Integrating the Standard Distribution CurveDate: 06/25/99 at 20:39:31 From: John Subject: Integrating the standard distribution curve Does anyone here know how to integrate the standard distribution curve for statistics? I need to find the integral from infinity to negative infinity of (e^((-x^2)/2))/((2pi)^(1/2)) This is a self-interest project. I may not even understand what the explanation means, but please write down everything needed to prove the answer as if I do. I will find some way to understand it. Thank you Date: 06/26/99 at 07:17:12 From: Doctor Jerry Subject: Re: Integrating the standard distribution curve Hi John, It only takes a change of variable to relate the above to the integral I = int[0,oo,e^(-x^2)*dx]. I'll work with the latter. One can write I^2 = int[0,oo,e^(-x^2)*dx]*int[0,oo,e^(-x^2)*dx] = int[0,oo,int[0,oo,e^(-x^2)*e^(-y^2)*dx]*dy] = int[0,oo,int[0,oo,e^(-(x^2+y^2))*dx]*dy] Now make a change of variable, to polar coordinates. This gives I^2 = int[0,pi/2,int[0,oo,r*e^(-r^2)*dr]*dt] The inner integral is easy and the outer one is trivial. This is the stardard way of evaluating I. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ Date: 06/26/99 at 18:49:59 From: NYSE DOW JONES Subject: Re: Integrating the standard distribution curve Doctor Jerry, Thanks for doing this, but could you please tell me what "0,oo" is? Could you please explain this to me as if this is a 20-point final exam question and you need to explain to your professor as clearly as possible? In other words, would you please show me more steps in between so that it is more comfortable for a person who only knows first year calculus to dig through. That way, I can look in books and figure out all this by myself more easily. Thanks Date: 06/27/99 at 07:58:19 From: Doctor Jerry Subject: Re: Integrating the standard distribution curve Hi John, Short of trying to type out a complete course on multi-dimensional integration, I'll try to fill in some details. The "oo" is meant to suggest the infinity symbol. So, when I wrote I = int[0,oo,e^(-x^2)*dx] I meant the integral, from 0 to infinity, of e^(-x^2). Of course, this is shorthand for the limit as a -> oo of int[0,a,e^(-x^2)*dx]. Here's a copy of what I wrote, with some added comments. I^2 = int[0,oo,e^(-x^2)*dx]*int[0,oo,e^(-x^2)*dx] The above, a key step, assumes that the improper integral Int[0,oo,e^(-x^2)*dx] converges to a finite number, called I. This is not hard to prove. The guts of the argument are: for a>1, int[0,a,e^(-x^2)*dx] = int[0,1,e^(-x^2)*dx]+int[1,a,e^(-x^2)*dx] < 1+int[1,a,e^(-x)*dx] = 1+1/e-1/e^a. In the second step = int[0,oo,int[0,oo,e^(-x^2)*e^(-y^2)*dx]*dy] we take advantage of the fact that int[0,oo,e^(-x^2)*dx] = int[0,oo,e^(-y^2)*dy] and the latter can be taken inside the other factor of the product, because it doesn't contain x. The next step, = int[0,oo,int[0,oo,e^(-(x^2+y^2))*dx]*dy] is clear. Now we change variable, specifically, X = r*cos(t) y = r*sin(t), At this point, you need to calculate the Jacobian of this change of variable. You need to look this up. It amounts to replacing dx*dy by r*dr*dt. And the region of integration changes from the first quadrant, that is, x from 0 to oo and y from 0 to oo, to r from 0 to oo and t from 0 to pi/2. I^2 = int[0,pi/2,int[0,oo,r*e^(-r^2)*dr]*dt] The rest is one-dimensional integration (first-year calculus). I hope this helps a bit. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
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