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A Simple Problem in Statistical Physics

Posted in Cute Problems with tags , , , , on July 11, 2011 by telescoper

In physics we often have to resort to computer simulations in which continuously varying quantities are modelled on a discrete lattice. We also have recourse from time to time to model physical properties of a system as random quantities with some associated probability distribution. The following problem came up in a conversation recently, and I think it’s rather cute so thought I’d post it here.

Consider a regular three-dimensional Cartesian grid, at each vertex of which is defined a continuous variable x which varies from site to site with the same probability distribution function F(x) at each location. The value of x at any vertex can be assumed to be statistically independent of the others.

Now define a local maximum of the fluctuating field defined on the lattice to be a point at which the value of x is higher than the value at all surrounding points, defined so that in D dimensions there are 3^{D}-1 neighbours.

What is the probability that an arbitrarily-chosen point is a local maximum?

Solution

Well, the most popular answer is in fact the correct one but I’m quite surprised that a majority got it wrong! Like many probability-based questions there are quick ways of solving this, but I’m going to give the laborious way because I think it’s quite instructive (and because I’m a bit slow).

Pick a point arbitrarily. The probability that the associated value lies between x and x+dx is f(x)dx, where f(x)=dF(x)/dx is the probability density function. According to the question there are 3^3-1=26 neighbours of this point. The probability that all of these are less than x isF(x)\times F(x)\times \ldots F(x) 26 times, i.e. [F(x)]^{26}  becauses they are independent. Note that this is a continuous variable so the probability of any two values being equal is zero. The probability of the chosen point being a local maximum with a given value of x is therefore f(x)dx\times [F(x)]^{26}. The probability of it being a local maximum with any value of x is obtained by integrating this expression over all allowed values of x, i.e. \int f(x) dx [F(x)]^{26} . But the integrand can be re-written

\int f(x) dx [F(x)]^{26} = \int dF \times F^{26} = \frac{1}{27} \int d\left(F^{27}\right) = \frac{1}{27},

because  F=1 at the upper limit of integration and F=0  at the bottom.

So you don’t need to know the form of F(x) – but the calculation does rely on it being a continuous distribution.

This long-winded method demonstrates the applicability of the product rule and the process of marginalising over variables, but the answer should tell you a much quicker way of getting there.  The central point and the 26 neighbours constitute a set of 27 points. The probability that any particular one is the largest of the set is just 1/27, as each is equally likely to be the largest. This goes for the central value too, hence the answer.

 

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