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...
I have a PDE that can be interpreted as basically an exit time problem for a certain stochastic process. I would like to use this to verify an analytical solution I've found. If I start the stochastic process at (x,y), then the average exit time from a certain region will be equal to the value...
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...
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.
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.