cdf of the normal distribution

[hobz]

I know that the cdf of the normal distribution can not be evaluated analytically, since e^(-t/2) has no antiderivative. Why is that?

[JaKiri]

I know that the cdf of the normal distribution can not be evaluated analytically, since e^(-t/2) has no antiderivative. Why is that?

 

e^-t/2 has an indefinite integral, it's -2e^-t/2. Did you mean e^-t2? (Although you can use various tricks to create definite intregrals of it)

[ajb]

e^-t/2 has an indefinite integral, it's -2e^-t/2. Did you mean e^-t2? (Although you can use various tricks to create definite intregrals of it)

 

This was my observation also!

 

For [math]e^{-t^{2}}[/math] you will need the error function, which is related to the normal distribution.

[hobz]

Yes. I meant e^(0.5*t^2). Sorry.

 

This was my observation also!

 

For [math]e^{-t^{2}}[/math] you will need the error function, which is related to the normal distribution.

 

But what happens if I try to integrate [math]f(t)=e^{-t^{2}}[/math]? Do I end up with something where I need to integrate the [math]f(t)[/math] again? My calculus is rusty, but I would expect some rules (chain rule if I am not mistaken) to allow an expression of [math]\int f(t)dt[/math] where integration of [math]f(t)[/math] again is needed?

[Country Boy]

Yes. I meant e^(0.5*t^2). Sorry.

 

 

 

But what happens if I try to integrate [math]f(t)=e^{-t^{2}}[/math]? Do I end up with something where I need to integrate the [math]f(t)[/math] again? My calculus is rusty, but I would expect some rules (chain rule if I am not mistaken) to allow an expression of [math]\int f(t)dt[/math] where integration of [math]f(t)[/math] again is needed?

Actually almost all integrable functions have do not have anti-derivatives that can be written in a simple form.

[Bignose]

ajb answered your question. You have to use the error function.

 

The error function is just another function that sits in your toolbox ready to come out when needed. It is like sine and cosine and sinh and cosh, but it is the error function. It is the integral of e^-(t^2) terms.

 

If you study diffusion problems, it will come up all the time (not coincidently, because the normal distribution and diffusion are closely related).

[hobz]

I understand. But it does not clarify why it does not have an antiderivative.

[Bignose]

It has an antiderivative. d erf(x) / dx = e^-(x^2) (there is a 2 over the square root of pi in there, too). It just may not be in a nice looking form, but erf is a function just like all the other ones. Is it unsatisfactory that d sin (x) / dx = cos(x) ?

 

Similarly to the error function, the gamma function ( http://mathworld.wolfram.com/GammaFunction.html ) is the antiderivative of terms that look like t^(z-1)e^(-t)

 

It isn't going to come out in terms of e's and polynomials -- that's why the error function and the gamma function were invented. But, they are both functions just like sine and cosine. They are well defined, and obey lots of rules and are exceptionally well-studied. There are a whole host of "other" functions out there. Confluent hypergeometric functions, Bessel functions, elliptic integrals, Mathieu functions, etc. In general, whenever a problem wasn't amenable by a straightforward approach, someone invented a function to solve the problem. That's where erf came from.

[D H]

As Halls noted, most functions do not have an antiderivative that can be written terms of elementary functions. That does not mean these functions don't have an antiderivative, period.

 

Take the standard normal distribution probability density function, [math]f(x) = \frac 1 {\sqrt{2\pi}} \exp\left(-\,\frac {t^2} 2\right)[/math] This is obviously a widely-used mathematical function, as is its antiderivative. The antiderivative exists, and since it is an important function in and of itself, it has a name:

 

[math]\Phi(x) =

\int_{-\infty}^x

\frac 1 {\sqrt{2\pi}}

\exp\left(-\,\frac {t^2} 2 \right)dt

[/math]

 

As ajb and Bignose have already noted, the error function [math]\mathrm{erf}(x)[/math] and [math]\Phi(x)[/math] are closely allied:

 

[math]\Phi(x) = \frac 1 2 \left(1+\mathrm{erf}\left(\frac x {\surd 2}\right)\right)[/math]

[hobz]

I see. Brought me closer to understanding the nature of the error function.

 

But I still do not quite understand why I cannot employ the usual techniques for evaluating the integral. What exactly will go wrong, if I try finding the [math]\int e^{\left( -\frac{t^2}{2}\right)}[/math]. And here I would use some product rules, chain rules or what ever, as is taught in calculus introduction courses.

[Bignose]

hobz, do it and try. The best way to learn is to try to do it. I'd suggest trying to implement a change in variable, something like x=t^2 and go from there. Feel free to ask any further questions along the way.

[hobz]

I have tried the substitution, with [math]u=t^2[/math] I get [math]\frac{du}{dt}=2t \Rightarrow du=2t~dt[/math].

This doesn't really help, since there are no [math]2t~dt[/math] to substitute for [math]du[/math]. Is this correct?

[Bignose]

Right, so you see where difficulties come in? You can multiply and divide by 2t to get the du term, and then try some integration by parts.

[hobz]

I indeed do. I will try that. Thanks for the help! Thank you all.