Need help solving CDF (Cumulative distribution function)

  • Thread starter invictor
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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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).
 
  • #3
34
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Ok so now you have the CDF in terms of the error function erf(/frac{1}{/sqrt{2}}) what is the associating PDF?
 
  • #4
mathman
Science Advisor
7,891
460
The density function for the error function is the normal density function, which is the integrand you started with.
 
  • #5
6
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Thanks i got it now :)
 

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