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Need help solving CDF (Cumulative distribution function)

  1. Sep 19, 2011 #1
    Hello, im new. I been looking on the net for a guide how to solve the CDF by hand, i know the answer and im about to crack this baby but I got stuck....

    Im trying to calculate Cumulative distribution function by hand:
    [itex]\int^{1}_{-1}\frac{1}{2\pi} e^{\frac{-z^{2}}{2}} dz[/itex] or wolfram alpha: integrate 1/sqrt(2*pi) * e^(-z^2 /2) dz from -1 to 1

    Anyway, this is the tricky part, how do this? (I left out the lefthand part above part for easier readability):

    [itex]\int e^{\frac{-z^{2}}{2}} dz = [/itex]

    [itex]u = \frac{-z^{2}}{2} [/itex]

    [itex]du = -z dz[/itex]

    [itex]\frac{du}{-z} = dz[/itex]

    [itex]\int e^{u} \frac{du}{-z} = [/itex]

    then? How do i need to do?. Can any friendly soul here show me step by step how to solve this?

    best regrads
    invictor
     
  2. jcsd
  3. Sep 19, 2011 #2

    CompuChip

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    You can define
    [tex]\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} dt[/tex]
    and express the result in terms of that, getting [itex]\operatorname{erf}(1/\sqrt{2})[/itex] as the answer.

    You cannot give a (more) exact answer than that (otherwise we wouldn't be needing erf in the first place).
     
  4. Sep 20, 2011 #3
    Ok so now you have the CDF in terms of the error function erf(/frac{1}{/sqrt{2}}) what is the associating PDF?
     
  5. Sep 20, 2011 #4

    mathman

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    The density function for the error function is the normal density function, which is the integrand you started with.
     
  6. Sep 22, 2011 #5
    Thanks i got it now :)
     
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