User - MathOverflow

time-shifted ODEs/volume of polytopes

2011-07-31T04:25:35Z

Hello,

I'm looking for help with the following ODE:

f'(t) = x f(1 - at)

for 0 < a < 1, x in any interval (though 0 < x < 1 would be best), and f(0) = 1. There should be a solution for $0 \leq t \leq 1$...

My rather weak repertoire of techniques from undergrad & an introductory textbook on time-shifted ODE's hasn't gotten me very far at all.

I don't know if the origins will be helpful, but just in case any combinatorics people are drifting through, this comes from calculating the volume of polytopes related to alternating permutations. Stanley has the answer when a = 1 in his survey of the subject, but none of the 3-4 easy ways of getting the answer in that case seem to generalize very well.

Thanks!

Stochastic Integrals and Cauchy Variables

2011-07-09T14:53:44Z

I hope there is a straighforward literature-pointer here.

If I were interested in $\sum_{t=1}^{n} f(t) X_{t}$, where $X_{t}$ consists of independent normal random variables, I could approximate the sum as an Ito integral, and then (if $f(t)$ is reasonably nice) get a good answer for the resulting approximation. Also, my impression is that this is really the 'best approach' as long as $n$ is getting big and $f(t)$ isn't too wildly spiky.

Is there an analogous theory when $X_{t}$ is Cauchy?

I'm aware that there are lots of 'infinity issues' around adding up Cauchy variables, e.g. that sums with equal weights are dominated by their biggest term and so on... but I'm still hoping that there is a somewhat unified approach for looking at this type of problem.

Thanks!

Marginals and Convex Sets

2011-06-24T17:31:57Z

I am looking for a weak convergence of marginals result, in the following type of situation. References to 'related' situations are also very much appreciated.

I have a collection of affine hyperplanes $H_{1}, H_{2}, \ldots$ and variables $X_{1}, X_{2}, \ldots$ such that each variable appears in at most, say, 10 hyperplanes and each hyperplane has at most 10 variables. In my situation, all of the coefficients for the variables were negative while the offset was always positive, which made things easier.

Now, look at the finite-dimensional convex body $K_{n}$ in $\mathbb{R}^{n}$ containing the origin, bounded by $1 \geq X_{i} \geq 0$ and all of the hyperplanes that only contain variables $X_{i}$ with $i \leq n$. If I draw uniformly at random from $K_{n}$, I get some marginal distribution on $X_{1}$.

My question is, under what conditions does this marginal distribution have a limit, even when the limit of volume($K_{n}$) is 0?

2 EDITS: As pointed out by Ricky Demer in the comment below, we sometimes have convergence to deterministic limiting distributions by forcing some of the values to e.g. eventually be 0. I think I'm mostly interested in cases where the limiting marginals are all non-deterministic..

Answer by unknown (google) for 'Focusing' the mass of the Probability Density Function for a Random Walk

2009-10-27T16:18:51Z

I have no idea about the continuous case, which is presumably subtle, but the discrete case has an easy answer ('not very well'). In particular, if the random walk is simple random walk on a finite collection of vertices in the integer lattice in R^{d} (possibly with many self-loops on the boundary to simulate reflection), the stationary distribution is proportional to the degree at every vertex. In other words, the points away from the boundary don't really feel its presence at all.

If you have quantitative questions, a sufficiently clever person may be able to answer them using, say, Billingsley's book on convergence and the fact that many of these questions will be easy in the finite case.

Answer by unknown (google) for Minkowski sum of small connected sets

2009-10-27T16:04:49Z

In dimension 1, why are the sets {-100}, {80},{120} not a counterexample?

(In particular, they are compact connected sets with 'diameter' 0; their Minkowski sum, going by the wikipedia definition, is {-20,20,200}, which does not contain 0, but whose convex hull [-20,200] contains a large ball around 0)

This example of course has nothing to do with dimension, and you can easily flush out the points into tiny balls, if you like. I suspect that this is based on a misreading of the question.

Answer by unknown (google) for Expectation of the product of almost independent Gaussians

2009-10-27T05:47:23Z

As a side note, it seems that we get the opposite inequality for free. If the X_{i} are independent, and we are looking at it for 1 to n, we get

$E[\Pi_{i=1}^{n} \vert X_{i} \vert] = E[\Pi_{i} exp(log(|X_{i}|))] = E[exp(\Sigma_{i} log(|X_{i}|)] \geq exp(E[\Sigma_{i} log(|X_{i}|)]) = exp(nb)$.

Also, you might notice that this doesn't depend at all on the independence of the $X_{i}$... or on the exponent being 1, since we are only taking the expectation of a sum, never a product.

I realize this doesn't answer your original question at all, so I was curious as to where the hypothesis came from. In particular, could you post a proof in that direction when the $X_{i}$ are independnt? Where does C come from?

Comment by

2011-08-01T18:27:50Z

(Finally, please let me know if you'd like a mention if/when the note using this is put up. The series seems to be enough for what I need)

Comment by

2011-08-01T18:23:32Z

Actually, a little confused by the result. If $t_{0} = 0$ (as we might as well assume it is), the sum above is completely symmetric in x and t, but the equation below certainly isn't. Strange.

Comment by

2011-08-01T18:22:21Z

This is quite nice, thank you! Accepting the answer, and I'd be interested in hearing how you recognized the (slightly complicated) explicit form below from the sum above.

Comment by

2011-07-31T16:42:03Z

Thanks for the comment. I think that you have shown that there is at most one solution for a given initial condition. However, I'm interested in the initial condition f(0) = 1, not f(0) = 0. For f(0) = 1, there is certainly a unique nonzero solution when a = 1, and your proof seems to not depend on a < 1 (which is really the only thing that initially made me suspicious, I have little intuition).

Comment by

2011-07-31T15:01:00Z

Sorry, yes x and a are both constants.

Comment by

2011-06-24T23:36:06Z

That's certainly true! I think I'll try again, this time restricting to say cases where the diameter is at least 0.5 throughout. Alternatively, maybe I should say that I want the marginal distribution to have a nondeterministic limit.