Sure. It's a little bit lengthy though, so it might take some work to read it:
P(m)=\frac{\sqrt{2 \pi n}n^ne^{-n}}{\sqrt{2 \pi m}m^me^{-m}*\sqrt{2 \pi (n-m)}(n-m)^{(n-m)}e^{-(n-m)}}p^m(1-p)^{n-m}=
\frac{n^{n+1}}{\sqrt{2 \pi n}*m^{m+\frac{1}{2}}*(n-m)^{(n-m)+\frac{1}{2}}}p^m(1-p)^{n-m}...
I want to show that the binomial distribution:
P(m)=\frac{n!}{(n-m)!m!}p^m(1-p)^{n-m}
using Stirling's formula:
n!=n^n e^{-n} \sqrt{2\pi n}
reduces to the normal distribution:
P(m)=\frac{1}{\sqrt{2 \pi n}} \frac{1}{\sqrt{p(1-p)}}
exp[-\frac{1}{2}\frac{(m-np)^2}{np(1-p)}]...
The link I'm watching from is:
http://www.uccs.edu/~math/vidarchive.html
It requires free registration, but is available to anyone. The course for some reason is called "real analysis", but it's entirely a course on measure theory.
It's kind of boring: basically the professor just writes...
Wait a second. How does that work? Some subsets of the Cantor set are not measurable. But the Cantor set is itself measurable (with zero measure). The problem is the fact that the Borel sigma algebra is not complete. So if you go to a complete one, like the Lebesgue sigma algebra, you introduce...
That's intuitive enough. The intersection of B with A-compliment, which is in the sigma algebra, must have zero measure if A and B have the same measure. So if you take all subsets of regions of zero measure, then you can build any subset inside B (but containing all of A).
I'm not exactly...
I just saw a demonstration of a subset on the real line that is non-measurable.
The proof relied on the fact that the measure of the union of a countable number of disjoint sets is additive, and also that the measure is unchanged by translation.
But I thought the Banach-Tarski paradox said...
What's the point of defining \sigma algebras, i.e., why must we assign measures only to elements of a \sigma algebra?
Second, can you give an example of a set contained in the Lebesgue \sigma algebra, but not the Borel \sigma algebra? Also, is there a set not contained even in the...