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A pdf question

  1. Nov 24, 2009 #1
    Could anyone help me figure out the the probability density function (pdf) of |X|^(1/2)+|Y|^(1/2)+|Z|^(1/2) if X, Y and Z are distributed normally with mean 0 and variance 1, N(0,1) ?

    Thanks in advance.
     
    Last edited: Nov 24, 2009
  2. jcsd
  3. Nov 24, 2009 #2

    statdad

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    Homework Helper

    Are your variables independent? If so, first work out the distribution of [tex] |X|^{1/2} [/tex]. The other distributions will be identical and you can use standard procedures to find the distribution of their sum.
     
  4. Nov 24, 2009 #3
    Yes, the variables are independent. But what are the standard procedures? It there a easier way to get the pdf if one has more random variables than three?
     
    Last edited: Nov 24, 2009
  5. Nov 24, 2009 #4
    If X, Y, and Z, are distributed normally, just make them a normal (gausian) PDF of a random variable.

    X=Ae^(ax). You're lucky that you can use little x,y, and z for your random variables to match up with their distributions.
     
  6. Nov 25, 2009 #5

    statdad

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    "X=Ae^(ax). " That isn't the form of a normal distribution.

    Okay, suppose [tex] X \sim n(0,1) [/tex]. Think this way.

    1) You should be able to write down the distribution of [tex] |X| [/tex] - it's a pretty
    standard result, and if you're working on this problem I'm guessing you know this.
    2) Use a standard transformation (if [tex] W = |X|, find the distribution of square root of W). This gives the distribution of [tex] |X|^{1/2}[/tex].
    3) Since [tex] X, Y[/tex] and [tex] Z [/tex] are i.i.d, the same is true for

    [tex]
    |X|^{1/2} + |Y|^{1/2} + |Z|^{1/2}
    [/tex]

    so the distribution of their sum should be relatively easy to obtain.
     
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