# Random points on some hemisphere

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! # Circular or spherical data, and density estimation I few years ago, while I was working on kernel based density estimation on compact support distribution (like copulas) I went through a series of papers on circular distributions. By that time, I thought it was something for mathematicians working on weird spaces…. but during the past weeks, I saw several potential applications of those estimators. • circular data density estimation Consider the density of an angle say, i.e. a function such that with a circular relationship, i.e. . It can be seen as an invariance by rotation. von Mises proposed a parametric model in 1918 (see here or there), assuming that where is Bessel modified function of order 1, (which is simply a normalization parameter). There are two parameters here, (some concentration parameter) and mu a direction. From a series of observed angles , the maximum likelihood estimator for kappa is solution of where and and where , where those functions are modified Bessel functions. Well, that estimator is biased, but it is possible to improve it (see here or there). This can be done easily in R (actually Jeff Gill – here – used that package in several applications). But I am not a big fan of that technique…. • density estimation for hours on simulated data A nice application can be on the estimation of the daily density of a temporal events (e.g. phone calls as we’ll see later on, or email arrival time). Let is the time (in hours) for the th observation (the th phone call received). Then set The time is now seen as an angle. It is possible to consider the equivalent of an histogram, set.seed(1) library(circular) X=rbeta(100,shape1=2,shape2=4)*24 Omega=2*pi*X/24 Omegat=2*pi*trunc(X)/24 H=circular(Omega,type="angle",units="radians",rotation="clock") Ht=circular(Omegat,type="angle",units="radians",rotation="clock") plot(Ht, stack=FALSE, shrink=1.3, cex=1.03, axes=FALSE,tol=0.8,zero=c(rad(90)),bins=24,ylim=c(0,1)) points(Ht, rotation = "clock", zero =c(rad(90)), col = "1", cex=1.03, stack=TRUE ) rose.diag(Ht-pi/2,bins=24,shrink=0.33,xlim=c(-2,2),ylim=c(-2,2), axes=FALSE,prop=1.5)  or a kernel based estimation of the density (the gray line on the right). circ.dens = density(Ht+3*pi/2,bw=20) plot(Ht, stack=TRUE, shrink=.35, cex=0, sep=0.0, axes=FALSE,tol=.8,zero=c(0),bins=24, xlim=c(-2,2),ylim=c(-2,2), ticks=TRUE, tcl=.075) lines(circ.dens, col="darkgrey", lwd=3) text(0,0.8,"24", cex=2); text(0,-0.8,"12",cex=2); text(0.8,0,"6",cex=2); text(-0.8,0,"18",cex=2) The c 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 The code is simply the following U=seq(0,1,by=1/250) O=U*2*pi U12=seq(0,1,by=1/24) O12=U12*2*pi X=rbeta(100,shape1=2,shape2=4)*24 OM=2*pi*X/24 XL=c(X-24,X,X+24) d=density(X) d=density(XL,bw=d$bw,n=1500)
I=which((d$x>=6)&(d$x<=30))
Od=d$x[I]/24*2*pi-pi/2 Dd=d$y[I]/max(d$y)+1 plot(cos(O),-sin(O),xlim=c(-2,2),ylim=c(-2,2), type="l",axes=FALSE,xlab="",ylab="") for(i in pi/12*(0:12)){ abline(a=0,b=tan(i),lty=1,col="light yellow")} segments(.9*cos(O12),.9*sin(O12),1.1*cos(O12),1.1*sin(O12)) lines(Dd*cos(Od),-Dd*sin(Od),col="red",lwd=1.5) text(.7,0,"6"); text(-.7,0,"18") text(0,-.7,"12"); text(0,.7,"24") R=1/24/max(d$y)/3+1 lines(R*cos(O),R*sin(O),lty=2)

Note that it is possible to stress more (visually) on hours having few phone calls, or a lot (compared with an homogeneous Poisson process), e.g.

plot(cos(O),-sin(O),xlim=c(-2,2),ylim=c(-2,2),
type="l",axes=FALSE,xlab="",ylab="")
for(i in pi/12*(0:12)){
abline(a=0,b=tan(i),lty=1,col="light yellow")}
segments(2*cos(O12),2*sin(O12),1.1*cos(O12),1.1*sin(O12), col="light grey")
segments(.9*cos(O12),.9*sin(O12),1.1*cos(O12),1.1*sin(O12))
text(.7,0,"6")
text(-.7,0,"18")
text(0,-.7,"12")
text(0,.7,"24")
R=1/24/max(d$y)/3+1 lines(R*cos(O),R*sin(O),lty=2) AX=R*cos(Od);AY=-R*sin(Od) BX=Dd*cos(Od);BY=-Dd*sin(Od) COUL=rep("blue",length(AX)) COUL[R<Dd]="red" CM=cm.colors(200) a=trunc(100*Dd/R) COUL=CM[a] segments(AX,AY,BX,BY,col=COUL,lwd=2) lines(Dd*cos(Od),-Dd*sin(Od),lwd=2) We get here those two graphs,  To be honest, I do not really like that representation – even if it looks nice. If we compare that circular representation to a more classical one (from 0:00 till 23:59 one the graph on the left, below), I do have a problem to interpret the areas in blue and pink.  density of wind direction On the left, we compare two densities, so the area in pink is the same as the area in blue. But here, it is no longer the case: the area in pink is always larger to the one in blue. So it might help so see when we have a difference, but there is a scaling issue that we cannot discuss further… But less us see if we can use that estimation technique to several problems. 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.  density of 911 phone calls Note that our estimate is consistent with several graphs that can be found on meteorological websites (e.g. the one above on the right, that was found here). In a recent post (here) I wanted to check about the “midnight crime” myth, using hours of 911 phone calls in Montréal. That was for all phone calls. But if we look more specifically, for burglaries, we have the distribution on the left, and for conflicts the one on the right  We do clearly observe that gun shots occur a bit before midnight. See also here for another study, but this time in NYC (thanks @PAC for the link).while for gun shots, we have the distribution on the left, and for “troubles” (basically people making too much noisy in parties) or “noise” the one on the right • density of earth temperatures, or earthquakes Of course it is also possible to work in higher dimension. Before, we went from densities on 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, The idea being that point on the left above are extremely close to the one on the right. An application can be e.g. on earthquakes occurrence. Data can be found here. library(ks) X=cbind(EQ$Longitude,EQ\$Latitude)
Hpi1 = Hpi(x = X)
DX=kde(x = X, H = Hpi1)
library(maps)
map("world")
cbind(X[,1]+360,X[,2]+180),cbind(X[,1]-360,X[,2]+180), cbind(X[,1],X[,2]-180),cbind(X[,1]+360, X[,2]-180),cbind(X[,1]-360,X[,2]-180)) DY=kde(x = Y, H = Hpi1) library(maps) plot (DY,add=TRUE,col="purple") 