Computing normal probability analyticallyis this possible?

[Applejacks01]

Hey guys. I'm sure some of you are aware of how to analytically integrate e^(-x^2) dx from - infinity to infinity using polar coordinates.

I have taken that logic and showed that the integral of the normal distribution( not necessarily standard) integrates to 1 over the entire domain.

However, now I am trying the case where we integrate from - infinity to some positive number, and I am having trouble.

So let's say I know the integral of the standard normal from - infinity to 1.96 = .975. Okay, well what I am trying to do is convert the integral to polar coordinates but I am having a hard time determining the limits of integration. Is this something that just can't be done analytically?

I thought I was on to something when I had my r limits from 0 to c( and thus I would solve for c, which in turn yields a nice conversion from Cartesian to polar for limit purposes) and my theta limits from 0 to 2pi, but then I realized that the theta limit can't be 2pi because we are not. integrating to infinity wrt y.

 

[A. Bahat]

There is no expression of this integral in terms of elementary function; see http://en.wikipedia.org/wiki/Error_function

 

[Ray Vickson]

Hey guys. I'm sure some of you are aware of how to analytically integrate e^(-x^2) dx from - infinity to infinity using polar coordinates.

I have taken that logic and showed that the integral of the normal distribution( not necessarily standard) integrates to 1 over the entire domain.

However, now I am trying the case where we integrate from - infinity to some positive number, and I am having trouble.

So let's say I know the integral of the standard normal from - infinity to 1.96 = .975. Okay, well what I am trying to do is convert the integral to polar coordinates but I am having a hard time determining the limits of integration. Is this something that just can't be done analytically?

I thought I was on to something when I had my r limits from 0 to c( and thus I would solve for c, which in turn yields a nice conversion from Cartesian to polar for limit purposes) and my theta limits from 0 to 2pi, but then I realized that the theta limit can't be 2pi because we are not. integrating to infinity wrt y.

The functions exp(+-x^2) are among those which do not possesses an "elementary" antiderivative. That means that there is no finite, closed-form formula for the antiderivative in terms of elementary functions---powers, roots, exponentials, logarithms, trigonometric and inverse trigonometric functions, etc. Note that there are *impossibility theorems* that justify this: it is a provable fact that it is impossible to write down a finite formula for the integral. It is not just that nobody has been smart enough to find the formula; it is that many people have been smart enough to prove that it can't be done. (Of course, you can write an infinite series for the integral, but that does not count as a finite formula.) Other examples that do not have elementary integrals are exp(x)/x, sin(x)/x, etc. Google "integration in finite terms" for more information.

RGV

 

[Applejacks01]

I know that the indefinite antiderivative doesn't have a closed form solution in terms of elementary functions, but what about definite integrals in polar coordinates?

For example, let's take the standard normal. Let's say I want to find P(Z<c), where c is a positive real #.

So we have Integral (1/sqrt(2pi))*e^(-(x^2)/2)) dx from -infinity to c = P(Z<c)

In polar coordinates, we could say Integral ( Integral (1/sqrt(2pi))*e^(-(r^2)/2) * r dr from 0 to R) dtheta from 0 to theta) = P(Z<c)^2

So using this logic, I've gotten this far: -((-1+e^(-R^2/2)) θ)/(2 π) = P(Z<c)^2
Either this logic doesn't go anywhere useful, or I'm just not seeing something that can turn this into something good.

 

[HallsofIvy]

Your integral is incorrect. If you integrate \int_0^R e^{-x^2}dxfollowing the same method as for \int_0^\infty e^{-x^2}dx], then you would have to say &amp;quot;Let I= \int_0^R e^{-x^2}dx= \int_0^R e^{-y^2}dy.&lt;br /&gt; &lt;br /&gt; Now we have &lt;br /&gt; I= \left(\int_0^R e^{-x^2}dx\right)\left(\int_0^R e^{-y^2}dy\right)= \int_{x=0}^R\int_{y= 0}^R e^{-(x^2+ y^2)}dydx&lt;br /&gt; &lt;br /&gt; It&amp;#039;s easy to change the integrand to polar coordinates:e^{-r^2}r drd\theta. But the region of integration is a &lt;b&gt;rectangle&lt;/b&gt; in the xy-plane, not a &amp;quot;sector&amp;quot; which your &amp;quot;dr from 0 to R&amp;quot; and &amp;quot;d\theta from 0 to \theta[/tex]&amp;amp;quot; would give.

 

[Applejacks01]

If I said A = Integral of e^(-x^2) dx from -infinity to infinity = Integral of e^(-y^2) dy from -infinity to infinity

Then A^2 = (Integral of e^(-x^2) dx from -infinity to infinity ) * (Integral of e^(-y^2) dy from -infinity to infinity ) [due to independence of the variables]
A^2 = Double Integral(-inf to inf for both limits) e^-(x^2+y^2)
And at this point we convert to polar coordinates, which does integrate over the limits theta = 0 to 2pi, and r = 0 to infinity.

But isn't this integral over a rectangular region as well? Yet it evaluates to sqrt(pi) just fine with a "sector" integral? Or is this different because since R goes to infinity, we map out the entire R^2 plane?

 

[Ray Vickson]

If I said A = Integral of e^(-x^2) dx from -infinity to infinity = Integral of e^(-y^2) dy from -infinity to infinity

Then A^2 = (Integral of e^(-x^2) dx from -infinity to infinity ) * (Integral of e^(-y^2) dy from -infinity to infinity ) [due to independence of the variables]
A^2 = Double Integral(-inf to inf for both limits) e^-(x^2+y^2)
And at this point we convert to polar coordinates, which does integrate over the limits theta = 0 to 2pi, and r = 0 to infinity.

But isn't this integral over a rectangular region as well? Yet it evaluates to sqrt(pi) just fine with a "sector" integral? Or is this different because since R goes to infinity, we map out the entire R^2 plane?

There is a difference between a finite rectangle (actually, a square) and the infinite rectangle corresponding to the whole first quadrant. Draw a picture if you don't believe it.

RGV