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Algebraic Integration of Standard Normal Distribution Function


Date: 02/11/98 at 05:00:54
From: julian keites
Subject: Algebraic integration of standard normal distribution 
         function

I've tried inspection and partial integration and don't seem to get 
the right results - I wonder if this is a function whose integral 
cannot be solved algebraically? Is numeric integration the only option 
or is there a known approximation?

f(Z)=(1/(srqt(2*PI))*e^((-Z^2)/2)


Date: 02/11/98 at 16:01:22
From: Doctor Anthony
Subject: Re: Algebraic integration of standard normal distribution 
             function

You need special techniques for handling this integral. Below is the 
method of integrating under the complete curve. Areas at intermediate 
values of z will be even more difficult.

INTEGRATING THE NORMAL CURVE

I will carry out the integral from 0 to infinity, and doubling the 
result (since the graph is symmetrical about the y axis) will give the 
total area from -infinity to +infinity.

Let  I = INT(0 to infinity)[e^(-x^2/2).dx]

This cannot be evaluated using elementary methods, so we proceed as 
follows:

  I^2 = [INT(0 to infinity)(e^(-x^2/2).dx]^2

  = INT(0 to infinity){e^(-x^2/2)dx) INT(0 to infinity)(e^(-y^2/2).dy)

  = INT(dx INT[e^(-(x^2+y^2)/2).dy]   all integrals are from 0 to 
                                      infinity.

  = INT.INT[e^(-(x^2+y^2)/2).dx.dy]

where the region of integration is the whole of the positive quadrant 
of the xy plane.

If we transform to polar coordinates, x^2+y^2 = r^2, also the element 
of area dx.dy is now given in polar coordinates by element of area 
r.d(theta).dr  The limits of integration will be 0 to infinity for r, 
and 0 to pi/2 for theta.

So our integral now becomes

  I^2 = INT.INT[e^(-r^2/2)r.dr.d(theta)]

      = INT[d(theta)INT[e^(-r^2/2)r.dr]

make the substitution  r^2/2 = u    then r.dr = du and the inner 
integral becomes

           INT[e^(-u).du]   for u from 0 to infinity.

           = - e^(-u)  = -(0 - 1) = 1

So now we have  

     I^2 = INT[d(theta)] from 0 to pi/2

         =  [theta] from 0 to pi/2

         =  pi/2

and so I = sqrt(pi/2)

and   2I = sqrt(2pi)  giving area from -infinity to +infinity.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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