Normal Distribution Integral

  • Thread starter wombat4000
  • Start date
  • #1
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Homework Statement



I'm having difficulty integrating something,
click http://en.wikipedia.org/wiki/Normal_distribution
and under Cumulative distribution function, there is an integral - how do you get to the next line?


Homework Equations





The Attempt at a Solution




i have tried the substitution of [tex]t=u-\mu[/tex] which gives
[tex]\int_{-\infty}^{x}e^{-t^{2}/2}dt[/tex]

but can't integrate this.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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see error functions

i have tried the substitution of [tex]t=u-\mu[/tex] which gives
[tex]\int_{-\infty}^{x}e^{-t^{2}/2}dt[/tex]
but can't integrate this.

Hi wombat! Welcome to PF :smile:

All you need is in http://en.wikipedia.org/wiki/Error_function

Good luck!

[size=-2](if you're happy, don't forget to mark thread "solved"!)[/size]​
 
  • #3
36
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Still don't really understand what's what - this to clarify this is what i was originally referring to
f26a8934f9cfafa33024e7ade8201463.png

and this is what i got from the error function page
7777288be8057a4c26904e48e168915c.png

but i need to integrate from [tex]-\infty[/tex] to x.
 
Last edited:
  • #4
36
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Thanks for your help by the way!
 
  • #5
tiny-tim
Science Advisor
Homework Helper
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An even function!

Ah! You didn't take account of the fact that the integrand is the same for -t as for t (an "even function"), so you can mutiply the limits by -1:

[tex]\int_{-\infty}^{x}e^{-t^{2}/2}dt[/tex]​

is the same as:

[tex]\int_{-x}^{\infty}e^{-t^{2}/2}dt\qquad,[/tex]​

which is (a multiple of) erfc(-x).

Then the top of the wikipedia page gives you the answer! :smile:

[size=-2](if you're happy, don't forget to mark thread "solved"!)[/size]​
 
  • #6
36
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i still don't get it.
 
  • #7
36
0
i know that
4ecb9cb151969361ae1b2357fea5d66f.png

and
066ed14d31d7ea03d453237b18111eba.png

but i don't end up with
f26a8934f9cfafa33024e7ade8201463.png
 
  • #8
36
0
should i being using this?

[tex]\int_{-x}^{\infty}e^{-t^{2}/2}dt\qquad[/tex] = [tex]\int_{-x}^{\infty}{e^{-t^{2}}e^{t^{2}/2}}dt\qquad[/tex]
 
  • #9
36
0
what should [tex]\phi[/tex] equal?

003dabb870f6a1fc0521a85000ea8090.png
?
 
  • #10
9
0
I think in this case, x should be greater than zero.
 

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