Recent content by kai_sikorski

I've been playing around with some MD simulations, a field not really familiar to me. I put together a code in LAMMPS to simulate a Lennard-Jones fluid and compute the RDF. I get the oscillations one would expect, which is good, but what is surprising is that the minimum of g(r) after the first...

Think I found what I need in Grigorius stochastic calculus book, it looks like my intuition was right.

You've found a particular solution to the problem, but I think you're forgetting about the two homogenous solutions that you can use to match your boundary conditions.

Consider the following PDE. A lot of this is from "Numerical Analysis of an Elliptic-Parabolic Partial Differential Equation" by J. Franklin and E. Rodemich. \frac{1}{2} \frac{\partial^2 T}{\partial y^2} + y \frac{\partial T}{\partial x} = -1 With |x|<1, |y| < \infty and we require...

Good to hear!

I did the calculation for the second game, because I was curious. If stephen goes .25, .5 and .25 in 1,2 and 3 fingers respectively, then no matter what Maude does the expectation is 0. You can see this easily by doing a weighted average of the rows.

Let p and q be the probability for Maude and Stephen to put up 1 finger respectively. Maude's expected payoff is given by E[M] = q (p *2 + (1 - p)*(-3)) + (1 - q) (p*(-3) + (1 - p)*4) = 4 - 7 q + p (-7 + 12 q) Suppose Stephen puts 1 finger up with probability 7/12 E[M] = - 1/12 So you're...

Hmm, yeah I see my problem. I'm trying to condition on there being a specific number of empty urns, but I'm calculating the probability to have more than or equal to that many.

Applying your formula to just the case k = 3 , we get 3(2/3)M - 3(1/3)M ≈ 3 (2 M )/(3 M ) But above doesn't go to 1. The probability that every urn has a ball must clearly go to 1 if you keep adding more and more balls.

I got 1 - \sum_{i=1}^{k-1} \left( \begin{array}{c} k \\ i \\ \end{array} \right) \left(\frac{k-i}{k}\right)^M

Maybe this will work better p0(M) = 1 - p1(M) - p2(M) - ... - pk-1(M) = 1 - (k-M)((k choose 1) * (k-1)M + ( k choose 2) * (k-2)M + ... +(k choose k-1)1M)

EDIT: Err this was wrong

Let pi(n) be the probability to have i empty urns at time n. pi(n+1) = pi(n)(k-i)/k + pi+1(n)(i+1)/k The initial condition is pi(1) = δk-1 Seems like you have to solve this recurrence.

That's not really the right approach because the urns are not independent and the OP is asking a question not about a given urn but about ALL the urns.

I haven't checked your calculations or anything, but you're saying the right things so at least the approach should be correct.