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How to generally express a shifted PDF ?

  1. Dec 16, 2013 #1
    Hello,

    i am trying to solve the following.

    Given a general PDF (i.e., fx(x), where x ≥ 0), how can i express the PDF of y = c x in terms of fx(x)?

    I think that goes like this: fy(y) = fx(y/c)/c, but i 'm not sure.

    Any help would be useful.

    Thanks in advance
     
  2. jcsd
  3. Dec 16, 2013 #2

    Office_Shredder

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    nikozm, it's best to work with the CDF and then convert to the PDF. Let
    [tex] F(x) = \int_{0}^{x} f_x(t) dt [/tex]

    Then
    [tex] P(y<M) = P(cx < M) = P(x < M/c) = F(M/c) [/tex]
    So
    [tex] P(y<M) = \int_{0}^{M/c} f_x(t) dt [/tex]
    To find the pdf you just need to do some manipulations to the integral so that you have an [itex] \int_{0}^{M}[/itex], alternatively you can differentiate with respect to M to get the PDF.
     
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