In my previous post, I tried to answer the following question Consider points uniformly distributed on a sphere. What is the probability that the
points lie on a same hemisphere, for some hemisphere (there is no south or north here) ?
If I have been able to use Monte Carlo simulations in dimension 2 (on a circle, not on a sphere), I could not get it in dimension 3. Hopefully, my colleague Simon gave me a nice solution (much more efficient than my previous code). Unfortunately, it was in Maple,
clear all
taillesim=5000;
for N=3:10
nb=0;
for k=1:taillesim
X=randn(N,1);
Y=randn(N,1);
Z=randn(N,1);
pts=[X Y Z];
for i=1:N
pts(i,:)=pts(i,:)/norm(pts(i,:),2);
end
tol=1e-07;
A=-pts;
f=[0;0;0];
b=zeros(N,1)-tol;
%options=optimset('Display','off');
[x,val,exitflag]=linprog(f,A,b);
if(exitflag==1)
nb=nb+1;
end
end
probapprox(N-2)=nb/taillesim;
end
ns=[3:10];
for i=3:10
probhs(i-2)=probh(3,i);
end
scatter(ns,probhs)
hold on
plot(ns,probapprox);
The idea is very clever (I did not know how to code it, but as we will see, it is actually simple – or say not too difficult – to code actually). The idea is based on the idea that all the points lie on the same hemisphere if there is some such that
for all
.
That is not a big improvement, compared with the previous post, because we have to find such a vector. The idea suggested by Simon is to use some constraint optimization algorithm. We can try to optimize (maximize, or minimize, actually, we do not care) something like
If the set of constraint is empty, then there is no solution, while we can find one (maybe more, but we do not care) if there is such a vector . I wanted to add some constraint on
, like
(for the Euclidean norm), but that would be a non-linear constraint. So here, I chose to assume that
was such that
. With R, the optimization routine would be like
library(lpSolve)
A=rbind(pts,c(1,1,1))
C=c(1,2,3)
B=c(rep(1e-10,N),1)
slp=lp("min",C,A,c(rep(">=",N),"="),B)
So that I can solve here
and then, we simply have to check for the value of
slp$status
0 means that we found a vector . But when you run the code, it does not work. Because actually, what is solved here is not exactly the program above, but the one below,
(yes, it is mentioned in the help of the optimization function that only positive values are considered). So, a simple strategy, since we focus on an orthant here, is to consider all possible orthants. For instance, using
test=FALSE
x1=0:(2^3-1)%/%(2^(3-1))
x2=((0:(2^3-1))-(2^2)*x1)%/%(2^(3-2))
x3=((0:(2^3-1))-(2^2)*x1-2*x2)
for(u in 1:(2^3)){
pts2=cbind(pts[,1]*(-1)^x1[u],
pts[,2]*(-1)^x2[u],
pts[,3]*(-1)^x3[u])
A=rbind(pts2,c(1,1,1))
C=c(1,2,3)
B=c(rep(1e-10,N),1)
slp=lp("min",C,A,c(rep(">=",N),"="),B)
if(slp$status==0) test=TRUE}
if(test==TRUE) nb0=nb0+1
So here, the code would be something like
taillesim=5000
probaap=rep(NA,10)
probath=rep(NA,10)
p=function(d,n) .5^(n-1) * sum(choose(n-1,0:(d-1)))
for(N in 3:10){
nb0=0
for(k in 1:taillesim){
MZ=matrix(rnorm(n*3),n,3)
d=apply(MZ,1,function(z) sqrt(sum(z^2)))
X=MZ[,1]/d; Y=MZ[,2]/d; Z=MZ[,3]/d;
pts=cbind(X,Y,Z)
test=FALSE
x1=0:(2^3-1)%/%(2^(3-1))
x2=((0:(2^3-1))-(2^2)*x1)%/%(2^(3-2))
x3=((0:(2^3-1))-(2^2)*x1-2*x2)
for(u in 1:(2^3)){
pts2=cbind(pts[,1]*(-1)^x1[u],
pts[,2]*(-1)^x2[u],
pts[,3]*(-1)^x3[u])
A=rbind(pts2,c(1,1,1))
C=c(1,2,3)
B=c(rep(1e-10,N),1)
slp=lp("min",C,A,c(rep(">=",N),"="),B)
if(slp$status==0) test=TRUE
}
if(test==TRUE) nb0=nb0+1
}
probaap[N]=nb0/taillesim
probath[N]=p(3,N)
}
and this time, the probability obtained using Monte Carlo simulation is extremely close to the theoretical value.
And the algorithm is extremely fast! So using optimization routines to see if sets defined by linear constraints are empty – or not – is truly a great idea!
such that
. It can be seen as an invariance by rotation.
is Bessel modified function of order 1,
(some concentration parameter) and mu a direction.
, the maximum likelihood estimator for kappa is solution of
, where those functions are modified Bessel functions. Well, that estimator is biased, but it is possible to improve it (see 
is the time (in hours) for the 
th observation (the 
ode looks rather simple. But I am not very comfortable using codes that I do not completely understand. So I did my own. The first step was to get a graph similar to the one we have on the right, except that I prefer my own kernel based estimator. The idea is that instead of estimating the density on 
, we estimate it on the sample 
. Then we multiply by 3 to get the density only on 
. For the bandwidth, I took the same as the one that we would have taken on 
A standard application when studying angles is wind direction. For instance, in Montréal, it is possible to find hourly observations, starting in 1974 (we just need a R robot to pick up the information, but I’ll tell more about that in another post, someday). Here, we have directly an angle. So we can use a code rather similar to the one used above to estimate the distribution of wind direction in Montréal.
to densities on the unit circle 
. But similarly, it is possible to go from 
to the unit sphere 
. A nice application being global climate studies,
