -∞ part of the integral doesn't matter …

    f(x) is only defined, in the question, for 1 ≤ x ≤ +∞ …

    in other words, f(x) = 0 for -∞ < x ≤ 1. :wink:

    Try again! :smile:
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