Algebraic Integration of Standard Normal Distribution FunctionDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/