[math-fun] one-line/proof without words/elegant proof that coeffs of (1+x)^n approx a Gaussian?
Is there a simple/elegant/one-line proof that the coefficients of (1+x)^n approximate a Gaussian curve? I know that it is the convolution of n boxes, and that a large convolution of non-impulses approaches a Gaussian, but is there a quick/obvious proof of this that also gives you the appropriate mean & std deviation parameters?
Does everyone know about the very simple purely geometric proof that a section of a cone is an ellipse? It's not easy to explain in words. If there's any interest I could find an illustration on the Web. Steve Gray Henry Baker wrote:
Is there a simple/elegant/one-line proof that the coefficients of (1+x)^n approximate a Gaussian curve? I know that it is the convolution of n boxes, and that a large convolution of non-impulses approaches a Gaussian, but is there a quick/obvious proof of this that also gives you the appropriate mean & std deviation parameters?
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On 2/14/07, Henry Baker <hbaker1@pipeline.com> wrote:
Is there a simple/elegant/one-line proof that the coefficients of (1+x)^n approximate a Gaussian curve? I know that it is the convolution of n boxes, and that a large convolution of non-impulses approaches a Gaussian, but is there a quick/obvious proof of this that also gives you the appropriate mean & std deviation parameters?
I think that if you had an easy proof, people would get pretty excited. I know I would! All the intro stats texts I've ever seen talk about the central limit theorem (and even the fact that binomial coefficients approach a normal curve, which they also usually justify by the CLT) as being something beyond the scope of the course, requiring too much mathematical machinery and yielding not enough intuition. So if anyone sends you something, Henry, I'd love to see it! Or if you come up with something reasonably clean on your own. Thanks, --Joshua
Quoting Henry Baker <hbaker1@pipeline.com>:
Is there a simple/elegant/one-line proof that the coefficients of (1+x)^n approximate a Gaussian curve? I know that it is the convolution of n boxes, and that a large convolution of non-impulses approaches a Gaussian, but is there a quick/obvious proof of this that also gives you the appropriate mean & std deviation parameters?
Why talk about convolutions? - The expansion of (1+x)^n has the same coefficients as (a+B)^n and you get them by figuring out how many ways you can get so many a's and the rest b's, having applied the distributive law to the product. But that is the number of (average plus epsilon)'s relative to the number of (average minus epsilons)'s you get while deriving the Gaussian distribution. QED! - hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos
Aren't you assuming what I'm trying to prove? Why should a discrete distribution (the binomial coefficients) approach the continuous distribution ~ exp(-(x/sigma)^2/2), and where does the sigma come from in the binomial distribution? http://en.wikipedia.org/wiki/Normal_distribution The proof given in this article still doesn't seem to me to be transparent enough: http://en.wikipedia.org/wiki/Central_limit_theorem At 12:27 AM 2/15/2007, mcintosh@servidor.unam.mx wrote:
Quoting Henry Baker <hbaker1@pipeline.com>:
Is there a simple/elegant/one-line proof that the coefficients of (1+x)^n approximate a Gaussian curve? I know that it is the convolution of n boxes, and that a large convolution of non-impulses approaches a Gaussian, but is there a quick/obvious proof of this that also gives you the appropriate mean & std deviation parameters?
Why talk about convolutions? - The expansion of (1+x)^n has the same coefficients as (a+B)^n and you get them by figuring out how many ways you can get so many a's and the rest b's, having applied the distributive law to the product. But that is the number of (average plus epsilon)'s relative to the number of (average minus epsilons)'s you get while deriving the Gaussian distribution. QED!
- hvm
Quoting Henry Baker <hbaker1@pipeline.com>:
Why should a discrete distribution (the binomial coefficients) approach the continuous distribution ~ exp(-(x/sigma)^2/2), ...
from taking limits, specifically, Stirling's approximation of factorials.
... and where does the sigma come from in the binomial distribution?
the epsilon in my original posting. The x^2 in the Gaussian comes from the pairing (ave + eps)(ave - eps). By the way, some careless editing on my part of the "reply" function has resulted in a doubling of these postings.
http://en.wikipedia.org/wiki/Normal_distribution
The proof given in this article still doesn't seem to me to be transparent enough:
Although Weisstein's pages, wikipedia, and similar are quite useful and informative, they are not always reliable, and often propagate common misconceptions and inaccurate folklore. Anyway, I was just trying to answer the original question about an interesting coincidence with a simple and plausible explanation. One can fancy up the details, but I sure wish somebody had used this explanation when I wes first studying statistics. I have used it in my own classes, somplete with a derivation of Stirling's formula, so I have confidence in its accuracy. Sorry I can't relate the epsilon to sigma; as I recall, it is essentially the standard deviation according to how much you keep missing the average, but there are some pi's and factors of 2 and the like, so as I recall there is some scaling involved. - hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos
participants (4)
-
Henry Baker -
Joshua Zucker -
mcintosh＠servidor.unam.mx -
Steve Gray