Also, my formula shows that $n^2(p_n(0)−p_n(x))\to\frac{25}{16\pi}\vert x\vert^2$. This characterizes the Euclidean distance in terms of $p_n$.

What you are suggesting implies that the probability of being at (3,4) is the same as being at (5,0) for all large n. That seems unlikely, and would guess that $C_n=5$ for n large.

Why would anyone want to consider such a limit?

Thanks Santker, that's a nice reference which answers your original question. I'm not sure what the significance of your comment about Radon measures is though (the example wasn't Radon).

Gerald - Does the example of a Gaussian measure on $\ell^\infty$ prove the original question either way? My guess is that the cylindrical sigma algebra on $\ell^\infty$ is not the same thing as the Borel sigma algebra (otherwise, why even use the cylindrical sigma algebra in the example). But I'm not sure that the example implies this.

I assume that you mean joint normal, in which case x_1-x_2 is normal. The second qu. asks if the set of joint normal rv's x_a-x_i are all positive, which is more difficult (requires some numerical integration, I think).

Sorry typo. Should have said C, not R.

Does the ultrafilter lemma imply the existence of such automorphisms of R. Or, just the ultrafilter lemma restricted to filters on N? That would clearly be necessary for there to be a positive answer to your question.

Actually, quadratic variation of BM is neither of those definitions ( they both give infinity). For BM you have convergence in probability for a sequence of deterministic partitions with mesh tending to zero. You also have almost sure convergence if each sequence is a refinement of the previous one.

Although you can define the Black-Scholes model by just specifying the distribution, and even prove some basic facts about it, that just doesn't seem very satisfactory. I mean, BS isn't even a very good model. And if you try to use something better, you're going to need at least some stochastic calculus. Besides, ideas such as martingale representation (Market completeness in math finance lingo) need some stochastic integration, even in BS.